Consider a bell-shaped symmetric distribution with mean of 16 and standard deviation of 1.5. Approximately what percentage of data lie between 13 and 19?

Answer: 95.45 %Step-by-step explanation:Given : The distribution is bell shaped , then the distribution must be normal distribution.Mean : [tex]\mu=\ 16[/tex]Standard deviation :[tex]\sigma= 1.5[/tex]The formula to calculate the z-score :-[tex]z=\dfrac{x-\mu}{\sigma}[/tex]For x = 13[tex]z=\dfrac{13-16}{1.5}=-2[/tex]For x = 19[tex]z=\dfrac{19-16}{1.5}=2[/tex]The p-value = [tex]P(-2<z<2)=P(z<2)-P(z<-2)[/tex][tex]0.9772498-0.0227501=0.9544997\approx0.9545[/tex]In percent, [tex]0.9545\times100=95.45\%[/tex]Hence, the percentage of data lie between 13 and 19 = 95.45 %