Question #171192

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.


1
Expert's answer
2021-03-16T07:13:37-0400

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

and standard deviation σ\text{and standard deviation }\sigma

μ=38.6\mu = 38.6

σ=10.4\sigma =10.4

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

P(25<x<45)P(25<x<45)

standardize\text{standardize}

z=xμσz=\frac{x-\mu}{\sigma}

P(25μσ<z<45μσ)=P(1.308<z<0.615)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(-1.308<z<0.615)= P(z<0.615)-P(z<-1.308)

P(z<0.615)P(z<1.308)=0.732370.09510=0.63727P(z<0.615)-P(z<-1.308)=0.73237-0.09510=0.63727

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

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

P(z<Z)=0.1P(z<Z)=0.1

z=aμσ;a=zσ+μz=\frac{a-\mu}{\sigma};a=z\sigma+\mu

z=.53983z= .53983

b=0.5398310.4+38.644.2b=0.53983*10.4+38.6\approx44.2

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


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

P(z>Z)=0.1P(z>Z)=0.1

P(z<Z)=0.1P(-z<Z)=0.1

z=0.46017-z = 0.46017

b=0.4601710.4+38.633.8b = - 0.46017*10.4+38.6\approx33.8

P(x<33.8)=10%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










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