Full-time college students report spending a mean of 29 hours per week on academ

Question

Full-time college students report spending a mean of 29 hours per week on academic? activities, both inside and outside the classroom. Assume the standard deviation of time spent on academic activities is 5 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 28 hours per? week? (Round to four decimal places as? needed.) b. If you select a random sample of 16 ?full-time college? students, there is an 86?% 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 normally distributed. D.The population is symmetrically? distributed, such that the Central Limit Theorem will likely hold for samples of size 16 d. If you select a random sample of 100 ?full-time college? students, there is an 86?% chance that the sample mean is less than how many hours per? week? (Round to two decimal places as? needed.)

Explanation / Answer

a)as z score =(X-mean)/std error

also std error =std deviation/sqrt(n)=5/sqrt(4)

P(X>28)=P(Z>(28-29)/(5/sqrt(16))=P(Z>-0.8)=0.7881

b)

here for top 86%; at 14th percentile critical value of z=-1.08

hence corresponding value =mean+z*std error=29-1.08*5/sqrt(16)=27.65 Hours

c). C.The population is normally distributed.

d)

corresponding value =mean+z*std error=29-1.08*5/sqrt(100)=28.46 Hours

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