3 Descriptive Statistics
3.01: Measures of Center
Both graphical and numerical methods of summarizing data make up the branch of statistics known as descriptive statistics. Later, descriptive statistics will be used to estimate and make inferences about population parameters using methods that are part of the branch called inferential statistics. This section introduces numerical measurements to describe sample data.
This section focuses on measures of central tendency. Many times, you are asking what to expect “on average.” Such as when you pick a career, you would probably ask how much you expect to earn in that field. If you are trying to buy a home, you might ask how much homes are selling for in your area. If you are planting vegetables in your garden, you might want to know how long it will be until you can harvest. These questions, and many more, can be answered by knowing the center of the data set. The three most common measures of the “center” of the data are called the mode, mean, and median.
3.1.1 Mode
To find the mode, you count how often each data value occurs, and then determine which data value occurs most often.
The mode is the data value that occurs the most frequently in the data.
There may not be a mode at all, or you may have more than one mode. If there is a tie between two values for the greatest number of times then both values are the mode and the data is called bimodal (two modes). If every data point occurs the same number of times, there is no mode. If there are more than two numbers that appear the most times, then usually we write there is no mode. When looking at grouped data in a frequency distribution or a histogram then the largest frequency is called the modal class.
Below is a dotplot showing the height of some 3-year-old children in cm and we would like to answer the question, “How tall are 3-year-olds?”
Figure 3-1
From the graph, we can see that the most frequent value is 95 cm. This is not exactly the middle of the distribution, but it is the most common height and is close to the middle in this case. We call this most frequent value the mode.
For larger data sets, use software to find the mode or at least sort the data so that you can see grouping of numbers. Excel reports a mode at the first repetitive value, so be careful in Excel with bimodal data or data with many multiples that would really have no mode at all.
Note that zero may be the most frequent value in a data set. The mode = 0 is not the same as “no mode” in the data set.
The mode is the observation that occurs most often.
- Example 3-1: -5 4 8 3 4 2 0 mode = 4
- Example 3-2: 3 -6 0 1 -2 1 0 5 0 mode = 0
- Example 3-3: 18 25 15 32 10 27 no mode (Excel writes N/A)
- Example 3-4: 15 23 18 15 24 23 17 modes = 15, 23 (bimodal)
- Example 3-5: 100 125 100 125 130 140 130 140 no mode (Excel gives 100)
Summation Notation
Throughout this course, we will be using summation notation, also called sigma notation. The capital Greek letter Σ “sigma” means to add. For example, Σx means to sum up all of the x values where X is the variable name.
A random sample of households had the following number of children living at home 4, –3, 2, 1, and 3. Calculate Σx.
Solution
Let x1 = 4, x2 = –3, x3 = 2, x4 = 1, x5 = 3. Start with the first value i = 1 up to the nth value i = 5 to get \(\sum_{i=1}^{n} x_{i}\) = 4 + –3 + 2 + 1 + 3 = 7.
To make things simpler we will drop the subscripts and write \(\sum_{i=1}^{n} x_{i}\) as Σxi or Σx.
The order of operations is important in summation notation.
For example, Σx2 = (4)2 + (–3)2 + (2)2 + (1)2 + (3)2 = 39.
When we insert parentheses (Σx)2 = (4 + –3 + 2 + 1 + 3)2 = (7)2 = 49.
Note that Σx2 ≠ (Σx)2.
“‘One of the interesting things about space,’ Arthur heard Slartibartfast saying to a large and voluminous creature who looked like someone losing a fight with a pink duvet and was gazing raptly at the old man's deep eyes and silver beard, ‘is how dull it is.’
‘Dull?’ said the creature, and blinked her rather wrinkled and bloodshot eyes.
‘Yes,’ said Slartibartfast, ‘staggeringly dull. Bewilderingly so. You see, there's so much of it and so little in it. Would you like me to quote some statistics?’
‘Er, well…’
‘Please, I would like to. They, too, are quite sensationally dull.’” (Adams, 2002)
3.1.2 Mean
The mean is the arithmetic average of the numbers. This is the center that most people call the average.
Distinguishing between a population and a sample is very important in statistics. We frequently use a representative sample to generalize about a population.
A statistic is any characteristic or measure from a sample. A parameter is any characteristic or measure from a population. We use sample statistics to make inferences about population parameters.
The sample mean = \(\overline{ x }\) (pronounced “x bar”) of a sample of n observations x1, x2, x3,…,xn taken from a population, is given by the formula:
\(\overline{ x }\) = \(\frac{\text { ∑x }}{\text { n }}\) = \(\frac{\text { x1+x2+x3+⋯+xn }}{\text { n }}\).
The population mean = μ (pronounced “mu”) is the average of the entire population, is given by the formula:
μ = \(\frac{\text { ∑x }}{\text { N }}\) = \(\frac{\text { x1+x2+x3+⋯+xN }}{\text {N }}\).
Most cases, you cannot find the population parameter, so you use the sample statistic to estimate the population parameter. Since μ cannot be calculated in most situations, the value for ��̅is used to estimate μ. You should memorize the symbol μ and what it represents for future reference.
Find the mean for the following sample of house prices ($1,000): 325, 375, 385, 395, 420, and 825.
Solution
Before starting any mathematics problem, it is always a good idea to define the unknown in the problem. In this case, you want to define the variable. The symbol for the variable is x. The variable is x = price of a house in $1,000.
\(\overline{ x }\) = \(\frac{\text { ∑x }}{\text { n }}\) = \(\frac{\text { 325+375+385+395+420+825 }}{\text { 6 }}\) = 454.1\(\overline{6}\)
The sample mean house price is $454,166.67.
We can use technology to find the mean. Directions for the TI calculators are in the next section. In Excel, use the cell function AVERAGE(array). For this example, we can type the data into column A and then in a blank cell =AVERAGE(A1:A6).
3.1.3 Weighted Mean
Weighted averages are used quite often in real life. Some teachers use them in calculating your grade in the course, or your grade on a project. Some employers use them in employee evaluations. The idea is that some components of a mean are more important than others are. As an example, a full-time teacher at a community college may be evaluated on their service to the college, their service to the community, whether their paperwork is turned in on time, and their teaching. However, teaching is much more important than whether their paperwork is turned in on time. When the evaluation is completed, more weight needs to be given to the teaching and less to the paperwork. This is a weighted average.
Weighted Mean = \(\frac{\text { sum of the scores times their weights }}{\text { sum of all the weights }}\) = \(\frac{\Sigma(x w)}{\sum w}\), where w is the weight of the data value x.
In your biology class, your final grade is based on several things: a lab score, scores on two major tests, and your score on the final exam. There are 100 points available for each score. The lab score is worth 15% of the course, the two exams are worth 25% of the course each, and the final exam is worth 35% of the course. Suppose you earned scores of 95 on the labs, 83 and 76 on the two exams, and 84 on the final exam. Compute your weighted average for the course.
Solution
Variable: x = score
The weighted mean is \(\frac{\Sigma(x w)}{\Sigma w}\) = \(\frac{95(0.15)+83(0.25)+76(0.25)+84(0.35)}{0.15+0.25+0.25+0.35}\) = \(\frac{83.4}{1.00}\) = 83.4
The course average is 83.4 %.
A faculty evaluation process at Portland State University rates a faculty member on the following activities: teaching, publishing, committee service, community service, and submitting paperwork in a timely manner. The process involves reviewing student evaluations, peer evaluations, and supervisor evaluation for each teacher and awarding them a score on a scale from 1 to 10 (with 10 being the best). The weights for each activity are 20 for teaching, 18 for publishing, 6 for committee service, 4 for community service, and 2 for paperwork.
a) One faculty member had the following ratings: 8 for teaching, 9 for publishing, 2 for committee work, 1 for community service, and 8 for paperwork. Compute the weighted average of their evaluation.
b) Another faculty member had ratings of 6 for teaching, 8 for publishing, 9 for committee work, 10 for community service, and 10 for paperwork. Compute the weighted average of their evaluation.
c) Which faculty member had the higher average evaluation?
