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The College Board reported the following mean scores for the three parts of the Scholastic Aptitude Test (SAT) (The World Almanac, 2009): Critical Reading 502 Mathematics 515 Writing 494 Assume that the population standard deviation on each part of the test is σ=100. a. What is the probability a random sample of 90 test takers will provide a sample mean test score within 10 points of the population mean of 502 on the Critical Reading part of the test?b. What is the probability a random sample of 90 test takers will provide a sample mean test score within 10 points of the population mean of 515 on the Mathematics part of the test? Compare this probability to the value computed in part (a).c. What is the probability a random sample of 100 test takers will provide a sample mean test score within 10 of the population mean of 494 on the writing part of the test? Comment on the differences between this probability and the values computed in parts (a) and (b).

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Answer:

Explanation:

Hello!

Xi: scores of one of three parts of the Scholastic Aptitude Test. (i= critical reading, mathematics, writing)

The average scores are

Critical reading 502

Mathematics 515

Writing 494

If the population standard deviation on each part of the test is δi= 100

a. What is the probability of a random sample of 90 test-takers will provide a sample mean test score within 10 points of the population means of 502 on the Critical Reading part of the test?

If the sample mean test score is within 10 points of the population mean, then (X[bar]-μ)±10 and the sample size is n=90

P(492≤X[bar]≤512)= P(X[bar]≤512)-P(X[bar]≤492)=

P(Z≤10/(100/√90))-P(Z≤-10/(100/√90))= P(Z≤0.95)-P(Z≤-0.95)= 0.82894-0.17106= 0.65788

b. What is the probability of a random sample of 90 test-takers will provide a sample mean test score within 10 points of the population means of 515 on the Mathematics part of the test? Compare this probability to the value computed in part (a).

If the sample mean test score is within 10 points of the population mean, then (X[bar]-μ)±10 and the sample size is n=90

P(505≤X[bar]≤525)= P(X[bar]≤525)-P(X[bar]≤505)=

P(Z≤10/(100/√90))-P(Z≤-10/(100/√90))= P(Z≤0.95)-P(Z≤-0.95)= 0.82894-0.17106= 0.65788

c. What is the probability of a random sample of 100 test-takers will provide a sample mean test score within 10 of the population means of 494 on the writing part of the test? Comment on the differences between this probability and the values computed in parts (a) and (b).

If the sample mean test score is within 10 points of the population mean, then (X[bar]-μ)±10 and the sample size is n=100

P(484≤X[bar]≤504)= P(X[bar]≤504)-P(X[bar]≤484)=

P(Z≤10/(100/√100))-P(Z≤-10/(100/√100))= P(Z≤1,00)-P(Z≤-1,00)= 0.84134-0.15866= 0.68268

The calculated probability is larger than the ones calculated in a) and b) because of the different sample sizes. The larger the sample, the greater the standard deviation of the sample mean and this leads to a greater width between the two values of the sample mean.

I hope it helps!

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