Assume that men's weights are normally distributed with a mean of 172 lb and standard deviation of 29 lb (national health survey) if 81 men are randomly selected, find the probability that they have a mean weight less than 167 lb.

Answer: 0.0606Step-by-step explanation:Given : The men's weights are normally distributed with a mean of 172 lb and standard deviation of 29 lb.i.e. [tex]\mu=172[/tex] and [tex]\sigma=29[/tex]and sample size : n= 81Let x be a random variable that denotes the men's weights.Formula : [tex]z=\dfrac{x-\mu}{\dfrac{\sigma}{\sqrt{n}}}[/tex]Then, the probability that they have a mean weight less than 167 lb will be :-[tex]P(x<167)=P(\dfrac{x-\mu}{\dfrac{\sigma}{\sqrt{n}}}<\dfrac{167-172}{\dfrac{29}{\sqrt{81}}})\\\\=P(z<\dfrac{-5}{\dfrac{29}{9}})\approx P(z<-1.55)\\\\=1-P(z<1.55)\ \ [\because\ P(Z<-z)=1-P(Z<z)]\\\\=1-0.9394\ \ [\text{Using z-value}]\\\\=0.0606[/tex]Hence, the  probability that they have a mean weight less than 167 lb = 0.0606