At a local company, the ages of all new employees hired during the last 10 years are normally distributed. The mean age is 38 years old, with a standard deviation of 10 years. Find the percent of new employees that are no more than 30 years old. Round to the nearest percent.

Answer: P = 21%Step-by-step explanation:We look for the percentage of employees who are not more than 30 years old.This is:[tex]P = \frac{x}{n} *100\%[/tex]Where x is the number of new employees who are not over 30 years old and n is the total number of new employees.We do not know the value of x or n. However, the probability of randomly selecting an employee that is not more than 30 years old is equal to [tex]P = \frac{x}{n}[/tex]Then we can solve this problem by looking for the probability that a new employee is not more than 30 years old.This is:[tex]P(X< 30)[/tex]Then we find the z-score[tex]Z = \frac{X - \mu}{\sigma}[/tex]We know that:μ = 38 years[tex]\sigma = 10[/tex] yearsSo[tex]Z = -0.8[/tex]Then[tex]P (X<30) = P (\frac{X- \mu}{\sigma} < \frac{30-38}{10})\\\\P (X<30) = P(Z<-0.8)[/tex]By symmetry of the distribution[tex]P(Z<-0.8)=P(Z>0.8)[/tex]Looking in the normal standard tables[tex]P(Z>0.8)=0.211[/tex]Finally P = 21%