Suppose the driving distance to work for residents of a certain community has a mean of 21 miles and a standard deviation of 3.6 miles. what is the probability that an individual drives between 19 and 26 miles to work? assume that the distribution is normal

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

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.

Φ(a) given a certain value ''a''.

For the problem :

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

whats the

the answer is simply just 6n add the two together as they have an n