Solution
a) Variable: x = rating
The weighted average is \(\frac{\Sigma(x w)}{\sum w}\) = \(\frac{8(20)+9(18)+2(6)+1(4)+8(2)}{20+18+6+4+2}\) = \(\frac{354}{50}\) = 7.08
The average evaluation score is 7.08.
b) The weighted average is \(\frac{\Sigma(x w)}{\sum w}\) = \(\frac{6(20)+8(18)+9(6)+10(4)+10(2)}{20+18+6+4+2}\) = \(\frac{378}{50}\) = 7.56
The average evaluation score is 7.56.
c) The second faculty member has a higher average evaluation.
3.1.4 Median
Another statistic that measures the center of a distribution is the median.
The median is the data value in the middle of the ordered data that has 50% of the data below that point and 50% of the data above that point. The median is also referred to as the 50th percentile and is the midpoint of a distribution.
To find the median:
- Arrange the observations from smallest to largest.
- If the number of observations n is odd, the middle observation is the median.
- If the number of observations n is even, the mean of the two middle observations is the median.
Find the median for the following sample of ages: 15, 23, 18, 15, 24, 23, and 17.
Solution
First, sort the data: 15, 15, 17, 18, 23, 23, and 24. The sample size is odd so the median will be the middle number. Use your fingers to cover outside numbers, one pair at a time until you get to 18. Median = 18 years old.
Find the median for the following sample of house prices (in $1,000): 325, 375, 385, 395, 420, and 825.
Solution
The data is already ordered from smallest to largest. The sample size is even so take the average of the two middle values \(\frac{\text { 385+395 }}{\text { 2 }}\) = 390. The median house price is $390,000.
We can use technology to find the median. Directions for the TI calculators are in the next section. In Excel the median is found using the cell function MEDIAN(array). For this example, we can type the data into column A and then in a blank cell =MEDIAN(A1:A6).
Recall that the sample mean house price is $454,167. Note that the median is much lower than the mean for this example. The observation of 825 is an outlier and is very large compared to the rest of the data. The sample mean is sensitive to unusual observations, i.e. outliers. The median is resistant to outliers.
3.1.5 Outliers
An outlier is a data value that is very different from the rest of the data and is far enough from the center. If there are extreme values in the data, the median is a better measure of the center than the mean. The mean is not a resistant measure because it is moved in the direction of the outlier. The median and the mode are resistant measures because they are not affected by extreme values.
As a consumer, you need to be aware that people choose the measure of center that best supports their claim. When you read an article in the newspaper and it talks about the “average,” it usually means the mean but sometimes it refers to the median. Some articles will use the word “median” instead of “average” to be more specific. If you need to make an important decision and the information says “average,” it would be wise to ask if the “average” is the mean or the median before you decide.
As an example, suppose that a company administration wants to use the mean salary as the average salary for the company. This is because the high salaries of the administration will pull the mean higher. The company can say that the employees are paid well because the average is high. However, the employees’ union wants to use the median since it discounts the extreme values of the administration and will give a lower value of the average. This will make the salaries seem lower and that a raise is in order.
Why use the mean instead of the median? When multiple samples are taken from the same population, the sample means tend to be more consistent than other measures of the center. The sample mean is the more reliable measure of center.
3.1.6 Distribution Shapes
Remember that there are varying levels of skewness and symmetry. Sample data is rarely exactly symmetric, but is approximately symmetric. Outliers will pull the mean in the direction of the outlier. If the distribution has a skewed tail to the left, the mean will be smaller than the median. If the distribution has a skewed tail to the right, the mean will be larger than the median. The mode, or modal class, is the tallest point(s), highest frequency, of the distribution. The following show examples of different distribution shapes. Figures 3-2 to 3-5 show example distribution shapes.
Figure 3-2
Figure 3-3
Figure 3-4
'
Figure 3-6
Comparing the mean and the median provides useful information about the distribution shape.
- If the mean is equal to the median, the data is symmetric, see Figure 3-6.
Figure 3-6
If the mean is larger than (to the right of) the median, the data is right skewed or positively skewed, see Figure 3-7.
If the mean is smaller than (to the left of) the median, the data is left skewed, or negatively skewed, see Figure 3-8.
Figure 3-7
Figure 3-8
The following is a histogram for a random sample of student rent prices. Comment on the distribution shape.
Figure 3-9
Figure 3-10
Solution
If we were to use Excel to find the mean and median we would get that the mean house rental price is $1,082.08 and the median house rental price is $1,030. The mean is larger than the median and is being pulled to the right by the outlier of $2,550.
If you were to draw a curve around the bars as in Figure 3- 10, you would get a tail for the one data point on the right. The outlier on the right is the direction of the skewness.
This distribution is skewed to the right, or positively skewed.
Which measure of center is used on which type of data?
- Mode can be found on nominal, ordinal, interval, and ratio data, since the mode is just the data value that occurs most often. You are just counting the data values.
- Median can be found on ordinal, interval, and ratio data, since you need to put the data in order. As long as there is order to the data, you can find the median.
- Mean can be found on interval and ratio data, since you must have numbers to add together.
3.02: Measures of Spread
Variability is an important idea in statistics. If you were to measure the height of everyone in your classroom, every student gives you a different value. That means not every student has the same height. Thus, there is variability in people’s heights. If you were to take a sample of the income level of people in a town, every sample gives you different information. There is variability between samples too. Variability describes how the data are spread out. If the data are very close to each other, then there is low variability. If the data are very spread out, then there is high variability. How do you measure variability? It would be good to have a number that measures it. This section will describe some of the different measures of variability, also known as variation.
Numerical statistics for variation can show how spread out data is. The variation of data is relative, and is usually used when comparing two sets of similar data. When we are making inferences about an average, we can make better estimates when there is less variation in the data. The four most common measures of the “spread” of data are called the range, variance, standard deviation, and coefficient of variation.
A sample of house prices (in $1,000): 325, 375, 385, 395, 420, and 825, found the mean house price of $454,167. How much does this tell you about the price of all houses? Can you tell if most of the prices were close to the mean or were the prices really spread out? What is the highest price and the lowest price? All you know is that the center of the price is $454,167. What if you were approved for only $400,000 for a home loan, could you buy a home in this area? You need more information.
3.2.1 Range
The range of a set of data is the difference between the highest and the lowest data values (or maximum and minimum values). Note in statistics we only report a single number which represents the spread from the lowest to highest value.
Range = Max – Min.
Look at the following three sets of data. Find the mean, median and range of each of data set.
1. 10, 20, 30, 40, 50
2. 10, 29, 30, 31, 50
3. 28, 29, 30, 31, 32
Solution
1. 
mean = 30, median = 30, range = 50 – 10 = 40
2. 
mean = 30, median = 30, range = 50 – 10 = 40
3. 
mean = 30, median = 30, range = 32 – 28 = 4
Based on the mean, median, and range, the first two distributions are the same, but you can see from the graphs that they are distributed differently. In part 1, the data are spread out equally. In part 2, the data has a clump in the middle and a single value at each end. The mean and median are the same for part 3, but the range is much smaller. All the data is clumped together in the middle.
3.2.2 Variance & Standard Deviation
The range does not really provide a very detailed picture of the variability. A better way to describe how the data is spread out is needed. Instead of looking at the distance as the highest value from the lowest, how about looking at the distance each value is from the mean? This spread is called the deviation.
Suppose a vet wants to analyze the weights of cats. The weights (in pounds) of five cats are 6.8, 8.2, 7.5, 9.4, and 8.2. Compute the deviation for each of the data values. The deviation is how far each data point is from the mean. To be consistent always subtract the data point minus the mean.
Solution
Variable: X = weight of a cat. First, find the mean for the data set. The mean is \(\overline{ x }\) = \(\frac{\Sigma(x)}{\text { n }}\) = \(\frac{\text { (6.8+8.2+7.5+9.4+8.2) }}{\text { 5 }}\) = 8.02 pounds. Subtract the mean from each data point to get the deviations.
