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) = 6.94 hours and SD(X) = 1.23 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 6.97. Incorrect: Your answer is incorrect. (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 6.97. (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 6.97. (use 4 decimal places in your answer)

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) = 6.94 hours and SD(X) = 1.23 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 6.97. X (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 6.97. (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 6.97. (use 4 decimal places in your answer)

Answer 1

Solution :

Given that,

mean = $\mu$ = 6.94

standard deviation = $\sigma$ = 1.23

a) n = 4

$\mu$$\bar x$ =  $\mu$ = 6.94

$\sigma$$\bar x$ = $\sigma$ / $\sqrt$ n = 1.23 / $\sqrt$ 4 = 0.615

P(6.79 < $\bar x$ < 6.97)  

= P[(6.79 - 6.94) / 0.615 < ($\bar x$ - $\mu$ $\bar x$) / $\sigma$ $\bar x$ < (6.97 - 6.94) / 0.615)]

= P(-0.24 < Z < 0.05)

= P(Z < 0.05) - P(Z < -0.24)

Using z table,  

= 0.5199 - 0.4052  

= 0.1147

b) n = 16

$\mu$$\bar x$ =  $\mu$ = 6.94

$\sigma$$\bar x$ = $\sigma$ / $\sqrt$ n = 1.23 / $\sqrt$ 16 = 0.3075

P(6.79 < $\bar x$ < 6.97)  

= P[(6.79 - 6.94) / 0.3075 < ($\bar x$ - $\mu$ $\bar x$) / $\sigma$ $\bar x$ < (6.97 - 6.94) / 0.3075)]

= P(-0.49 < Z < 0.10)

= P(Z < 0.10) - P(Z < -0.49)

Using z table,  

= 0.5398 - 0.3121

= 0.2277

c) n = 25

$\mu$$\bar x$ =  $\mu$ = 6.94

$\sigma$$\bar x$ = $\sigma$ / $\sqrt$ n = 1.23 / $\sqrt$ 25 = 0.246

P(6.79 < $\bar x$ < 6.97)  

= P[(6.79 - 6.94) / 0.246 < ($\bar x$ - $\mu$ $\bar x$) / $\sigma$ $\bar x$ < (6.97 - 6.94) / 0.246)]

= P(-0.61 < Z < 0.12)

= P(Z < 0.12) - P(Z < -0.61)

Using z table,  

= 0.5478 - 0.2709

= 0.2769