Question

Full-time college students report spending a mean of 26 hours per week on academics activities, both inside and outside the class room. Assume the standard deviation of time spent on academic activities is 5 hours. If you select a random sample of 16 Full-time college students, what is the probability that the mean tie spent on academics is at least 24 hours per week? b. If you select a random sample of 16 full-time college students, there is an 87% chance that the sample mean is less than how many hours per week?

Answer 1

Solution:

Given that,

= 26

= 5

n = 16

So,

= 26

=  ( /n) = ( 5 / 16 ) = 1.25

p (     24 )

= 1 - p (   24 )

= 1 - p ( -  /   ) ( 24 - 26 / 1.25)

= 1 - p ( z - 2 / 1.25 )

= 1 - p ( z -1.6 )

Using z table

= 1 - 0.0548

= 0.9452

Probability = 0.9452

Using standard normal table,

P(Z < z) = 87%

P(Z < z) = 0.87

P(Z < 1.126 ) = 0.87

z = 1.126

Using z-score formula,

= z * +

= 1.126 * 1.25 + 26 = 27.4075

= 27.41

27.41 hours per week.