Figure 3-11
Now average the deviations. Add the deviations together.
Figure 3-12
The average distance from the mean cannot be zero. The reason the deviations add to 0 is that there are some positive and negative values. The sum of the deviations from the mean will always be zero.
To get rid of the negative signs square each deviation.
Figure 3-13
Then average the total of the squared deviations. The only thing is that in statistics there is a strange average here. Instead of dividing by the number of data values, you divide by the number of data values minus one. This n – 1 is called the degrees of freedom and will be discussed more later in the text. When we divide by the degrees of freedom, this gives an unbiased statistic. In this case, you would have the following:
s2 = \(\frac{(x-\bar{x})^{2}}{n-1}\) = \(\frac{\text { 3.728 }}{\text { 5 − 1 }}\) = \(\frac{\text { 3.728 }}{\text { 4 }}\) = 0.932 pounds2
Notice that this statistic is denoted as s2 . This statistic is called the sample variance and it is a measure of the average squared distance from the mean. If you now take the square root, you will get the average distance from the mean. The square root of the variance is called the sample standard deviation, and is denoted with the letter s.
s= \(\sqrt{0.932}\) = 0.9654 pounds
The standard deviation is the average (mean) distance from a data point to the mean. It can be thought of as how much a typical data point differs from the mean.
The sample variance formula: s2 = \(\frac{∑(x-\bar{x})^{2}}{n-1}\).
Where \(\overline{ x }\) is the sample mean, n is the sample size, and Σ means to find the sum.
The sample standard deviation formula: s = \(\sqrt{s^{2}}=\sqrt{\frac{\sum(x-x)^{2}}{n-1}}\).
The n – 1 in the denominator has to do with a concept called degrees of freedom (df). Dividing by the df makes the sample standard deviation a better approximation of the population standard deviation than dividing by n.
We rarely will find a population variance or standard deviation, but you will need to know the symbols.
The population variance formula: \(\sigma^{2}=\frac{\sum(x-\mu)^{2}}{N}\).
The population standard deviation formula: \(\sigma=\sqrt{\frac{\sum(x-\mu)^{2}}{N}}\).
The lower-case Greek letter σ pronounced “sigma” and σ2 represents the population variance, μ is the population mean, and N is the size of the population.
Note: the sum of the deviations should always be zero. Try not to round too much in the calculations for standard deviation since each rounding causes a slight error.
Suppose that a manager wants to test two new training programs. They randomly select 5 people for each training type and measures the time it takes to complete a task after the training. The times for both trainings are in table below. Which training method is more consistent?
Solution
It is important that you define what each variable is since there are two of them.
Variable 1: X1 = productivity from training 1
Variable 2: X2 = productivity from training 2
The units and scale are the same for both groups. To answer which training method better, first you need some descriptive statistics. Start with the mean for each sample.
\(\overline{ x }\)1 = \(\frac{\text { 56 + 75 + 48 + 63 + 59 }}{\text { 5 }}\) = 60.2 minutes
\(\overline{ x }\)2 = \(\frac{\text { 60 + 58 + 66 + 59 + 58 }}{\text { 5 }}\) = 60.2 minutes
Since both means are the same values, you cannot answer the question about which is better.
Now calculate the standard deviation for each sample.
Figure 3-14
Figure 3-15
The variance for each sample is: \begin{aligned}
&s_{1}^{2}=\frac{394.8}{4}=98.7 \text { minutes }^{2} \\
&s_{2}^{2}=\frac{44.8}{4}=11.2 \text { minutes }^{2}
\end{aligned}
The standard deviations are: s1 = \(\sqrt{ 98.7 }\) = 9.9348 minutes
s2 = \(\sqrt{ 11.2 }\) = 3.3466 minutes.
Comparing the standard deviations, the second training method seemed to be the better training since the data is less spread out. This means it is more consistent. It would be better for the managers in this case to have a training program that produces more consistent results so they know what to expect for the time it takes to complete the task.
Descriptive statistics can be time-consuming to calculate by hand so use technology.
One Variable Statistics on the TI Calculator
The procedure for calculating the sample mean ( \(\overline{ x }\) ) and the sample standard deviation (sx) for the TI calculator are shown below. Note, the TI calculator also gives you the population standard deviation (σx) because it does not know whether the data you input is a population or a sample. You need to decide which value you need to use, based on whether you have a population or sample. In almost all cases you have a sample and will be using sx. In addition, the calculator uses the notation of sx instead of just s. It is just a way for it to denote the information.
TI-84: Enter the data in a list and then press [STAT]. Use cursor keys to highlight CALC. Press 1 or [ENTER] to select 1:1-Var Stats. Press [2nd], then press the number key corresponding to your data list. Press [Enter] to calculate the statistics. Note: the calculator always defaults to L1 if you do not specify a data list.
sx is the sample standard deviation. You can arrow down and find more statistics. Use the min and max to calculate the range by hand. To find the variance simply square the standard deviation.
TI-89: Press [APPS], select FlashApps then press [ENTER]. Highlight Stats/List Editor then press [ENTER]. Press [ENTER] again to select the main folder. To clear a previously stored list of data values, arrow up to the list name you want to clear, press [CLEAR], then press enter.
Press [F4], select 1: 1-Var Stats. To get the list name to the List box, press [2nd] [VarLink], arrow down to list1 and press [Enter]. This will bring list1 to the List box. Press [Enter] to enter the list name and then enter again to calculate.
Use the down arrow key to see all the statistics.
Sx is the sample standard deviation. You can arrow down and find more statistics. Use the min and max to calculate the range by hand. To find the variance simply square the standard deviation or take the last sum of squares divided by n – 1.
Excel: Type in the data into one column, select the Data tab, and choose Data Analysis. Select Descriptive Statistics, and then select OK.
Highlight the data for the Input Range, if you highlighted a label; check the Labels in first row box. Select the circle to the left of Output Range, then click into the box to the right of the Output Range and select one cell where you want the top left-hand corner of your summary table to start. Select the box next to Summary statistics, then select OK, see below.
We get the following summary statistics:
In general, a “small” standard deviation means the data are close together (more consistent) and a “large” standard deviation means the data is spread out (less consistent). Sometimes you want consistent data and sometimes you do not. As an example, if you are making bolts, you want the lengths to be very consistent so you want a small standard deviation. If you are administering a test to see who can be a pilot, you want a large standard deviation so you can tell whom the good and bad pilots are.
What do “small” and “large” mean? To a bicyclist whose average speed is 20 mph, s = 20 mph is huge. To an airplane whose average speed is 500 mph, s = 20 mph is nothing. The “size” of the variation depends on the size of the numbers in the problem and the mean. Another situation where you can determine whether a standard deviation is small or large is when you are comparing two different samples. A sample with a smaller standard deviation is more consistent than a sample with a larger standard deviation.
If we were to compare the variability between two histograms. The standard deviation and variance measure the average spread from left to right. Take a moment and see if you can order the following histograms from the smallest to the largest standard deviation.
Figure 3-16
Figure 3-17
FIgure 3-18
Figure 3-19
The histogram that has more of the data close to the mean will have the smallest standard deviation. The histogram that has more of the data towards the end points will have a larger standard deviation. Figure 3-16 will have the largest standard deviation since more of the data is grouped in the first and last class. Figure 3-17 will have the smallest standard deviation since more of the data is grouped in the center class which will be close to the mean in a symmetric distribution. Figures 3-18 and 3-19 are harder to compare without also having access to the mean and median to indicate skewness. However, Figure 3-19 does have smaller frequencies in the first and last three classes compared to Figure 3-18.
The correct order from smallest to largest standard deviation would be Figure 3-17, Figure 3-19, Figure 3-18, and then Figure 3-16.
One should not compare the range, standard deviation or variance of different data sets that have different units or scale.
3.2.3 Coefficient of Variation
The coefficient of variation, denoted by CVar or CV, is the standard deviation divided by the mean. The units on the numerator and denominator cancel with one another and the result is usually expressed as a percentage. The coefficient of variation allows you to compare variability among data sets when the units or scale is different.
