Question

​Full-time college students report spending a mean of 25 hours per week on academic​ activities, both inside and outside the classroom. Assume the standard deviation of time spent on academic activities is 4 hours. Complete parts​ (a) through​ (d) below. a. If you select a random sample of 16 ​full-time college​ students, what is the probability that the mean time spent on academic activities is at least 24 hours per​ week? nothing ​(Round to four decimal places as​ needed.) b. If you select a random sample of 16 ​full-time college​ students, there is an 80​% chance that the sample mean is less than how many hours per​ week? nothing ​(Round to two decimal places as​ needed.) c. What assumption must you make in order to solve​ (a) and​ (b)? A. The population is symmetrically​ distributed, such that the Central Limit Theorem will likely hold for samples of size 16. B. The sample is symmetrically​ distributed, such that the Central Limit Theorem will likely hold. C. The population is normally distributed. D. The population is uniformly distributed. d. If you select a random sample of 100 ​full-time college​ students, there is an 80​% chance that the sample mean is less than how many hours per​ week? nothing ​(Round to two decimal places as​ needed.

Answer 1

(a)

= 25

= 4

n = 16

SE = /

= 25/

= 6.25

To find P(<24):

Z =(24 - 25)/6.25

= - 0.16

By Technology, Cumulative Area Under Standard Normal Curve = 0.4364

So,

Answer is:

0.4364

(b)

80% corresponds to area =0.80 - 0.50 = 0.30.

Table of Area Under Standard Normal Curve gives Z = 0.84

So,

Z = 0.84 = ( - 25)/6.25

So,

= 25 + (0.84 X 6.25)

= 30.25

So,

Answer is:

30.25

(c)

Correct option:

C. The population is normally distributed.

(d)

= 25

= 4

n = 100

SE = /

= 25/

= 2.5

80% corresponds to area =0.80 - 0.50 = 0.30.

Table of Area Under Standard Normal Curve gives Z = 0.84

So,

Z = 0.84 = ( - 25)/2.5

So,

= 25 + (0.84 X 2.5)

= 27.10

So,

Answer is:

27.10