Answer to Question #182599 in Statistics and Probability for Shivani Shivani

Question

 A new gas-electric hybrid car has recently hit the market. The distance traveled on 1 gallon of fuel is normally distributed with a mean of 65 miles and a standard deviation of 4 miles. Find the probability of the following events. (show the concerned region by z curve) 1. The car travels more than 70 miles per gallon. 2. The car travels less than 60 miles per gallon. 3. The car travels between 55 and 70 miles per gallon.

Answer 1

From the question, we obtain the following information.

$\mu$ = 65 miles (the mean distance traveled on 1 gallon of fuel by the new gas-electric hybrid car).

$\sigma$ = 4 miles (the standard deviation of the mean distance traveled by the new gas-electric hybrid car).

We then proceed to answer the following questions as follows

1. The car travels more than 70 miles per gallon.

To find the probability that the car travels more than 70 miles per gallon, we transform the variable to standard normal. This is given by

$$P(x>70)=P(z>\frac{x-\mu}{\sigma})=P(z>\frac{70-65}{4})=P(z>1.25).$$

The above can be illustrated in the following diagram.

Since the standard normal distribution has a total area (probability) equal to 1 and it is also symmetrical about the mean, we can compute the above probability as follows

$$P(z>1.25) =1-P(z<1.25)=1-\Phi (1.25),\\=1-0.89435=0.10565.$$

2. The car travels less than 60 miles per gallon.

Similarly, to find the probability that the car travels less than 60 miles per gallon, we transform the variable to standard normal. This is given by

$$P(x<60)=P\left(z<\frac{x-\mu}{\sigma}\right),\\=P\left(z<\frac{60-65}{4}\right)=P(z<-1.25).$$

The above can be illustrated in the following diagram.

Thus, the required probability from the normal table is given by

$$P(z<-1.25) =\Phi(-1.25)=0.10565.$$

3. The car travels between 55 and 70 miles per gallon.

To find the probability that the car travels between 55 and 70 miles per gallon, we transform the variable to standard normal. This is given by

$$P(55<x<70)=P\left(\frac{x-\mu}{\sigma}<z<\frac{x-\mu}{\sigma}\right),\\ =P\left(\frac{55-65}{4}<z<\frac{70-65}{4}\right),\\ P\left(-\frac{10}{4}<z<\frac{5}{4}\right)=P(-2.5<z<1.25).$$

This can be illustrated in the following diagram

The required probability from the normal table is then given by

$$P(-2.5<z<1.25)=\Phi(1.25)-\Phi(-2.5),\\ =0.89435-0.00621=0.88814.$$