Coefficient of Variation = CVar = ( \(\frac{\text { s }}{\text { \(\overline{ x }\) }}\)
∙ 100) %
The following is a sample of the alcohol content and calories for 12 oz. beers. Is the alcohol content (alcohol by volume ABV) or calories more variable?
3.03: Measures of Placement
3.3.1 Z-Scores
A z-score is the number of standard deviations an observation x is above or below the mean. Z-scores are used to compare placement of a value compared to the mean.
If x is an observation from a sample then the standardized value of x is the z-score where z = \(\frac{x-\bar{x}}{s}\)
If x is an observation from a population then the standardized value of x is the z-score where z = \(\frac{x-\mu}{\sigma}\)
If the z-score is negative, x is less than the mean. If the z-score is positive, x is greater than the mean.
There are no shortcuts on the calculator in Excel for z-score, but you can find s and \(\overline{ x }\) then simply subtract and divide.
The number of standard deviations that a data value is from the mean is frequently used when comparing position of values. If a z-score is zero, then the data value is the same as the mean. If the z-score is one, then the data value x is one standard deviation above the mean. If the z-score is –3.5, then the data value is three and a half standard deviations below the mean. The shaded area in Figure 3-20 represents one standard deviation from the mean.
Figure 3-20
For a random sample, the mean time to make a cappuccino is 2.8 minutes with a standard deviation of 0.86 minutes. Find the z-score for someone that makes their cappuccino in 4.95 minutes.
Solution
z = \(\frac{x-\bar{x}}{s}\) = \(\frac{4.95-2.8}{0.86}\) = 2.5. Their time is 2.5 standard deviations above average.
On a math test, a student scored 45. The class average was 50 with a standard deviation of 3. The same student scored an 80 on a history test, and the class average was 85 with a standard deviation of 2.5. Which exam did the student perform better on compared with the rest of the class?
Solution
\(z_{m}\) = \(\frac{45-50}{3}\) = -1.67 \(\quad\) \(z_{h}\) = \(\frac{80-85}{2.5}\) =-2
Test scores are “better” when they are larger, so whichever has the largest z-score did better. The student did better on the math test than the history test, compared to the rest of the class.
Be careful with the word “better,” depending on the context, better may be smaller rather than larger. For example, golf scores, time running a race, and cholesterol levels would be better if they were smaller values.
The length of a human pregnancy has a mean of 272 days. A pregnancy lasting 281 days or more has a z-score of one. How many standard deviations above the mean is a pregnancy lasting 281 days or more?
Solution
One, since by definition the z-score is the number of standard deviations from the mean.
The length of a human pregnancy has a mean of 272 days. A pregnancy lasting 281 days or more has a z-score of one. What is the standard deviation of human pregnancy length?
Solution
We know the z-score = 1 and mean = 272. Replace these two numbers in the z-score formula then solve for the standard deviation.
1 = \(\frac{281-272}{\sigma}\) \(\quad\) \(\Rightarrow\) \(\quad\) 1 = \(\frac{9}{\sigma}\) \(\quad\) \(\Rightarrow\) \(\quad\) \(\sigma\) = 9
3.3.2 Percentiles
Along with the center and variability, another useful numerical measure is the ranking of a number. A percentile is a measure of ranking. It represents a location measurement of a data value to the rest of the values. Many standardized tests give the results as a percentile. Doctors use percentiles graphs to show height and weight standards.
Interpreting Percentiles
The p th percentile is the value that separates the bottom p% from the upper (100 – p)% of the ordered (smallest to largest) data. For example, the 75th percentile is the value that separates the bottom 75% from the upper 25% of the data. There are several methods used to find percentiles. You may get different percentile values depending on which software or calculator you use. For example, Excel has two methods, both of which are not the same method as the TI calculators.
What does a score of the 90th percentile represent?
Solution
This means that 90% of the scores were at or below this score. (A person did the same as or better than 90% of the test takers.)
What does a score of the 70th percentile represent?
Solution
This means that 70% of the scores were at or below this score.
Percentile versus Score
If the test was out of 100 points and you scored at the 80th percentile, what was your score on the test? You do not know! All you know is that you scored the same as or better than 80% of the people who took the test. If all the scores were low, you could have still failed the test. On the other hand, if many of the scores were high you could have gotten a 95% or so.
Note there is more than one method to find percentiles. This rounding rule in Excel is not the same as used on your TI calculators.
Finding a Percentile:
Step 1: Arrange the data in order from lowest to highest.
Step 2: Substitute into the formula i= \(\frac{(n+1) \cdot p}{100}\) where n = sample size and p = percentile.
Step 3A: If i is a whole number, count out i places from the lowest number to find the percentile. For example, if you get i = 3, then the 3rd value is the percentile.
Step 3B: If i is not a whole number, then take the weighted average between the ith and ith +1 data value as the percentile. For example, if i = 3.25, this would be 25% of the distance between the 3rd and the 4th data values as the percentile. Percentile = ith data value + (ith + 1 data value – ith data value)*(0.##) where ## is the remainder percent.
Compute the 10th percentile of the random sample of 13 ages: 15, 18, 22, 25, 26, 31, 33, 35, 38, 46, 51, 53, and 95.
Solution
Data is already ordered, so next find i= \(\frac{(n+1) \cdot p}{100} = \frac{14 \cdot 10}{100}\) = 1.4. Since i is not a whole number use Step 3B, take the weighted average of 40% of the way between the 1st and 2nd values. This would be 15 + (18 – 15)∙0.4 = 16.2 and this is your 10th percentile, P10 = 16.2.
In Excel use =PERCENTILE.EXC(array, k) where array is the cell reference to where the data is located and k is the percentile as a decimal between 0 and 1. Note you do not have to sort the data prior to typing it in to Excel.
For this example, if you type in the data into column A, then use the formula =PERCENTILE.EXC(A1:A13, 0.1) = 16.2.
3.3.3 Quartiles
There are special percentiles called quartiles. Quartiles are numbers that divide the data into fourths. One fourth (or a quarter) of the data falls between consecutive quartiles. There are three quartiles Q1, Q2, and Q3 that subsequently divide the ordered data into the 4 pieces of approximately equal size, or 25% each. Thus, 25% of the values are less than Q1, 25% of the data values are between Q1 and Q2, 25% of the data values are between Q2 and Q3, and 25% are of the data values are greater than Q3.
Use the dollar as an example. If we make change for a dollar, we would get four quarters to make one dollar. Hence, quarter for quartiles.
To find the quartiles use the same rules as percentiles where we:
1. Arrange the observations from smallest to largest and use the previous percentile rule.
2. Then find all three quartiles.
- Q1 = first quartile = 25th percentile
- Q2 = second quartile = median = 50th percentile
- Q3 = third quartile = 75th percentile
Compute all three quartiles for the random sample of 13 ages: 15, 18, 22, 25, 26, 31, 33, 35, 38, 46, 51, 53, and 95.
Solution
For the first quartile i= \(\frac{(n+1) \cdot p}{100}\) = \(\frac{14 \cdot 25}{100}\) = 3.5. Since i is not a whole number take the weighted average of half way between the 3rd and 4th data values 22 + (25 – 22)∙0.5 = 23.5, so Q1 = 23.5.
In Excel you could use the percentile formula, but there is also a quartile formula: =QUARTILE.EXC(array, quartile), where array is the cell reference to the data and quartile is either 1, 2 or 3 for the 3 possible quartiles.
In this example we would have =QUARTILE.EXC(A1:A13, 1) = 23.5.
To find the second quartile: i = \(\frac{(n+1) \cdot p}{100}=\frac{14 \cdot 50}{100}\) = 7. Since i is a whole number use the 7th value for Q2, so Q2 = 33.
Or use the Excel formula =QUARTILE.EXC(A1:A13, 2) = 33.
