Question

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

Answer 1

Answer:

$P(19$

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 $P(19$

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.

$P(Z\leq a)=$ Φ(a) given a certain value ''a''.

For the problem :

$P(19$

$P(\frac{19-21}{3.6}$

$P(-0.5555$

$P(19$ 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