AP Review Questions 61-120

Question 1

Find the mean and median from the datasets below. Which measure of central tendency would be the appropriate measure to use for each? Why? Dataset 1: {7,7,2,8,5,10,9,1,10,3} Dataset 2: {6,2,6,5,3,8,5,7,15,5,} (A) Dataset 1 and 2: mean: 6.2; because there are no outliers. (B) Dataset 1: Mean: 6.2; because there are no outliers. Dataset 2: Median: 5.5; because 15 is an outlier. (C) Dataset 1: Median: 7; because there are no outliers. Dataset 2: Mean: 6.2; because 15 is an outlier. (D) Datasets 1 and 2: Median: 6.2; because 10 and 15 are outliers. (E) Datasets 1 and 2: Mean: 6.2; because 10 and 15 are outliers.

Answer 1

(B) The interquartile ranges for the datasets are 6 and 2, respectively. There are no outliers in Dataset 1 so the mean of 6.2 would be the best measure of central tendency. 15 is an extreme outlier in Dataset 2, so the median 5.5 would be the best measure of central tendency since the mean is more sensitive to outlier.

Question 2

A score at or above the 90th percentile is how many standard deviations from the mean? (A) 1.28 standard deviations or more above the mean. (B) 1.64 standard deviations or more above the mean. (C) 2.58 standard deviations or more above the mean. (D) 1.28 standard deviations or more below the mean. (E) 1.64 standard deviations or more below the mean.

Answer 2

(A) The z-score with area of 0.9 to the left of it is 1.28 which means that the 90th percentile or above is 1.28 standard deviations above the mean.

Question 3

A score at or above the 70th percentile is how many standard deviations form the mean? (A) 0.52 standard deviations or more above the mean. (B) 1.04 standard deviations or more above the mean. (C) 1.64 standard deviations or more above the mean. (D) 0.52 standard deviations or more below the mean. (E) 1.04 standard deviations or more below the mean.

Answer 3

(A) The z-score with area of 0.7 to the left of it is 0.52 with means that the 70th percentile or above is 0.52 standard deviations above the mean.

Question 4

Test scores are normally distributed with a mean of 72 and a standard deviation of 4.2. What is the percentile of a score of 65 (round to the nearest whole percentile). (A) 95th percentile (B) 90th percentile (C) 85th percentile (D) 5th percentile (E) 1st percentile

Answer 4

(D) The z-score associated with a test score of 65 is 65-72/4.2=-1.67. The area to the left of a z-score of -1.67 is 0.0475 or approximately the 5th percentile.

Question 5

What data point is indicated on the distribution. With a mean of 67.0 and a standard deviation of 5.1. (A) 61.9 (B) 64.5 (C) 67.0 (D) 72.1 (E) 77.2

Answer 5

(D) The data point is at the point of inflection on the bell curve which is one standard deviation above the mean or X=67+5.1=72.1.

Question 6

The wage gap between men and women’s earnings was 23 cents on the dollar in 2009, down from a low of 22.2 cents in 2007. Some regions in the Untied States are more progressive and pay is more equitable between genders. “According to WebMD, male specialists earn a median salary of $225,000 a year. Female specialists, on the other hand, take home a median annual salary o f$160,000– a difference of $65,000 a year. This difference in pay means that in the course of a 35-year career, female doctors lose a total of $2.3 million on average, noted Forbes.” (a) What might explain the reporting of median salaries rather than average salaries? (b) A sample of eight specialist salaries for male and female doctors was collected in one region of the United States. A summary is shown. On average, how does this sample compare with findings by WebMD? How does it compare to the wage gap reported for 2009 of 23 cents on the dollar? Men Women Category Mean Median Mean Median Anesthesiologist 267.9 247 189.3 194 Obstetricians/ 158.8 148 143.7 143 Gynecologists pediatricians 141.4 146 112.8 112 psychiatrists 157.4 171 114.1 126 surgeons 196 200 149.1 164 podiatrists 97.8 89 74.3 74

Answer 6

(a) If there are outliers, particularly extreme outliers, the median is less sensitive and more resistant than the mean. That may explain why median salaries would be reported rather than average salaries (b) From the summary data, extracting the specialist and finding the median shows that, for men, it is 148+171/2=159.5. For women, it is 143+126/2=134.5. This is lower than what was reported; additionally, the wage gap for these two medians is about 16 cents, which is less than the reported 23 cents for 2009. 159.5-134.5/159.5=X/100, x=16.