For the third quartile, i = \(\frac{(n+1) \cdot p}{100}=\frac{14 \cdot 75}{100}\) = 10.5. Since i is not a whole number use the weighted average of half way between the 10th and 11th values 46 + (51 – 46)∙0.5 = 48.5.
Or use the Excel formula =QUARTILE.EXC(A1:A13, 3) = 48.5, so Q3 = 48.5.
The high school graduating class of 2016 in Oregon had the following ACT quartile scores. Interpret what the number 26 under the Composite column represents.
https://www.act.org/content/dam/act/unsecured/documents/P_38_389999_S_S_N00_ACT-GCPR_Oregon.pdf
Solution
From the report we can see that the third quartile for composite score is 26, this means that 75% of Oregon students that took the ACT exam scored 26 or below.
Other Types of Percentiles
Quintiles break a data set up into five equal pieces. We will not be using these, but be aware that percentiles come in different forms.
Deciles break a data set up into ten equal pieces and are found using the percentile rule. For example, the 6th decile = D6 = 60th percentile.
Use the dollar as an example. If we make change for a dollar, we would get ten dimes to make one dollar. Hence, a dime might help you remember deciles.
Earlier in Example 2-14, we made an ogive using Excel with the following sample of 35 ages. Use the ogive to find the age for the 8th decile.
3.3.4 Five Number Summary & Outliers
If you record the quartiles together with the minimum and maximum values from a data set, you have five numbers. These five numbers are known as the five-number summary consisting of the minimum, the first quartile (Q1), the median (Q2), the third quartile (Q3), and the maximum (in that order).
The interquartile range, IQR, is the difference between the first and third quartiles, Q1 and Q3. Half of the data (50%) falls in the interquartile range. If the IQR is “large,” the data is spread out and if the IQR is “small,” the data is closer together.
The interquartile range (IQR) = Q3 – Q1
Not only does the IQR give a range of the middle 50% of the data, but is also used to determine outliers in a sample.
To find these outliers we first find what are called a lower and upper limit sometimes called fences.
The lower limit, or inner fence, is Q1 – (1.5·IQR). Any values that are less than the lower limit is considered an outlier.
Similarly, the upper limit, or outer fence, is Q3 + (1.5·IQR). Any values that are more than the upper limit are considered outliers.
If all the numbers in the sample fall between the lower and upper limit, including the endpoints, then there are no outliers in the sample. Any values outside these limits would be considered outliers.
3.3.5 Modified Box-and-Whisker Plot
A boxplot (or box-and-whisker plot) is a graphical display of the five-number summary. A boxplot can be drawn vertically or horizontally. The modified boxplot shows outliers, whereas a regular boxplot does not show outliers. The basic format of the plot is a box drawn from Q1 to Q3, a vertical line drawn inside the box for the median, and horizontal lines (called whiskers) extending out of the middle of each end of the box to the minimum and maximum. The box should not touch the number line. The modified boxplot extends the left line to the smallest value greater than the lower fence, and extends the right line to the largest value less than the upper fence. Dots, circles or asterisks represent any outlier. We will make modified boxplots for this course. Like always, label the tick marks on the number line and give the graph a title.
A boxplot is a graph of the 5-number summary, see Figure 3-23.
Figure 3-23
It is important to note that when you are making the boxplot the limits for finding outliers are not graphed in the plot, they were only used to find the outliers. The whiskers would go to the next largest (or smallest) value in the data set after you removed the outlier(s).
If the sample has a symmetrical distribution, then the boxplot will be visibly symmetrical. If the data distribution has a left skew or a right skew, the line on that side of the boxplot will be visibly long in the direction of skewness. If the four quartiles are all about the same distance apart, then the data are likely a near uniform distribution. If a boxplot is symmetrical, and both outside lines are noticeably longer than the Q1 to median and median to Q3 distance, the distribution is then probably bell-shaped.
Make a modified box-and-whisker plot for the random sample of 13 ages: 15, 18, 22, 25, 26, 31, 33, 35, 38, 46, 51, 53, and 95.
Solution
Use Excel to compute the three quartiles as:
Q1 =QUARTILE.EXC(A1:A13, 1) = 23.5
Q2 =QUARTILE.EXC(A1:A13, 2) = 33
Q3 =QUARTILE.EXC(A1:A13, 3) = 48.5.
The 5-number summary values are 15, 23.5, 33, 48.5 and 95. Each of these numbers will need to be incorporated into the box-and-whisker plot, and any outliers to graph the modified box-and-whisker plot.
To find the outliers, first find the IQR, and then find the lower and upper limits. The IQR = Q3 – Q1 = 48.5 – 23.5 = 25. The lower limit is Q1 – (1.5· IQR) = 23.5 – 1.5(25) = –14. The upper limit is Q3 + (1.5· IQR) = 48.5 + 1.5(25) = 86.
Any value in our data set that is not between the lower and upper limits [–14, 86] is an outlier. By observation, we have one number that is outside the range so the outlier is 95.
The whiskers would be drawn to the next largest (or smallest) value in the data set after you removed the outlier(s). For this example, the next largest value in the data set is 53.
Now put that all together to get the following graph in Figure 3-24
Figure 3-24
The TI-calculator and newer versions of Excel will make a modified boxplot. Note that the quartile rules used in the TI calculators are slightly different then in Excel and what is presented in this content. They do not use a weighted mean between values, just half way between values.
TI-84: First, enter your data in to list 1. Next, press 2nd > STAT PLOT, then choose the first plot. Note that your calculator may say Plot1…Off or show a different type of graph then the screenshot. Using your arrow keys, turn the plot on. Choose the modified boxplot which is the first of the two boxplot options with the small dots to the right of the whiskers representing outliers. Make sure your Xlist: is on L1, keep frequency as a one, and any mark will work, but the square shows up best.
Here is screen shot from the calculator for the last example. You can use Trace on the boxplot from the TI-84 calculator below to see where each quartile, whisker and outlier are.
TI-89: Enter the data into the Stat/List editor under list 1. Press [APP] then scroll down to Stat/List Editor; on the older style TI-89 calculators, go into the Flash/App menu, and then scroll down the list. Make sure the cursor is in the list, not on the list name, and type the desired values pressing [ENTER] after each one. To clear a previously stored list of data values, arrow up to the list name you want to clear, press [CLEAR], and then press enter. After you enter the data, select Press [F2] Plots, scroll down to [1: Plot Setup] and press [Enter]. Then select [F1] Define.
Use your arrow keys to select Mod Box Plot for Type, and then scroll down to the x-variable box. Press [2nd] [VarLink] this key is above the + sign. Then arrow down until you find your List1 name under the Main file folder.
Then press [Enter] and this will bring the name List1 back to the menu. You will now see that Plot1 has a small picture of a boxplot. To view the boxplot, press [F5] Zoom Data.
Select [F3:Trace] to see the five-number summary and any outliers. Use the left and right arrow keys to move to the other values.
Excel: Note this example is on a PC running Excel 2019. Older versions of Excel may not have a boxplot option. First, type your sample data into column A in any order. Highlight the data, and then select the Insert tab. Under the graphing options, the picture shaped liked a histogram called statistical charts, select Box and Whisker.
You can change the formatting options and add the chart title as needed.
Below is the finished Excel boxplot. Note that Excel does a vertical boxplot rather than the traditional horizontal number line.
Excel marks an × just above the median where the mean would fall. Usually one would not include the mean on a boxplot. Remember that when the mean is greater than the median the distribution is usually skewed to the right.
When a boxplot has outliers only on one side, then we can also say the distribution is skewed in the direction of the outlier, which also indicates that these ages are skewed to the right.
Side by side boxplots are great at comparing quartiles and distribution shapes for several samples using the same units and scale.
There are four franchises in different parts of town. Compare the weekly sales over a year for each of the four franchises. Compare the boxplots shown in Figure 3-25.
