Chapter 2 Solutions. a) Construct a stem-and-leaf display for the distribution of ages.

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1 Chapter 2 Solutions 20. A health club is interested in the distribution of ages for its members that use the club on Friday evenings. The club s personnel director surveys the membership one Friday evening and collects the following data: Club Members Ages in Years a) Construct a stem-and-leaf display for the distribution of ages. Solution: To construct a stem-and-leaf display, use the procedure outlined on page 38 of the text. Step 1. The stem portion of the ages is the first digit, or tens place for the two digit numbers and last digit, or units place is the leaf value. Step 2. Since the smallest data value is 19 and the largest data value is 84, then the stem values range from 1 to 8. Notice the listing of the stem digits in the graphical solution are listed to the left of the vertical line. Step 3. For each age value, the leaf within the corresponding stem row and to the right of the vertical line is recorded. Notice the leaf values in the graphical solution. Step 4. Leaf values within each stem row are arranged in increasing order from left to right. Stem-and leaf display of Club Members Ages in Years b) The shape is skewed right or positively skewed. c) Since there are more leaf values in the 3 stem, most people are in the 30 s or 3 rd age decade. d) Age 28 to 43 contains more people (16 individuals) than any other fifteen year age difference.

2 # 22 a. The ones place value will be used as the leafs. Exp STEM Control b. Yes, it appears the new drug did lower the blood pressures of the participants. c. The experimental group is skewed right, and the control group is skewed left. Pg 73 #39 a. First calculate the class width using the class width formula. range CW = = = 2.857= 3 (this result will always be rounded up) # of classes 7 Next, construct the classes by starting at the smallest data value which is 20, and successively adding the class width 6 times in order to create the 7 lower class boundaries. Class Next, we find the upper class limits. To find the first upper class limit, you must subtract 1 from the second lower class limit. This will be 23 1 giving 22 as the first upper class limit. Once you have the first upper class limit, you can successively add the class width to create 6 the remaining upper class limits. Class

3 *Next, add in a column for the frequencies by counting up the number of data values that fall in each class. Remember that the endpoints of the classes get counted as well. Class Frequency *To add the class boundaries you need to remember that a class boundary is the midpoint between an upper class limit and the next lower class limit. For example the boundary between the upper limit of 22 in the first class & the lower limit of 23 in the second class is the midpoint of 22 and 23, which is This process would be repeated to calculate the class boundaries for all of the classes. (Hint: For whole number data, like we have above, a quick method to get a class boundary is to subtract.5 from the lower limit and add.5 to the upper limit of each class.) Class Frequency Class Boundaries *A class mark is just the midpoint of a class. To find it, just add the upper and lower class limits and divide by 2. For example, the class mark of the first class is found by the following formula: ( )/2 = 42/2 = 21 Class Frequency Class Boundaries Class Mark *The idea behind a relative percentage is it is a number that represents the percent of data values that fall in a particular class. This data set has 60 data values. The first class

4 contains 2 of the 60 data values. This is represented as a percent by 2. As a decimal, %. So the formula for relative frequency can be summarized 60 frequency as RP= *100. This would have to be repeated for all classes. sample size Class Frequency Class Boundaries Class Mark Relative Percentage % % % % % % % b) This question only wants you to use the table above to answer the questions. The first class is 20-22, and contains 2 data values. Now we have the data values, but if we didn t, we would not know what the two data values were. They could both be 20, or one 21 & one 22. or some other combination. Because of this some of these questions can be answered with only the table and some can t. i. At least 26 means 26 or more. This would include all of the 3 rd, 4 th, 5 th, 6 th & 7 th classes. So YES we can answer this question. ii. Less than 31 means 30 or fewer. Since 30 lands in the middle of a class, there is no way of telling how many 30 s there are, and we can t answer this question. iii. More than 33 means 34 or higher. Again, 34 is in the middle of a class so NO we can t answer this question. iv. Exactly 29 can t be answered either because we have no way of knowing how many of the 20 data values in that class happen to be The following data set represents the IQ Scores of 120 high school students. 145,126,118,95,110,150,108,128,107,123,100,113,109,130,98,118,110,121,10 5,109,136,123,112,117,113,92,118,104,117,97,127,108,122,105,117,121,109, 138,125,118,99,97,149,92,113,100,109,117,113,116,116,98,125,123,124,129, 94,148,144,127,119,101,127,95,107,112,121,125,113,99,120,122,123,127,128,117,114,129,95,113,145,112,122,106,119,93,132,122,110,117,98,121,110,97, 121,131,109,145,103,135,98,142,127,116,111,104,112,116,125,109,134,110,1 07,97,101,131,96,113,107,115 a) Construct a frequency distribution table using 10 classes. Within the table include classes, class marks, class boundaries, relative frequencies and percentages.

