Question

1. The time needed to complete a final examination in a particular college course is normally distributed with a mean of 83 minutes and a standard deviation of 13 minutes. Answer the following questions.

a. What is the probability of completing the exam in one hour or less (to 4 decimals)?

b. What is the probability that a student will complete the exam in more than 60 minutes but less than 75 minutes (to 4 decimals)?

2. According to the Sleep Foundation, the average night's sleep is 6.8 hours (Fortune, March 20, 2006). Assume the standard deviation is .7 hours and that the probability distribution is normal.

a. What is the probability that a randomly selected person sleeps more than 8 hours (to 4 decimals)?

b. What is the probability that a randomly selected person sleeps 6 hours or less (to 4 decimals)?

c. Doctors suggest getting between 7 and 9 hours of sleep each night. What percentage of the population gets this much sleep (to the nearest whole number)?

Answer 1

According to the given information;

1. The time needed to complete a final examination in a particular college course is normally distributed with a mean of 83 minutes and a standard deviation of 13 minutes. Therefore;

Let us define $x$ be the random variable that the time needed to complete a final examination in a particular college course.

$x\sim N(\mu =83,\sigma =13)$

a. Therefore the probability of completing the exam in one hour or less is determined as:

$P(x\leq 60)=P(\frac{x-\mu }{\sigma }\leq \frac{60-83}{13})$

$=P(z\leq -1.769)$

$=0.0384$

(to 4 decimals)

(from standard normal table)

Hence the required probability is $0.0384$

b. The probability that a student will complete the exam in more than 60 minutes but less than 75 minutes :

$P(60\leq x\leq 75)=P(\frac{60-83}{13}\leq \frac{x-\mu }{\sigma }\leq \frac{75-83}{13})$

$=P(-1.769\leq z\leq -0.6154)$

$=0.2307$

(to 4 decimals)

(from standard normal table)

Hence the required probability is $0.2307$

2. According to the Sleep Foundation, the average night's sleep is 6.8 hours (Fortune, March 20, 2006). Assume the standard deviation is 0 .7 hours and that the probability distribution is normal.

Let us define $x$ be the random variable that the time needed to complete a final examination in a particular college course.

$x\sim N(\mu =6.8,\sigma =0.7)$

a. The probability that a randomly selected person sleeps more than 8 hours is determined as:

$P(x>8)=P(\frac{x-\mu }{\sigma }> \frac{8-6.8}{0.7})$

$=P(z>1.7143)$

$=0.0432$

(to 4 decimals)

(from standard normal table)

Hence the required probability is $0.0432$

b. The probability that a randomly selected person sleeps 6 hours or less is determined as:

$P(x\leq 6)=P(\frac{x-\mu }{\sigma }\leq \frac{6-6.8}{.07})$

$=P(z\leq -1.143)$

$=0.1265$

(to 4 decimals)

(from standard normal table)

Hence the required probability is $0.1265$

c. Doctors suggest getting between 7 and 9 hours of sleep each night. Therefore the percentage of the population gets this much sleep is determined as:

$P(7\leq x\leq 9)=P(\frac{7-6.8}{0.7}\leq \frac{x-\mu }{\sigma }\leq \frac{9-6.8}{0.7})$

$=P(0.2857\leq z\leq 3.1429)$

$=0.3867$

(from standard normal table)

Therefore the percentage of the population gets this much sleep is $39%$ % ((to the nearest whole number).