Z-scores and Normal Distributions: Comparing ACT and SAT Scores, and Weight of Field Mice, Quizzes of Probability and Statistics

Solutions to two problems involving z-scores and normal distributions. The first problem compares act and sat scores of two students, while the second problem determines the proportion of field mice with weights above a certain threshold. Additionally, the document includes questions to test understanding of concepts related to z-scores, normal distributions, and data summaries.

Typology: Quizzes

Pre 2010

Uploaded on 07/29/2009

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Quiz 2a Name
Math 243 GTF
April 9, 2009 Disc. Time
1. The ACT is normally distributed with mean µ= 20.9 and standard deviation σ= 4.8, and the
SAT is normally distributed with mean µ= 1026 and standard deviation σ= 209. Student A
gets a score of 1240 on the SAT and student B gets a score of 26.1 on the ACT. Given that
each test supposedly measures the same thing, which student’s score is better? Show your
work.
Student A has a z-score of: 12401026
209 1.0239
Student B has a z-score of: 26.120.9
4.81.0833
Student B has the greater z-score and so we can presume student B has the better score.
2. Given that the body weight in grams of field mice is normally distributed with mean µ= 43.4
and standard deviation σ= 8.9, find the proportion of field mice with weight above 31.
A weight of 31 grams corresponds to a z-score of: 3143.4
8.9 1.39326.
We round this to two decimal places, 1.39, and use Table A to find that the proportion of
the field mice below a z-score of 1.39 is approximately 0.0823. Therefore, the proportion of
the field mice with weight above 31 grams is approximately 1 0.0823 = 0.9177.
Using a calculator we get a (slightly) better approximation of 0.9182
3. Answer the following questions about an arbitrary data set:
(a) Which of the following can be used to determine some measure of how widely spread the
data is? (circle one)
i. The mean and standard deviation can but the 5-number summary can not.
ii. The 5-number summary can but the mean and standard deviation can not.
iii. Both the mean and standard deviation and the 5-number summary can.
iv. Neither the mean and standard deviation nor the 5-number summary can.
(b) Which of the following provides some information about the skewness of the distribution?
i. The mean and standard deviation does but the 5-number summary does not.
ii. The 5-number summary does but the mean and standard deviation does not.
iii. Both the mean and standard deviation and the 5-number summary do.
iv. Neither the mean and standard deviation nor the 5-number summary do.
(c) Which of the following can be used to determine the range of possible values in the data
set?
i. The mean and standard deviation can but the 5-number summary can not.
ii. The 5-number summary can but the mean and standard deviation can not.
iii. Both the mean and standard deviation and the 5-number summary can.
iv. Neither the mean and standard deviation nor the 5-number summary can.
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Quiz 2a Name Math 243 GTF April 9, 2009 Disc. Time

  1. The ACT is normally distributed with mean μ = 20.9 and standard deviation σ = 4.8, and the SAT is normally distributed with mean μ = 1026 and standard deviation σ = 209. Student A gets a score of 1240 on the SAT and student B gets a score of 26.1 on the ACT. Given that each test supposedly measures the same thing, which student’s score is better? Show your work.

Student A has a z-score of: 1240209 −^1026 ≈ 1. 0239

Student B has a z-score of: 26.^14 −. 820.^9 ≈ 1. 0833

Student B has the greater z-score and so we can presume student B has the better score.

  1. Given that the body weight in grams of field mice is normally distributed with mean μ = 43. 4 and standard deviation σ = 8.9, find the proportion of field mice with weight above 31.

A weight of 31 grams corresponds to a z-score of: 31 − 8.^439.^4 ≈ − 1 .39326. We round this to two decimal places, − 1 .39, and use Table A to find that the proportion of the field mice below a z-score of − 1 .39 is approximately 0.0823. Therefore, the proportion of the field mice with weight above 31 grams is approximately 1 − 0 .0823 = 0.9177.

Using a calculator we get a (slightly) better approximation of 0. 9182

  1. Answer the following questions about an arbitrary data set:

(a) Which of the following can be used to determine some measure of how widely spread the data is? (circle one) i. The mean and standard deviation can but the 5-number summary can not. ii. The 5-number summary can but the mean and standard deviation can not. iii. Both the mean and standard deviation and the 5-number summary can. iv. Neither the mean and standard deviation nor the 5-number summary can. (b) Which of the following provides some information about the skewness of the distribution? i. The mean and standard deviation does but the 5-number summary does not. ii. The 5-number summary does but the mean and standard deviation does not. iii. Both the mean and standard deviation and the 5-number summary do. iv. Neither the mean and standard deviation nor the 5-number summary do. (c) Which of the following can be used to determine the range of possible values in the data set? i. The mean and standard deviation can but the 5-number summary can not. ii. The 5-number summary can but the mean and standard deviation can not. iii. Both the mean and standard deviation and the 5-number summary can. iv. Neither the mean and standard deviation nor the 5-number summary can.

  1. Which of the following scatterplots best shows the greatest positive correlation? (circle one)

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  1. For each of the following situations, identify the explanatory and response variables, or state that the relationship is unclear.

(a) Yield per acre Y of bushels of tomatoes and number of tomato plants per acre N.

Explanatory: N

Response: Y

(b) Cost per barrel of oil C and price per gallon of gasoline P.

Explanatory: C

Response: P

(c) Number of art galleries G per 10,000 people and crime rate R in U.S. cities.

Explanatory:

Response:

relationship unclear