A certain car model has a mean gas mileage of 31 miles per gallon (mpg) with astandard deviation 3 mpg. A pizza delivery company buys 43 of these cars. What is theprobability that the average mileage of the fleet is greater than 30.7 mpg?

Probability that the average mileage of the fleet is greater than 30.7 mpg is 0.7454.

Step-by-step explanation:

We are given that a certain car model has a mean gas mileage of 31 miles per gallon (mpg) with a standard deviation 3 mpg.

A pizza delivery company buys 43 of these cars.

Let = sample average mileage of the fleet

The z-score probability distribution of sample average is given by;

                   Z =  ~ N(0,1)

where, = mean gas mileage = 31 miles per gallon (mpg)

            = standard deviation = 3 mpg

           n = sample of cars = 43

So, probability that the average mileage of the fleet is greater than 30.7 mpg is given by = P( > 30.7 mpg)

     P( > 30.7 mpg) = P( > ) = P(Z > -0.66) = P(Z < 0.66)

                                                                 = 0.7454

Because in z table area of P(Z > -x) is same as area of P(Z < x). Also, the above probability is calculated using z table by looking at value of x = 0.66 in the z table which have an area of 0.7454.

Therefore, probability that the average mileage of the fleet is greater than 30.7 mpg is 0.7454.

16 km due west

Step-by-step explanation:

The bearing of the school p from school q is 16 km due west.

To find the bearing of school q from school p, we have to find the direction that the school q is with respect to school p.

Since p is directly west of q, then it implies that q must be directly east of p.

We now know the direction.

Since the distance from q to p is exactly the same as the distance from p to q, then, the distance from p to q is 16 km.

Hence, the bearing of q from p is 16 km due west.