Figure 3-25
Solution
We can see that Store 2 in Figure 3-25 has the highest sales since the median for this store is higher than the third quartile for all the other stores. Store 2 also has sales that are more consistent from week to week with the smaller range and has a symmetric distribution. The lowest performing store, Store 1, has the lowest median sales and is skewed to the right. Both Stores 3 and 4 have moderate sales and are skewed left
3.3.6 Empirical Rule
Before looking at the process for computing probabilities, it is somewhat useful to look at the Empirical Rule which gives the approximate proportion of data points under a bell-shaped curve between two points. The Empirical Rule is just an approximation, more precise methods for finding these proportions will be demonstrated in later sections.
The Empirical Rule should only be used with bell-shaped data.
The Empirical Rule (also called the 68-95-99.7 Rule)
In a bell-shaped distribution with mean μ and standard deviation σ,
- Approximately 68% of the observations fall within 1 standard deviation (σ) of the mean μ.
- Approximately 95% of the observations fall within 2 standard deviations (2σ) of the mean μ.
- Approximately 99.7% of the observations fall within 3 standard deviations (3σ) of the mean μ.
Note that we are using notation for the population mean μ and the population standard deviation σ, but the rule would also work using the sample mean and sample standard deviation.
Figure 3-26
In 2009 the average SAT mathematics score was 501, with a standard deviation of 116. Assume that SAT scores are bell-shaped.
a) Approximately what proportion of students scored between 269 and 733 on the 2009 SAT mathematics test?
b) Approximately what proportion of students scored between 385 and 617 on the 2009 SAT mathematics test?
c) Approximately what proportion of students scored at least 617 on the 2009 SAT mathematics test?
Solution
a) The key word is bell shaped so we can use the Empirical Rule. Start by finding the z-scores z = \(\frac{x-\mu}{\sigma}\) for both endpoints given in the question.
A z-score by definition is the number of standard deviations a data value is from the mean.
Figure 3-27
The two z-scores show that the test scores of 269 and 733 are two standard deviations from the mean. Using the second bulleted item in the Empirical Rule the answer would be approximately 95% of the math SAT scores will fall between 269 and 733.
b) Take the z-scores of the endpoints to get: z=\(\frac{385-501}{116}=-1, z=\frac{617-501}{116}=1\)
Figure 3-28
The two z-scores show that the test scores of 385 and 617 are one standard deviation from the mean. Using the first bulleted item in the Empirical Rule the answer would be approximately 68% of the math SAT scores will fall between 385 and 617.
c): Start by taking the z-score of 617, we get z = \(\frac{617-501}{116}\) = 1.
Since a bell-shaped curve is symmetric and we can assume that 100% of the population is represented then we can subtract the middle from the whole to get 100% – 68% = 32%. If we divide this outside area by two, \(\frac{32%}{2}\) = 16%, we would expect 16% in each tail area. See Figure 3-29.
Figure 3-29
The answer would be approximately 16% of students scored at least 617 on the 2009 SAT mathematics test.
If you were to get a z-score that is not –3, –2, –1, 1, 2 or 3 then you would not be able to apply the Empirical Rule. We also need to ensure that our population has a bell-shaped curve before using the Empirical Rule.
3.3.7 Chebyshev’s Theorem
One way of estimating the proportion of values from any data set within a certain number of standard deviations is Chebyshev’s Theorem. Pafnuty Chebyshev (Чебышёва) was a Russian mathematician who proved several important theorems. One that we will use for this chapter is called Chebyshev’s Inequality. The Empirical Rule only works for bell-shaped distributions. However, we can use Chebyshev’s Inequality for any distribution shape.
Chebyshev’s Inequality: The proportion (percent or fraction) of values from a data set that will fall within k standard deviations of the mean will be at least \(\left(\left(1-\frac{1}{(z)^{2}}\right) \cdot 100\right)\) % , where z is a real number that has an absolute value greater than 1 (z is not necessarily an integer).
The average number of acres for farms in a certain country is 443 with a standard deviation of 42 acres. At least what percent of farms will have between 338 and 548 acres?
Solution
The question gives no indication for the distribution shape for number of farm acres so we will use Chebyshev’s Inequality instead of the Empirical Rule. The easiest way to start is to find the z-score of the lower and upper bounds given in the question where μ = 443 and σ = 42.
z = \(\frac{338−443}{42}\) = −2.5 and z = \(\frac{548−443}{42}\) = 2.5
Use either z-score, but it is easier to use the positive value, and substitute into the formula:
\(\left(\left(1-\frac{1}{(z)^{2}}\right) \cdot 100\right) \%=\left(\left(1-\frac{1}{(2.5)^{2}}\right) \cdot 100\right) \%=84 \%\)
At least 84% of the farms will have between 338 and 548 acres.
The average quiz score for a statistics course is 15.2 with a standard deviation of 3.15. What are the quiz scores that would have at least 75% of the scores between?
Solution
This question gives a percent and we need to work backward. We will use Chebyshev’s Inequality since there is no mention of a bell-shaped distribution. Use algebra to solve for z in the following formula
\(\left(\left(1-\frac{1}{(z)^{2}}\right) \cdot 100\right) \%\) = 75%.
Start by dividing both sides by 100% to get rid of the %. We then have 1 − \(\frac{1}{(z)^{2}}\) = 0.75.
Next, add \(\frac{1}{(z)^{2}}\) to both sides of the equation and subtract 0.75 from both sides of the equation to get 0.25 = \(\frac{1}{(z)^{2}}\).
Multiply both sides by z2 , divide both sides by 0.25 and simplify to get (z)2 = \(\frac{1}{0.25}\) \(\quad\) \(\Rightarrow\) \(\quad\) z2 = 4.
Take the square root of both sides \(\sqrt{z^{2}}\) = \(\sqrt{4}\) \(\quad\) \(\Rightarrow\) \(\quad\) = \(\pm 2\).
This means that according to Chebyshev’s Theorem at least 75% of the data will fall within two standard deviations from the mean. Next, we need to find what quiz scores are two standard deviations from the mean.
Find the mean \(\pm 2\) standard deviations, by computing: \(\begin{aligned}
&\mu-2 \cdot \sigma=15.2-2 \cdot 3.15=8.9 \\
&\mu+2 \cdot \sigma=15.2+2 \cdot 3.15=21.5
\end{aligned}\)
At least 75% of the students scored between 8.9 and 21.5 on the quiz.
The general formulas for finding the endpoints are: Lower endpoint: a = μ – z·σ
Upper endpoint: b = μ + z·σ
If the distribution of quiz scores were bell shaped we would have a larger percent (95%) between 8.9 and 21.5. Chebyshev’s Inequality assumes the distribution is skewed and says “at least” which would then also be correct if more students fell within the range.
Since Chebyshev’s Inequality works for any shape of a distribution we should only use if between two values, not strictly below or above a point. If we were interested in below a certain point, we would not know if we had the fat or skinny tail of a skewed distribution. The following picture shows a positively skewed distribution with 1.5 standard deviations from the mean shaded in red. Chebyshev’s Inequality would guarantee at least
\(\left(\left(1-\frac{1}{(1.5)^{2}}\right) \cdot 100\right) \%\) = 56% of the data in the shaded area would fall within 1.5 standard deviations from the mean.
Figure 3-30
Summary:
Use the mode for nominal, ordinal, interval, and ratio data, since the mode is just the data value that occurs most often. You are just counting the data values. The median can be found on ordinal, interval, and ratio data, since you need to put the data in order. As long as there is order to the data you can find the median. The mean can be found on interval and ratio data, since you must have numbers to add together. The mean is pulled in the direction of outliers. By comparing the mean to the median, you can decide if a distribution is symmetric or skewed. The range, variance, standard deviation and coefficient of variation are used to measure how spread out a data set is from the middle. When comparing two data sets with different units or scale use the coefficient of variation. Z-scores tell you how many standard deviations a data point is away from the mean. Quartiles are special percentiles that are used to find the interquartile range, identify outliers, and make a box-and-whisker plot. Use the Empirical Rule when finding the proportion of a sample or population that fall within 1, 2 or 3 standard deviations on a bell-shaped curve. If the distribution is bell shaped then the Empirical Rule states that approximately 68% of the data will fall within one standard deviation, 95% within two standard deviations and 99.7% within three standard deviations. If you do not know the distribution shape, then use Chebyshev’s Inequality to find the minimum proportion within |z| > 1 standard deviations from the mean.