Question 7

Researchers form the University of Pennsylvania, Philadelphia, found that incentives increased all IQ scores, but particularly for those of individuals with lower baseline IQ scores. They found that a financial incentive could raise IQ as much as 15 points. (a) Construct a graphical display for each variable shown in the following dataset. Boys IQ Scores No Incentive Incentive 108 105 89 132 93 116 94 81 115 83 96 90 73 85 85 131 103 136 (b) Analyze the data for outliers (c) Report an appropriate measure of central tendency for both groups. (d) what is the IQ score for the 75th percentile for both samples?

Answer 7

(a) Stem- and- leaf of No Incentive N=10 Leaf Unit= 10 1 7 3 3 8 59 (3) 9 346 4 10 38 2 11 5 1 12 3 stem- and-leaf of incentive N=10 Leaf Units= 1.0 4 8 0259 5 9 0 5 10 4 4 11 6 3 12 3 13 125 (b) The stem and leaf plots indicate that the neither group has any outliers. There are no data points in either group that are more than 1.5(IQR) above Q3 or more than 1.5(IQR) below Q1. (c) The mean would be an

Question 8

Answer 8

h

Question 9

Water consumption in gallons per day from a small town is listed below.

water consumed

167, 180, 192, 173, 145, 151, 174, 175, 178, 160, 195, 224, 244, 146, 162, 146, 177, 163, 149, 188

(a) construct a frequency distribution table using only five classes.
(b) construct a graphical display.
(c) report an appropriate measure of central tendency.
(d) analyze the data for outliers using a box plot.

Answer 9

(A) classes Frequency

144. 5-165 8
145. 5-186 7
146. 5-207 3
147. 5-228 1
148. 5-249 1

20

(b)

4. 14. 5669
5. 15. 1
6. 16. 0237
       (5) . 17. 34578
7. 18. 08
8. 19. 25
9. 20
10. 21
11. 22. 4
12. 23
13. 24. 4

(C) due to the outlier, the median– 173.5 gallons per day– would be an appropriate measure of central tendency.

Question 10

The following is a sample of butterfly wingspan measured in millimeters.

wingspand

32,31,29,46,37,38,40,30,28,33,36,39,37,39,41,29,32,38,39,35,37,33

(a) construct a steam and leaf plot.
(b) reporte an appropriate measure of central tendency.
(c) analyze the data for outliers.
(d) what wingspan is the 75th percentile.

Answer 10

(A)

3. 2899
4. 3012233
   (10) . 35677788999
5. 401
6. 46
   (b) with the absence of outliers, the mean– 35.4 mm– is an appropriate measure of central tendency. The median– 36.5 mm– is also an appropriate measure of central tendency.

(C) the box plot indicates that there are no outliers. There are no date points in either group that are more than 1.5(iqr) above q3 or more than 1.5(iqr) below q1.

(d) the third quartile is the same as the 75th percentile. 39 is the wingspan for the 75th percentile.

Question 11

Which of the following statements best describes the scatter plot, that has a positive correlation and association.

I. A linear model would fit the data well.

II. The bivariate data are positive correlated.

III. the linear relationship is weak.

(a) I only
(b) I and III only
(c) I and II only
(d) II ONLY
(e) III only

Answer 11

(C) the data points are trending linearly with a positive slope.

Question 12

What can be said about the relationship between time spent studying and test scores? Based on a positive correlation and association, scatter plot.

