Question

The amount of gasoline sold each month to customers at Bob's Exxon station in downtown Navasota is a random variable. This random variable can be described as having a normal distribution with mean 2500 gallons and a standard deviation of 200 gallons (and there is no seasonal variation in sales). Exxon will give Bob an all-expense paid trip to College Station, including a free dinner at Burger King, if his station pumps more than 2800 gallons in any one month.
What is the probability that Bob will win that wonderful trip on the basis of his gasoline sales this month?

Answer 1

Answer:

The probability that Bob will win that wonderful trip on the basis of his gasoline sales this month

P(X≥ 2800) = P(Z₁≥1.5) = 0.0768

Step-by-step explanation:

Step(i):-

Mean of the Population (μ) = 2500 gallons

Standard deviation of the population (σ) =200 gallons

Let 'X' be a random variable in Normal distribution

Given X = 2800

$Z = \frac{x-mean}{S.D} = \frac{2800-2500}{200} = 1.5$

Step(ii):-

The probability that Bob will win that wonderful trip on the basis of his gasoline sales this month

P(X≥ 2800) = P(Z₁≥1.5)

                  = 0.5 - A(Z₁)

                 =  0.5 - A(1.5)

                 =   0.5 -0.4232    ( from normal table)

                = 0.0768

Conclusion:-

The probability that Bob will win that wonderful trip on the basis of his gasoline sales this month

P(X≥ 2800) = P(Z₁≥1.5) = 0.0768