5 Solution: Use the Procedure for Constructing a Frequency Distribution Table for Numerical Data on page 47 of the text to construct a frequency distribution table for the numerical data. Step 1. In the problem statement, part a you are instructed to use 10 classes for the data set. Step 2. To determine the class width, you must identify both the largest and smallest data value and divide the difference by the number of classes. Using the formula: Approximate class width = l arg est data value smallest data value # of classes = = is rounded UP to 6. This is the class width of each class. Step 3. The first class begins by selecting the smallest data value within the data set. In this problem 92 is the smallest data value. Thus the first class begins at 92 and the upper limit for the first class is 97. The upper limit is found by adding the class width minus ONE to the lower limit. Or = 97. Therefore, the first class is from 92 to 97. The remaining classes are obtained by following the pattern of: Lower class limit = lower class limit of the previous class plus class width: =98 and, Upper class limit = upper class limit of the previous class plus class width: = 103. This continues until the largest data value is contained within a class. To see the remaining classes examine the solution for the entire frequency distribution table. Step 4 To determine the class marks for each class, use the formula: upper class lim it+ lower class limit Class Mark = 2 For the data in the problem, the first class mark = = The remaining class marks can be determined by adding the class width, 6, to the previous class mark. To see the remaining class marks examine the solution for the entire frequency distribution table that follows. Step 5 The Class Boundaries are determined by the formula:

6 Class Boundary = upper limit of one class+ lower limit of next class 2 The class boundaries for the classes are 0.5 subtracted from the lower class limit and 0.5 added to the upper class limit to 97.5 are the class boundaries for the first class. If the class width, 6, is added to each class boundary, the remaining class boundaries are determined as shown in the solution table for the entire frequency distribution table that follows. Step 6 To determine the frequency for each class you must find the data values that fall within that class. For the first class there are 12 data values that are between 92 and 97. They are 92,92,93,94,95,95,95,96.97,97,97,97. To see the remaining classes with the corresponding frequencies examine the solution for the entire frequency distribution table. Each percent frequency is determined by dividing the frequency for each class by 120, the number of data values for the data set and changing the decimal to a percentage. For the first class 92 97, there are 12 data values out of 120. So the percent frequency is 12/120 or 10% or the relative frequency is To see the remaining classes with the corresponding percentages and relative frequency examine the solution for the entire frequency distribution table. b) Solution: If you add the relative frequencies from the table the sum is 1 or 100%. c) Solution: If you exam the table above, you will be able to determine the percentage of students that have an IQ score which is: 1) Less than 116 From the table: The percentages that are less than the IQ score of 116 are: 15.83% % % + 10% = (NOTICE that less than means starting at the

7 class that is below the 116 class. Thus, you should begin at the class 110 to 115) 2) Between 128 to 133 inclusive From the table: There is one percentage that is associated with IQ scores between 128 to 133 inclusive is 6.77%.(NOTICE that inclusive means including the beginning and ending class values of 128 and 133.) 3) Greater than 121 From the table: The percentages that are greater than the IQ score of 121 are 15.83% % % % % = 32.5.(NOTICE that greater than means starting at the class that is above 116 to 121.Thus, you should begin at the class 122 to 127.) 4) At most 109 From the table: At most 109 means that the percentages for the classes that are less than and including the class ending with 109. This class is 104 to 109. Thus, the percentages for the class ending with 109 and those that are below the class 104 to 109 are: 14.17% % + 10% = 33.34%. 5) At least 134 From the table: At Least 134 are the classes that begin with 134 and are greater than 134. Thus, the percentages for the classes starting 134 to 139 and greater are: 3.33% % % = 10%. 76. A statistics instructor has recorded the amount of time students need to complete the final examination. The times, stated to the nearest minute, are: 50, 70, 62, 55, 38, 42, 49, 75, 80, 79, 48, 45, 53, 58, 64, 77, 48, 49, 50, 61, 72, 74, 10, 95, 120, 79, 75, 48, 72, 37, 35, 79, 72, 75, 77, 32, 30, 77, 75, 79, 45, 70, 39, 75, 72, 45, 47, 73, 72, 71 a) Construct a stem-and-leaf display using the hundreds and tens digits for the stem portion. Step 1. Under Stem, List the stem values starting from 1 to 12 to represent all the times from 10 minutes, 20 minutes, etc to 120 minutes. Step 2. Under Leaf, list in numerical order the values pertaining to the units digit for each stem value.