3.04: Chapter 3 Formulas
| Sample Mean: \(\bar{x} = \frac{\sum x}{n}\) | Population Mean: \(\mu = \frac{\sum x}{N}\) |
| Weighted Mean: \(\bar{x} = \frac{\Sigma(x w)}{\sum w}\) | Range = Max – Min |
| Sample Standard Deviation: s = \(\sqrt{\frac{\sum(x-\bar{x})^{2}}{n-1}}\) | Population Standard Deviation = σ |
| Sample Variance: s2 = \(\frac{\Sigma(x-\bar{x})^{2}}{n-1}\) | Population Variance = σ2 |
| Coefficient of Variation: CVar = \(\left(\frac{s}{\bar{x}} \cdot 100\right) \%\) | Z-Score: z = \(\frac{x-\bar{x}}{s}\) |
| Percentile Index: i = \(\frac{(n + 1) \cdot p}{100}\) | Interquartile Range: IQR = Q3 – Q1 |
| Empirical Rule: z = 1, 2, 3 \(\Rightarrow\) 68%, 95%, 99.7% | Outlier Lower Limit: Q1 – (1.5·IQR) |
| Chebyshev’s Inequality: \(\left(\left(1-\frac{1}{(z)^{2}}\right) \cdot 100\right) \%\) | Outlier Upper Limit: Q3 + (1.5·IQR) |
3.05: Chapter 3 Exercises
1. A sample of eight cats found the following weights in kg.
a) Compute the mode.
b) Compute the median.
c) Compute the mean.
2. Cholesterol levels in milligrams (mg) of cholesterol per deciliter (dL) of blood, were collected from patients two days after they had a heart attack (Ryan, Joiner & Ryan, Jr, 1985).
Retrieved from http://www.statsci.org/data/general/cholest.html.
a) Compute the mode.
b) Compute the median.
c) Compute the mean.
3. The lengths (in kilometers) of rivers on the South Island of New Zealand that flow to the Tasman Sea are listed below.
Data from http://www.statsci.org/data/oz/nzrivers.html.
a) Compute the mode.
b) Compute the median.
c) Compute the mean.
4. A university assigns letter grades with the following 4-point scale: A = 4.00, A– = 3.67, B+ = 3.33, B = 3.00, B– = 2.67, C+ = 2.33, C = 2.00, C– = 1.67, D+ = 1.33, D = 1.00, D– = 0.67, F = 0.00. Calculate the grade point average (GPA) for a student who took in one term a 4 credit math course and received a B+, a 1 credit seminar course and received an A, a 3 credit history course and received an A- and a 5 credit writing course and received a D.
5. A university assigns letter grades with the following 4-point scale: A = 4.00, A– = 3.67, B+ = 3.33, B = 3.00, B– = 2.67, C+ = 2.33, C = 2.00, C– = 1.67, D+ = 1.33, D = 1.00, D– = 0.67, F = 0.00. Calculate the grade point average (GPA) for a student who took in one term a 3 credit biology course and received a C+, a 1 credit lab course and received a B, a 4 credit engineering course and received an A- and a 4 credit chemistry course and received a C+.
6. An employee at Clackamas Community College (CCC) is evaluated based on goal setting and accomplishments toward the goals, job effectiveness, competencies, and CCC core values. Suppose for a specific employee, goal 1 has a weight of 30%, goal 2 has a weight of 20%, job effectiveness has a weight of 25%, competency 1 has a weight of 4%, competency 2 has weight of 3%, competency 3 has a weight of 3%, competency 4 has a weight of 3%, competency 5 has a weight of 2%, and core values has a weight of 10%. Suppose the employee has scores of 3.0 for goal 1, 3.0 for goal 2, 2.0 for job effectiveness, 3.0 for competency 1, 2.0 for competency 2, 2.0 for competency 3, 3.0 for competency 4, 4.0 for competency 5, and 3.0 for core values. Compute the weighted mean score for this employee. If an employee has a score less than 2.5, they must have a Performance Enhancement Plan written. Does this employee need a plan?
7. A statistics class has the following activities and weights for determining a grade in the course: test 1 worth 15% of the grade, test 2 worth 15% of the grade, test 3 worth 15% of the grade, homework worth 10% of the grade, semester project worth 20% of the grade, and the final exam worth 25% of the grade. If a student receives an 85 on test 1, a 76 on test 2, an 83 on test 3, a 74 on the homework, a 65 on the project, and a 61 on the final, what grade did the student earn in the course? All the assignments were out of 100 points.
8. A statistics class has the following activities and weights for determining a grade in the course: test 1 worth 15% of the grade, test 2 worth 15% of the grade, test 3 worth 15% of the grade, homework worth 10% of the grade, semester project worth 20% of the grade, and the final exam worth 25% of the grade. If a student receives a 25 out of 30 on test 1, a 20 out of 30 on test 2, a 28 out of 30 on test 3, a 120 out of 140 points on the homework, a 65 out of 100 on the project, and a 31 out of 35 on the final, what grade did the student earn in the course?
9. A sample of eight cats found the following weights in kg.
a) Compute the range.
b) Compute the variance.
c) Compute the standard deviation.
10. The following data represents the percent change in tuition levels at public, four-year colleges (inflation adjusted) from 2008 to 2013 (Weissmann, 2013). To the right is the histogram. What is the shape of the distribution?
11. The following is a histogram of quiz grades.
a) What is the shape of the distribution?
b) Which is higher, the mean or the median?
12. Cholesterol levels were collected from patients two days after they had a heart attack (Ryan, Joiner & Ryan, Jr, 1985). Retrieved from http://www.statsci.org/data/general/cholest.html.
a) Compute the range.
b) Compute the variance.
c) Compute the standard deviation.
13. Suppose that a manager wants to test two new training programs. The manager randomly selects five people for each training type and measures the time it takes to complete a task after the training. The times for both trainings are in table below. Which training method is more variable?
14. The lengths (in kilometers) of rivers on the South Island of New Zealand that flow to the Tasman Sea are listed below.
Data from http://www.statsci.org/data/oz/nzrivers.html.
a) Compute the range.
b) Compute the variance.
c) Compute the standard deviation.
15. Here are pulse rates before and after exercise. Which group has the larger range?
16. A midterm in a statistics course had a mean score of 70 with a standard deviation 5. A quiz in a biology course had an average of 20 with a standard deviation of 5.
a) Compute the coefficient of variation for the statistics midterm exam.
b) Compute the coefficient of variation for the biology quiz.
c) Aaliyah scored a 75 on the statistics midterm exam. Compute Aaliyah’s z-score.
d) Viannie scored a 25 on the biology quiz. Compute Viannie’s z-score.
e) Which student did better on their respect test? Why? f) Was there more variability for the midterm or the quiz scores? Justify your answer with statistics.
17. The following is a sample of quiz scores.
a) Compute \(\overline{ x }\).
b) Compute s2.
c) Compute the median.
d) Compute the coefficient of variation.
e) Compute the range.
18. The time it takes to fill an online order was recorded and the following descriptive statistics were found using Excel. What is the coefficient of variation?
19. The following is the height and weight of a random sample of baseball players.
a) Compute the coefficient of variation for both height and weight.
b) Is there more variation in height or weight?a) Compute the coefficient of variation for both height and weight.
20. A midterm in a statistics course had a mean score of 70 with a standard deviation 5. A quiz had an average of 20 with a standard deviation of 5. A student scored a 73 on their midterm and 22 on their quiz. On which test did the student do better on compared to the rest of the class? Justify your answer with statistics.
21. The length of a human pregnancy is normally distributed with a mean of 272 days with a standard deviation of 9.1 days. William Hunnicut was born in Portland, Oregon, at just 181 days into his gestation. What is the zscore for William Hunnicut’s gestation? Retrieved from: http://digitalcommons.georgefox.edu/cgi/viewcontent.cgi?article=1149&context=gfc_life.