(a) more time spent studying is related to lower test scores.
(b) less time spent studying is related to higher test scores.
(c) more time spent studying is related to higher test scores.
(d) higher test scores are caused by more time spent studying.
(e) there is no discernible relationship between study time and test scores.

Answer 12

(C) the more time spent studying is related to a higher test scores since there is a positive relationship.

Question 13

(B) y = 24.765+ 9.051x. There is positive linear correlation.

Answer 13

From the data, find the least-squares regression line. What type of correlation is there?

study(hrs). Test scores

0. 20
1. 35
2. 50
3. 47.5
4. 65
5. 72.5
6. 75
7. 67.5
8. 92.5
9. 85
10. 90
11. 95
12. 70
    (a) y = 24.765-9.051x. There is negative linear correlation.
    (b) y = 24.765+9.051x. There is positive linear correlation.
    (c) y = 9.051+24.765x. There is positive linear correlation.
    (d) y = 9.051-24.765x. There is negative linear correlation.
    (e) y = -24.765+9.051x. There is negative linear correlation.

Question 14

(D) y = 95.494-4.508x. There is negative linear correlation.

Answer 14

From the data, find the least-squares regression line. What type of correlation is there?

entertainment media. Test scores

0. 97.5
1. 85
2. 82.5
3. 75
4. 95
5. 67.5

5 75

5 82.5

6 57.5

7 65

7 75

10 45

(a) y = 95.494+4.508x. There is positive linear correlation.
(b) y = 4.508-95.494x. There is negative linear correlation.
(c) y = 4.508+95.494x. There is negative linear correlation.
(d) y = 95.494-4.508x. There is negative linear correlation.
(e) y = -95.494-4.508x. There is negative linear correlation.

Question 15

d

Answer 15

d

Question 16

d

Answer 16

d

Question 17

d

Answer 17

d

Question 18

d

Answer 18

d

Question 19

d

Answer 19

d

Question 20

d

Answer 20

d

Question 21

Using the data, which of the following statements best describes the correlation?

x secs. Y mm. X secs. Ymm

2. 5. 116 5.25. 153
3. 75. 116. 5.5. 140
4. 130. 5.75. 144
5. 5. 129. 6. 142
6. 75. 137 6.25. 140
7. 133. 6.5. 132
8. 25. 138. 6.75. 125
9. 5. 143. 7. 122
10. 75. 142. 7.25. 114

I. A linear model would fit the data well.

II. The data point y-value would need to be transformed to achieve linearity.

III. the bivariate data are negatively correlated.

(a) I only
(b) I and III only
(c) I and II only
(d) II only
(e) III only

Answer 21

(D) the scatter plot of the data shows that the data are clearly nonlinear.

Question 22

Using the dataset, what can be said about the relationship between earnings and dividends?

earnings. Dividends
2. 78. 2.16
1. 41. 0.88
2. 74. 1.04
0. 92. 1.1
2. 44. 0.96
3. 5. 2.18
3. 68. 1.54
1. 97. 1.39
1. 95. 1.41
2. 07. 2.88
(a) more earnings are related to more dividends.
(b) there is no discernible relationship between earnings and dividends.
(c) more earnings are related to less dividends.

(D) more dividends are caused by more earnings.

(e) fewer earnings are related to more dividends.

Answer 22

(B) the scatter plot and coefficient of determination indicate that earnings and dividends are unrelated.

Question 23

From the data, find the least-squares regression line. What type of correlation is there?

calories. Sodium
186. 495
181. 477
176. 425
149. 322
184. 482
190. 587
158. 370
139. 322
175. 479
148. 375
(a) y = -299.5+4.347x. There is negative linear correlation.
(b) y = -299.5+4.347x. There is positive linear correlation.
(c) y = 299.5-4.347x. There is positive linear correlation.
(d) y = 299.5-4.347x. There is negative linear correlation.
(e) y = -299.5+4,347x. There is no discernible relationship.

Answer 23

(B) y = -299.5+4.347x. Because slope is positive and the coefficient of determination is 0.871, there is strong positive linear correlation.

