Answer to Question #188766 in Statistics and Probability for Laura Leonard

Question

According to the Sleep Foundation, the average nights' sleep is 6.8 hours. Assume the standard deviation is 0.6 hours and that the probability distribution is normal.

1) What is the probability that a randomly selected person sleeps more than 8 hours?

2) What is the probability that a randomly selected person sleeps 6 hours or less?

3) What is the percentage of people sleeping between 6,2 and 7,4 hours ? 

4) Compute once again the above probabilities based on a standard normal distribtuion.

Answer 1

Given,$\mu=6.8 hrs, \sigma=0.6 hrs$

(i) $P(x>8)=P(z>\dfrac{8-6.8}{0.6})=P(z>2)=1-P(z<2)=1-0.9772=0.0228$

(b) $P(x\le 6)=P(z<\dfrac{6-6.8}{0.6})=P(z<-1.33)=0.0918$

(c) $P(6.2<x<7.4)=P(\dfrac{6.2-6.8}{0.6}<z<\dfrac{7.4-6.8}{0.6})=P(-0.1<z<0.1)=0.28696$

(d) The above probabilities from standard normal distribution is same as above-

$P(x>8)=0.0228\\P(x\le 6)=0.0918\\P(6.2<x<7.4)=0.28696$