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# Some questions on normal distribution

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(NOTES) calculate the z-value using the appropriate formula z=x−μ σ z=x−μ σ
and find the area under the curve for z-value.
Normal distribution is calculated by subtracting upper limit area from lower limit area
(Round all answers to three decimal places)

The mean gas mileage for a hybrid car is 57 miles per gallon. Suppose that the gasoline mileage is approximately normally distributed with a standard deviation of 3.5 miles per gallon.
a. What is the probability that a randomly selected hybrid gets more than 60 miles per gallon?
b. What is the probability that a randomly selected hybrid gets 51 miles per gallon or less?
c. What is the probability that a randomly selected hybrid gets between 57 and 62 miles per gallon?
d. What is the probability that a randomly selected hybrid gets less than 45 miles per gallon?

Before negotiating a long-term construction contract, building contractors must carefully estimate the total cost of completing the project. At a certain university, a contractor proposed a model for total cost of a long-term contract based on the normal distribution. For one particular construction contract, the university assumed total cost, x, to be normally distributed with mean 840,000 and standard deviation 180,000. The revenue, R, promised to the contractor is \$1,010,000.
a. The contract will be profitable if revenue exceeds total cost. What is the probability that the contract will be profitable for the contractor?
b. What is the probability that the project will result in a loss for the contractor?
c. Suppose the contractor has the opportunity to renegotiate the contract. What value of R should the contractor strive for in order to have a 0.98 probability of making a profit?

The ages of a group of 50 women are approximately normally distributed with a mean of 52 years and a standard deviation of 5 years. One woman is randomly selected from the group, and her age is observed.
a. Find the probability that her age will fall between 56 and 60 years.
b. Find the probability that her age will fall between 49 and 53 years.
c. Find the probability that her age will be less than 35 years.
d. Find the probability that her age will exceed 40 years.

Personnel tests are designed to test a job applicant's cognitive and/or physical abilities. A particular dexterity test is administered nationwide by a private testing service. It is known that for all tests administered last year, the distribution of scores was approximately normal with mean 79 and standard deviation 7.7
a. A particular employer requires job candidates to score at least 84 on the dexterity test. Approximately what percentage of the test scores during the past year exceeded 84?
b. The testing service reported to a particular employer that one of its job candidate's scores fell at the
95%percentile of the distribution (i.e., approximately 95% of the scores were lower than the candidate's, and only 5% were higher). What was the candidate's score?
a. Approximately ?% of the test scores during the past year exceeded 84.
(Round to one decimal place as needed.)
b. The candidate's score was
(Round to the nearest whole number as needed.)

https://brainmass.com/math/probability/some-questions-normal-distribution-623492

#### Solution Preview

Question 1:
The mean gas mileage for a hybrid car is 57 miles per gallon. Suppose that the gasoline mileage is approximately normally distributed with a standard deviation of 3.5 miles per gallon.
a. What is the probability that a randomly selected hybrid gets more than 60 miles per gallon?
Solution:

P(X > 60) = P(z > 0.8571) = 0.196
b. What is the probability that a randomly selected hybrid gets 51 miles per gallon or less?
Solution:

P( X ≤ 51) = P(z < -1.714) = 0.043

c. What is the probability that a randomly selected hybrid gets between 57 and 62 miles per gallon?
Solution:

P(57 ≤ X ≤ 62) = P(0<z<1.429) = P(z<1.429)-P(z<0) = 0.924-0.5 = 0.424

d. What is the probability that a randomly selected hybrid gets less than 45 miles per gallon?
Solution:

P( X < 45) = P(z < -3.429) = 0.0003