Answer:
[tex]P(19<X<26)=0.63[/tex]
Step-by-step explanation:
We start by defining the random variable X.
X : ''The driving distance to work for residents of a certain community''
X can be modeled as a normal random variable.
X ~ N (μ,σ)
Where μ is the mean and σ is the standard deviation.
For this problem :
X ~ N (21,3.6)
Where the unit for the mean and the standard deviation is miles.
We are looking for [tex]P(19<X<26)[/tex]
Given a normal random variable,we can subtract it the mean and then divide by the standard deviation in order to obtain a new random variable ''Z''.
Where ''Z'' is a new normal random variable.This is called standardizing.
Z ~ N (0,1)
The mean of Z is 0 and the standard deviation is 1.
We can find the cumulative probability distribution of Z in any table.
[tex]P(Z\leq a)=[/tex] Φ(a) given a certain value ''a''.
For the problem :
[tex]P(19<X<26)=[/tex]
[tex]P(\frac{19-21}{3.6}<\frac{X-21}{3.6}<\frac{26-21}{3.6})=[/tex]
[tex]P(-0.5555<Z<1.3888)=P(Z<1.3888)-P(Z<-0.5555)[/tex]
[tex]P(19<X<26)=[/tex] Φ(1.3888) - Φ(-0.5555) = 0.9177 - 0.2877 = 0.63
The probability that an individual drives between 19 and 26 miles to work is 0.63
