Answer to Question #171192 in Statistics and Probability for kmy

Question

1. Information from the Department of Motor Vehicles indicates that

the average age of licensed drivers is 38.6 years with a standard

deviation of 10.4 years. Assume that the distribution of the

driver’s ages is normal.  

a. What proportion of licensed drivers are from 25 to 45 years old?

b. Determine the ages of licensed drivers separating the upper 10% and

lower 10% percent of the population.

Answer 1

$N(\sigma,\mu) -\text{is the normal distribution with the mean } \mu$

$\text{and standard deviation }\sigma$

$\mu = 38.6$

$\sigma =10.4$

$a.)\text{find probability of hitting normal random}\newline \text{values in the interval }(25;45)$

$P(25<x<45)$

$\text{standardize}$

$z=\frac{x-\mu}{\sigma}$

$P(\frac{25-\mu}{\sigma}<z<\frac{45-\mu}{\sigma})=P(-1.308<z<0.615)$

$P(-1.308<z<0.615)= P(z<0.615)-P(z<-1.308)$

$P(z<0.615)-P(z<-1.308)=0.73237-0.09510=0.63727$

$P(25<x<45) \approx 0.637\text{ or } 63.7\%$

$b.) P(x>a)=10\%$

$P(z<Z)=0.1$

$z=\frac{a-\mu}{\sigma};a=z\sigma+\mu$

$z= .53983$

$b=0.53983*10.4+38.6\approx44.2$

$P(x>44.2)=10\%$

$P(x<b)=10\%$

$P(z>Z)=0.1$

$P(-z<Z)=0.1$

$-z = 0.46017$

$b = - 0.46017*10.4+38.6\approx33.8$

$P(x<33.8)=10\%$

Answer:a)63.7% drivers from 25 to 45 years old;

b)10% under 33.8

10% older 44.2