Answer to Question #184982 in Statistics and Probability for Mawmaw

Question

According to Nielsen Media Research, the (population) average number of hours of TV viewing per household per week in the United States is 50.4 hours. Suppose the (population) standard deviation is known to be 11.8 hours. A random sample of 42 U.S. households is taken. What is the probability that the sample mean is less than 47.5 hours?

Answer 1

$\mu=50.4,s=11.8 ,n=42$

Population Standard deviation $\sigma=\dfrac{s}{\sqrt{n}}=\dfrac{11.8}{\sqrt{42}}=1.82$

Let X denote the sample mean

$P(35<X<40)=P(\dfrac{35-50.4}{1.82}<\dfrac{x-\mu}{\sigma}<\dfrac{40-50.4}{1.82})\\$

$=P(-8.46<z<-5.71)=0.0001$