Question

Full-time college students report spending a mean of 25hours 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 25 ?full-time college? students, what is the probability that the mean time spent on academic activities is at least 24 hours per? week?
?(Round to four decimal places as? needed.)
b. If you select a random sample of 25 ?full-time college? students, there is an 89?% chance that the sample mean is less than how many hours per? week?
?(Round to two decimal places as? needed.)
c. What assumption must you make in order to solve? (a) and? (b)?
-A.The population is uniformly distributed.
B.The sample is symmetrically? distributed, such that the Central Limit Theorem will likely hold.
C.The population is symmetrically? distributed, such that the Central Limit Theorem will likely hold for samples of size 25.
D.The population is normally distributed.
d. If you select a random sample of 100 ?full-time college? students, there is an 89?% chance that the sample mean is less than how many hours per? week?
?(Round to two decimal places as? needed.)

Answer 1

a) P(> 24)

= P(( - )/() > (24 - )/())

= P(Z > (24 - 25)/(4/sqrt(25)))

= P(Z > -1.25)

= 1 - P(Z < -1.25)

= 1 - 0.1056

= 0.8944

b) P( < x) = 089

or, P(( - )/() < (x - )/()) = 0.89

or, P(Z < (x - 25)/(4/sqrt(25))) = 0.89

or, (x - 25)/(4/sqrt(25)) = 1.23

or, x = 1.23 * (4/sqrt(25)) + 25

or, x = 25.98

c) Option - D) The population is normally distributed.

d) P( < x) = 089

or, P(( - )/() < (x - )/()) = 0.89

or, P(Z < (x - 25)/(4/sqrt(100))) = 0.89

or, (x - 25)/(4/sqrt(100)) = 1.23

or, x = 1.23 * (4/sqrt(100)) + 25

or, x = 25.49