Assume that men's weights are normally distributed with a mean of 172 lb and a standard deviation of 29 lb. If 4 man is randomly selected, find the probability that they have a mean weight between 160 lb and 180 lb.(up to four decimal place, please)

Answer:  0.5056Step-by-step explanation:Let x be the random variable that represents the readings on scientific thermometers .Given : The men's weights are normally distributed,Population mean : [tex]\mu=172\ lb[/tex]Standard deviation : [tex]\sigma= 29\ lb[/tex]Sample size : n=4Z-score : [tex]z=\dfrac{x-\mu}{\dfrac{\sigma}{\sqrt{n}}}[/tex]Now, the z-value corresponding to 160  :  [tex]z=\dfrac{160-172}{\dfrac{29}{\sqrt{4}}}\approx-0.83[/tex] z-value corresponding to 180  :  [tex]z=\dfrac{180-172}{\dfrac{29}{\sqrt{4}}}\approx0.55[/tex]By using standard normal distribution table for z-values,P-value = [tex]P(160<x<180)=P(-0.83<z<0.55)=P(z<0.55)-P(z<-0.83)[/tex][tex]= 0.7088403-0.2032694=0.5055709\approx0.5056\text{ (Rounded to four decimal places)}[/tex]Hence, the probability that they have a mean weight between 160 lb and 180 lb = 0.5056Hence, the probability of the reading greater than -1.05 in degrees Celsius.= 0.8531