22. Arm span (sometimes referred to as wingspan) is the physical measurement of the length of an individual's arms from fingertip to fingertip. The average arm span of a man is 70 inches with a standard deviation of 4.5 inches. The Olympic gold medalists Michael Phelps has an arm span of 6 feet 7 inches, which is three inches more than his height. What is the z-score for Michael Phelps arm span?
23. The average time to run the Pikes Peak Marathon 2017 was 7.44 hours with a standard deviation of 1.34 hours. Rémi Bonnet won the Pikes Peak Marathon with a run time of 3.62 hours. Retrieved from: http://pikespeakmarathon.org/results/ppm/2017/.
The Tevis Cup 100-mile one-day horse race for 2017 had an average finish time of 20.38 hours with a standard deviation of 1.77 hours. Tennessee Lane won the 2017 Tevis cup in a ride time of 14.75 hours. Retrieved from: https://aerc.org/rpts/RideResults.aspx.
a) Compute the z-score for Rémi Bonnet’s time.
b) Compute the z-score for Tennessee Lane’s time.
c) Which competitor did better compared to their respective events?
24. Cholesterol levels were collected from patients two days after they had a heart attack (Ryan, Joiner & Ryan, Jr, 1985). Retrieved from http://www.statsci.org/data/general/cholest.html.
a) Compute the 25th percentile.
b) Compute the 90th percentile.
c) Compute the 5th percentile.
d) Compute Q3.
25. A sample of eight cats found the following weights in kg. Compute the 5-number summary.
26. The following data represent the grade point averages for a sample of 15 PSU students.
a) Compute the lower and upper limits.
b) Identify if there are any outliers.
c) Draw a modified box-and-whisker plot.
27. The lengths (in kilometers) of rivers on the South Island of New Zealand that flow to the Tasman Sea are listed below.
Data from http://www.statsci.org/data/oz/nzrivers.html.
a) Compute the 5-number summary.
b) Compute the lower and upper limits and any outlier(s) if any exist.
c) Make a modified box-and-whisker plot.
28. The following are box-and-whiskers plot for life expectancy for European countries and Southeast Asian countries from 2011. What is the distribution shape of the European countries’ life expectancy?
29. To determine if Reiki is an effective method for treating pain, a pilot study was carried out where a certified second-degree Reiki therapist provided treatment on volunteers. Pain was measured using a visual analogue scale (VAS) immediately before and after the Reiki treatment (Olson & Hanson, 1997). Higher numbers mean the patients had more pain.
a) Use the box-and-whiskers plots to determine the IQR for the before treatment measurements.
b) Use the box-and-whiskers plots of the before and after VAS ratings to determine if the Reiki method was effective in reducing pain.
30. The median household income (in $1,000s) from a random sample of 100 counties that gained population over 2000-2010 are shown on the left. Median incomes from a random sample of 50 counties that had no population gain are shown on the right.
(OpenIntro Statistics, 2016)
What is the distribution shape for the counties with no population gain?
31. Match the correct descriptive statistics to the letter of the corresponding histogram and boxplot. Choose the correct letter for the corresponding histogram and Roman numeral for the corresponding boxplot. You should only use the visual representation, the definition of standard deviation and measures of central tendency to match the graphs with their respective descriptive statistics.
i. 
ii. 
iii. 
a) 
b)
c) 
32. A sample has the following statistics: Minimum = 10, Maximum = 50, Range = 40, \(\overline{ x }\) = 20, mode = 32, Q1 =30, Q2 =35, Q3 =38, standard deviation = 5 and IQR = 8.
a) According to Chebyshev’s Inequality, what percentage of data would fall within the values 12.5 and 27.5?
b) If this sample were bell-shaped what percentage of the data would fall within the values 10 and 30?
33. The length of a human pregnancy is bell-shaped with a mean of 272 days with a standard deviation of 9 days (Bhat & Kushtagi, 2006). Compute the percentage of pregnancies that last between 245 and 299 days.
34. Arm span is the physical measurement of the length of an individual's arms from fingertip to fingertip. A man’s arm span is approximately bell-shaped with mean of 70 inches with a standard deviation of 4.5 inches. What percent of men have an arm span between 61 and 79 inches?
35. The size of fish is very important to commercial fishing. A study conducted in 2012 found the length of Atlantic cod caught in nets in Karlskrona to have a mean of 49.9 cm and a standard deviation of 3.74 cm (Ovegard, Berndt & Lunneryd, 2012).
a) According to Chebyshev’s Inequality, at least what percent of Atlantic cod should be between 44.29 and 55.51 cm?
b) Assume the length of Atlantic cod is bell-shaped. Approximately what percent of Atlantic cod are between 46.16 and 53.64 cm?
c) Assume the length of Atlantic cod is bell-shaped. Approximately what percent of Atlantic cod are between 42.42 and 57.38 cm?
36. Scores on the SAT for a certain year were bell-shaped with a mean of 1511 and a standard deviation of 194.
a) What two SAT scores that separated the middle 68% of SAT scores for that year?
b) What two SAT scores that separated the middle 95% of SAT scores for that year?
c) How high did a student need to score that year to be in the top 2.5%?
37. In a mid-size company, the distribution of the number of phone calls answered each day by each of the 12 employees is bell-shaped and has a mean of 59 and a standard deviation of 10. Using the empirical rule, what is the approximate percentage of daily phone calls numbering between 29 and 89?
38. The number of potholes in any given 1 mile stretch of pavement in Portland has a bell-shaped distribution. This distribution has a mean of 54 and a standard deviation of 5. Using the empirical rule, what is the approximate percentage of 1-mile long roadways with potholes numbering between 44 and 59?
39. A company has a policy of retiring company cars; this policy looks at number of miles driven, purpose of trips, style of car and other features. The distribution of the number of months in service for the fleet of cars is bellshaped and has a mean of 42 months and a standard deviation of 3 months. Using the Empirical Rule, what is the approximate percentage of cars that remain in service between 48 and 51 months?
40. The following is an infant weight percentile chart. What is the 50th percentile height in cm for a 10-month old boy?
Retrieved from: https://www.cdc.gov/growthcharts/data/set1clinical/cj41l017.pdf.
- Answer to Odd Numbered Exercises
-
1 a) mode = 4 b) median = 3.75 c) \(\overline{ x }\) = 3.725
3) a) 56 & 64 b) 64 c) 67.6818
5) 2.833
7) 72.25
9) a) Range = 0.9 b) s 2 = 0.0870 c) s = 0.2949
11) a) Negatively skewed b) The median is higher.
13) s1 = 9.9348, s2 = 3.3466; Training 1 is more variable
15) Before range = 42, after range = 42, both groups have the same range.
17) a) 30.35 b) 136.7 c) 32.7 d) 38.52% e) 28.4
19) a) CVheight = 3.04%; CVweight = 9.84% b) Weight, because it has a higher coefficient of variation.
21) -10
23) a) -2.8507 b) -3.1808 c) Tennessee Lane
25) Min = 3.2, Q1 = 3.45 (TI: 3.55), Q2 = 3.7, Q3 = 3.95 (TI: 3.85), Max = 4.1
27) a) Min = 32, Q1 = 46 (TI: 48), Q2 = 64, Q3 = 77 (TI: 76), Max = 177 b) lower limit = -0.5 (TI: 6), upper limit = 123.5 (TI: 118), outliers = 177 (TI: 121 & 177) c)
29) a) 4 b) Yes, the treatment was effective.
31) 1. c. ii; 2. a. iii; 3. b. i
33) 99.7%
35) a) 55.56% b) 68% c) 95%
37) 99.7%
39) 2.35%
From Mostly Harmless Statistics by Rachel L. Webb, adapted from LibreTexts. Licensed CC BY-SA 4.0. XYZ Homework OER web edition, adapted with 7 verified corrections (see errata).