Question 24

From the regression output below, which type of variation does R-squared pertain to?

the regression equation is: sodium = -299.5+4.35 calories

predictor Coefficient Se coefficient T P

constant -299.5 100.3 -2.99 0.017

calories 4.3469 0.5917 7.35 0.000

s = 32.6453 R-Sq = 87.1% R-Sq(adj) = 85.5%

(a) total variation
(b) unexplained variation
(c) explained variation
(d) least squares variation
(e) error variation

Answer 24

(C) R-Sq is a measure of the explained variation in the response variable by the predictor variable.

Question 25

From the regression output below, what percentage of variation is unexplained?

The regression equation is: sodium = -299+4.35 calories

predictor. Coefficient. Se coef. T. P
constant. -299.5. 100.3. -299. 0.017
calories. 4.3469. 0.5917. 7.35. 0.000

s = 32.6453. R-Sq = 87.1%. R-Sq(adj) = 85.5%

(a) 87.1%
(b) 85.5%
(c) 32.6%
(d) 12.9%
(e) all the variation is explained

Answer 25

(D) 12.9% of the variation is unexplained. It is calculated by subtracting the explained variation from 100% and the explained variation is R-Sq, 100%-87.1% = 12.9%.

Question 26

Find the slope from the least-squares regression line for the data below of final grade as a percent and days absent. What does it tell us about the relationship?

final absences

81. 1
82. 0
83. 2
84. 3
85. 6
86. 4
87. 7
88. 2
89. 0
90. 1
    (a) slope is 94.695. For each day absent, the student’s final grade increased by 0.947%.
    (b) slope is-94.695. For each day absent, the student’s final grade decreased by 0.947%.
    (c) slope is 6.767. For each day absent, the student’s final grade increased by 6.767%.
    (d) slope is 6.767. For each day absent, the student’s final grade decreased by 6.767%.
    (e) slope is-6.767. For each day absent, the student’s final grade decreased by 6.767%.

Answer 26

(E) slope is-6.767 from the linear regression model for these data y =94.695-6.767x. For each day absent, the student’s final grade decreased by 6.767%.

Question 27

Find the intercept from the least-squares regression line for the data below of final grade as a percent and days absent. What does it tell us about the relationship?

final. Absences
81. 1
90. 0
86. 2
76. 3
51. 6
75. 4
44. 7
81. 2
94. 0
93. 1
(a) intercept is 94.965. For no absences, the student’s predicted grade is 94.965%.
(b) intercept is 94.695. For no absences, the student’s predicted grade is 94.695%.
(c) intercept is 6.767. For no absences, the student’s predicted grade is 67.67%.
(d) intercept is-6.767. For no absences, the student’s predicted grade is 67.67%.
(e) intercept is 6.676. For no absences, the student’s predicted grade is 66.76%.

Answer 27

(B) the intercept is 94.695 from the linear regression model for this data y = 94.695-6.767x. If a student has no absences x=0, the model predicts a grade of 94.695%.

Question 28

d

Answer 28

d

Question 29

d

Answer 29

d

Question 30

From the regression output of final grade as a percent versus days absent, make a prediction for a student who is absent five days.

the regression equation is: final = 94.7-6.77absences

predictor. Coef. Se coef. T. P
constant. 94.695. 2.376. 39.85. 0.00
absences. -6.7672. 0.6860. -9.87. 0.00

s = 4.96563. R-Sq = 92.4%. R-Sq(adj) = 91.5%

(a) y = 46.673. A student with five absences will receive a final grade of 46.673%.
(b) y = 68.05. A student with five absences will receive a final grade of 68.05%.
(c) y = 63.55. A student with five absences will receive a final grade of 63.55%.
(d) y = 60.85. A student with five absences will receive a final grade of 60.85%.
(e) y = 60.58. A student with five absences will receive a final grade of 60.58%.

Answer 30

(D) evaluating the equation with = 5, y = 94.7-6.77(5) = 60.85. The predicted grade for a student with five absences is 60.85%.