More than 100 million people around the world are not getting enough sleep; the

Question

More than 100 million people around the world are not getting enough sleep; the average adult needs between 7.5 and 8 hours of sleep per night. College students are particularly at risk of not getting enough shut-eye.

A recent survey of several thousand college students indicated that the total hours of sleep time per night, denoted by the random variable X, can be approximated by a normal model with E(X) = 6.87 hours and SD(X) = 1.14 hours.

Question 1. Find the probability that the hours of sleep per night for a random sample of 4 college students has a mean x between 6.79 and 6.96.

(use 4 decimal places in your answer)

Question 2. Find the probability that the hours of sleep per night for a random sample of 16 college students has a mean x between 6.79 and 6.96.

(use 4 decimal places in your answer)

Question 3. Find the probability that the hours of sleep per night for a random sample of 25 college students has a mean x between 6.79 and 6.96.

(use 4 decimal places in your answer)

NOTE: please be descriptive with the z-score part. Thanks!

Explanation / Answer

In every part Z part is got by standarding variable.

Like Z = (X-Mu) / (Stdev/sqrt(n))

After that refer to Z tables over net to get transaltion of Z got here, to p value.

I have included Z calaulations and the all p values referring to these Z values in the answer.

1)

P(6.79<X<6.96)

= P( (6.79-6.87)/(1.14/sqrt(4))<Z<(6.96-6.87)/(1.14/sqrt(4))
= P(-.14<Z<.16)
=.5636- .4443
=.1193

2)

P(6.79<X<6.96)
= P( (6.79-6.87)/(1.14/sqrt(16))<Z<(6.96-6.87)/(1.14/sqrt(16))
= P(-.28<Z<.32)
=.6255-.3897
=.2358

3)

P(6.79<X<6.96)
= P( (6.79-6.87)/(1.14/sqrt(25))<Z<(6.96-6.87)/(1.14/sqrt(25))
= P(-.35<Z<.4)
=.6544-.3632
=.2912

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