Question

A professor selects a random sample of 50 Missouri adults (ages 18-64) and observes each of them for a day, recording the amount of time that they spend engaging in physical activity (sports or exercise). Suppose that, unknown to the professor, Missourians in this age group average 17 minutes of physical activity per day with a standard deviation of 12 minutes.

A: Find the probability that the mean of the professor’s sample will be 15 minutes or more

B:There is a 97% probability that the mean of the professor’s sample will be less than _______ minutes.

Answer 1

Given,

= 17, = 12

The central limit theorem is

P( < x) = P( Z < x - / / sqrt(n) )

A)

P( >= 15) = P( Z >= 15 - 17 / 12 / sqrt(50) )

= P( Z >= -1.1785)

= P( Z < 1.1785)

= 0.8807

B)

We have to calculate x such that P( < x) = 0.97

That is

P( Z < x - / / sqrt(n) ) = 0.97

From the Z table, z-score for the probability of 0.97 is 1.8815

Therefore,

x - / / sqrt(n) = 1.8815

Put the values of , and n in above expression and solve for x

(x - 17) / (12 / sqrt(50) ) = 1.8815

x - 17 = 1.8815 * ( 12 / sqrt(50) )

x = 20.193

There is 97% probability that the mean of the professor's sample will be less than 20.193

minutes.