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

hroncits8y 2021-11-21 Answered
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?

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Expert Answer

Parminquale
Answered 2021-11-22 Author has 818 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, \(\displaystyle\mu={36.8},\sigma={1.3}\).
The required probability is,
\(\displaystyle{P}{\left({X}{>}{37.5}\right)}={1}-{P}{\left({X}\le{37.5}\right)}\)
\(\displaystyle={1}-{P}{\left({\frac{{{X}-\mu}}{{\sigma}}}\le{\frac{{{37.5}-{36.8}}}{{{1.3}}}}\right)}\)
\(\displaystyle={1}-{P}{\left({z}\le{\frac{{{0.7}}}{{{1.3}}}}\right)}\)
\(\displaystyle={1}-{P}{\left({z}\le{0.54}\right)}\)
From the "standard normal table", the area to the left of \(\displaystyle{z}={0.54}\ {i}{s}\ {0.7054}\).
\(\displaystyle{P}{\left({X}{>}{37.5}\right)}={1}-{P}{\left({z}\le{0.54}\right)}\)
\(\displaystyle={1}-{0.7054}\)
\(\displaystyle={0.2946}\)
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,
\(\displaystyle{P}{\left({X}{<}{35}\right)}={P}{\left({\frac{{{X}-\mu}}{{\sigma}}}\le{\frac{{{35}-{36.8}}}{{{1.3}}}}\right)}\)
\(\displaystyle={P}{\left({z}\le{\frac{{-{1.8}}}{{{1.3}}}}\right.}\)
\(\displaystyle={P}{\left({z}\le-{1.38}\right)}\)
From the "standard normal table", the area to the left of \(\displaystyle{z}=-{1.38}\ {i}{s}\ {0.0838}\).
\(\displaystyle{P}{\left({X}{<}{35}\right)}={P}{\left({z}\le-{1.38}\right)}\)
\(\displaystyle={0.0838}\)
Thus, the probability that one of these cars will have an mpg of less than 35 is 0.0838.

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