An automobile manufacturer produces a certain model of car. The fuel e

hroncits8y

hroncits8y

Answered question

2021-11-21

An automobile manufacturer produces a certain model of car. The fuel economy figures of these cars are normally distributed with a mean mileage per gallon (mpg) of 36.8 and a standard deviation of 1.3
(a) What is the probability that one of these cars will have an mpg of more than 37.5?
(b) What is the probability that one of these cars will have an mpg of less than 35?
(c) What is the probability that one of these cars will have an mpg of less than 40?

Answer & Explanation

Parminquale

Parminquale

Beginner2021-11-22Added 17 answers

part a
Obtain the probability that one of these cars will have an mpg of more than 37.5.
The probability that one of these cars will have an mpg of more than 37.5 is obtained below:
Let X denotes the mileage of cars follows normal distribution with mean 36.8 mileage per gallon and standard deviation is 1.3 mile age gallon. That is, μ=36.8,σ=1.3.
The required probability is,
P(X>37.5)=1P(X37.5)
=1P(Xμσ37.536.81.3)
=1P(z0.71.3)
=1P(z0.54)
From the "standard normal table", the area to the left of z=0.54 is 0.7054.
P(X>37.5)=1P(z0.54)
=10.7054
=0.2946
Thus, the probability that one of these cars will have an mpg of more than 37.5 is 0.2946.
part b
Obtain the probability that one of these cars will have an mpg of less than 35.
The probability that one of these cars will have an mpg of less than 35 is obtained below:
The required probability is,
P(X<35)=P(Xμσ3536.81.3)
=P(z1.81.3
=P(z1.38)
From the "standard normal table", the area to the left of z=1.38 is 0.0838.
P(X<35)=P(z1.38)
=0.0838
Thus, the probability that one of these cars will have an mpg of less than 35 is 0.0838.

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