Question

The average price for a gallon of gasoline in the United States is $3.73 and in Russia it is $3.40. Assume these averages are the population means in the two countries and that the standard deviation of $.25 in the United States and a standard deviation of $.20 in Russia. What is the probability that a randomly selected gas station in the United States charges less than $3.50 per gallon

Answer 1

Answer:

0.17879

Step-by-step explanation:

This is a normal distribution problem

The mean rates of gasoline per gallon for the US is

μ = $3.73

Standard deviation = σ = $0.25

The probability that a randomly selected gas station in the United States charges less than $3.50 per gallon = P(x < 3.5)

We first normalize/standardize/take the z-score

The standardized score for any value is the value minus the mean then divided by the standard deviation.

z = (x - μ)/σ = (3.50 - 3.73)/0.25 = - 0.92

To determine the probability that a randomly selected gas station in the United States charges less than $3.50 per gallon = P(x < 3.50) = P(z < -0.92)

We'll use data from the normal probability table for these probabilities

P(x < 3.50) = P(z < -0.92) = 0.17879

Hope this Helps!!!

Answer 2

Following are the calculation to the P-value:      

Given:  

$\to \text{critical value}\ (x) = \ 3.5\\\\ \to \text{mean}\ \mu = \ 3.73 \\\\\to \text{standard deviation} \ (\sigma)= \ 0.25$

To find:

P - value=?

Solution:

Using formula:

$\to \bold{z = \frac{(x - \mu)}{\sigma}}$

       $\bold{ = \frac{(3.5 - 3.73)}{0.25}} \\\\ \bold{ = \frac{(-0.23)}{0.25}}\\\\ \bold{ = -0.92}$      

Therefore,  by using a table we calculate the left tailed area:        

$\to P(z < -0.92 ) = 0.17878638= 17.87\%$

Therefore, the final answer is "17.87%".

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