Answer to Question #218184 in Statistics and Probability for sai

Question

The average travel time to work for a person living and working in​ Kokomo, Indiana is 17 minutes. Suppose the standard deviation of travel time to work is 4.5 minutes and the distribution of travel time is approximately normally distributed.

Which of these statements is asking for a measurement​ (i.e. is an inverse normal​ question)?

A.

What percentage of people living and working in Kokomo have a travel time to work that is between thirteen and fifteen​minutes?

B.

If​ 15% of people living and working in Kokomo have travel time to work that is below a certain number of​ minutes, how many minutes would that​ be?

Use the empirical rule to solve the problem.

At one​ college, GPA's are normally distributed with a mean of 2.9 and a standard deviation of 0.5. What percentage of students at the college have a GPA between 2.4 and 3.4​? Round to the nearest percent.

A.

Almost all

B.

​68%

C.

​84%

D.

​95%

Answer 1

Problem 1:

$\mu=17 \\ \sigma= 4.5$

A.

$P(13<X<15) = P(X<15) -P(X<13) \\ =P(Z < \frac{15-17}{4.5}) -P(Z< \frac{13-17}{4.5}) \\ = P(Z< -0.444) -P(Z< -0.888) \\ = 0.3285 -0.1872 \\ = 0.1413 \\ = 14.13 \; \%$

14.13 % of people living and working in Kokomo have a travel time to work that is between thirteen and fifteen minutes.

B.

$P(X<x) = 0.15 \\ P(Z< \frac{x-17}{4.5}) = 0.15 \\ \frac{x-17}{4.5} = -1.036 \\ x-17 = 4.5 \times (-1.036) \\ x = 17 - 4.662 =12.338$

15% of people living and working in Kokomo have travel time to work that is below a 12.34 minutes.

Problem 2:

$\mu= 2.9 \\ \sigma= 0.5 \\ P(2.4<X<3.4) = P(X<3.4) -P(X<2.4) \\ = P(Z< \frac{3.4-2.9}{0.5}) -P(Z< \frac{2.4-2.9}{0.5}) \\ = P(Z< 1) -P(Z< -1) \\ = 0.8413 -0.1586 \\ = 0.6827 \\ = 68 \; \%$

Answer: option B. 68 %