8 Stem Leaf b) Describe the shape of the distribution. The distribution shape is bimodal. c) Are there any outlier(s) in the data set? If so, give an explanation for the outlier(s)? The outliers are 10 minutes and 120 minutes. For the 10 minute outlier, the student probably was not prepared to take the final and just handed back the exam after 10 minutes. For the 120 minute outlier, the student either works very slow or checked the work before submitting the exam. d) What percentage of the class required from 40 to 75 minutes to complete the test? Since there are 33 times out of a total of 50 times that fall between 40 and 75 minutes inclusive, then the percentage of the class that required from 40 to minutes is: *100 = 66%. 50 e) If the statistics instructor only allowed one hour and 20 minutes for the exam, what percentage of the class would not have had enough time to complete the exam? One hour and 20 minutes equals 80 minutes. From the stem-and-leaf display, only two students out of the 50 students would require more than 80 minutes. Thus the percentage of class that would not have enough time is: 2 *100 4% 50 =. 77. A medical researcher surveyed one hundred mothers who recently gave birth. The researcher determined whether the mothers smoked during pregnancy and the babies' birth weights. The birth weights were separated into two groups, smoking mothers and non-smoking mothers. The following data represents the researcher s findings. Smoking Mothers 5 lbs 7 ozs, 6 lbs 6 ozs, 7 lbs 2 ozs, 9 lbs 2 ozs, 7 lbs 3 ozs, 8 lbs 2 ozs, 7 lbs 5 ozs, 6 lbs 2 ozs, 5 lbs 3 ozs, 3 lbs 4 ozs, 5 lbs 3 ozs, 7 lbs 6 ozs, 6 lbs 1 oz, 5 lbs 4 ozs, 5 lbs 2 ozs, 6 lbs 8 ozs, 7 lbs 4 ozs, 7 lbs 2 ozs, 9 lbs 3 ozs, 6 lbs 5 ozs, 7 lbs 7 ozs, 8 lbs 2 ozs, 7 lbs 6 ozs, 3 lbs 1 oz, 5 lbs 12 ozs, 6 lbs 2 ozs, 7 lbs 5 ozs, 6 lbs 5 ozs, 8 lbs 8 ozs, 9 lbs 1 oz,

9 8 lbs 3 ozs, 7 lbs 8 ozs, 6 lbs 1 oz, 5 lbs 9 ozs, 6 lbs 5 ozs, 6 lbs 6 ozs, 8 lbs 3 ozs, 7 lbs 8 ozs, 6 lbs 6 ozs, 5 lbs 12 ozs, 5 lbs 9 ozs, 6 lbs 9 ozs, 6 lbs 8 ozs, 5 lbs 2 ozs, 7 lbs 4 ozs, 6 lbs 9 ozs, 7 lbs 5 ozs, 7 lbs 0 ozs, 8 lbs 0 ozs, 8 lbs 0 ozs Non-smoking Mothers 8 lbs 2 ozs, 7 lbs 1 oz, 6 lbs 6 ozs, 5 lbs 9 ozs, 9 lbs 4 ozs, 7 lbs 0 ozs, 8 lbs 5 ozs, 6 lbs 9 ozs, 5 lbs 13 ozs, 8 lbs 4 ozs, 7 lbs 6 ozs, 6 lbs 6 ozs, 5 lbs 11 ozs, 9 lbs 0 ozs, 8 lbs 9 ozs, 7 lbs 1 oz, 6 lbs 8 ozs, 7 lbs 9 ozs, 9 lbs 3 ozs, 8 lbs 8 ozs, 6 lbs 6 ozs, 7 lbs 3 ozs, 7 lbs 14 ozs, 8 lbs 1 oz, 7 lbs 5 ozs, 6 lbs 5 ozs, 7 lbs 6 ozs, 6 lbs 12 ozs, 7 lbs 8 ozs, 7 lbs 9 ozs, 9 lbs 3 ozs, 8 lbs 4 ozs, 7 lbs 11 ozs, 6 lbs 14 ozs, 5 lbs 11 ozs, 8 lbs 6 ozs, 5 lbs 15 ozs, 6 lbs 10 ozs, 7 lbs 3 ozs, 9 lbs 8 ozs, 7 lbs 15 ozs, 7 lbs 2 ozs, 6 lbs 14 ozs, 7 lbs 12 ozs, 7 lbs 10 ozs, 6 lbs 8 ozs, 6 lbs 14 ozs, 11 lbs 2 ozs, 10 lbs 1 oz, 10 lbs 5 ozs a) Construct a back-to-back stem-and-leaf display for the birth weights of the babies for the smoking and nonsmoking mothers using lbs for the stem portion. b) Describe the shape of each distribution of birth weights for smoking and non-smoking mothers. c) Does it appear from the stem-and-leaf display that smoking may have an effect on the birth weight of a newborn? Support your conclusion. 76. a) The stem values represent pounds and are listed in the middle of the display. The leaves pertaining to the ounces of the baby weights of the smoking mothers are listed on the left of the stem values while the leaves of the ounces pertaining to the baby weights of the non-smoking mothers are listed on the right. smoking non-smoking b) Both distributions are skewed. c) Yes because there are a greater percentage of lower baby weights for the smoking mothers. There are also two extremely low baby weights for the smoking mothers that could be outliers.

10 #84 a. The ones place are the leafs and the tens place are the stems. Stem Leaf b. Class frequency Class Boundaries Class Mark Relative Percentage % % % % % % % %

11 c. Histogram of Age Frequency Age Frequency Polygon of Ages Freq AGE d. The shape of the distribution is skewed right because there is a long tail on the right of the graph.