The manufacturer of a new compact car claims the miles per gallon (mpg) for the gasoline consumption is mound-shaped and symmetric with a mean of 24.6 mpg and a standard deviation of 11.2 mpg. if 30 such cars are tested, what is the probability the average mpg achieved by these 30 cars will be greater than 27

Mean (μ)= 24.6 mpg, Standard deviation(S.D) = 11.2 mpg and Sample size (n) = 30 , Average greater than 27

z(27) = (27 -μ)/(S.D/sqrt n)
z(27) = 27- 24.6/(11.2/sqrt30)
z(27) = 1.17
P(x-bar >27) = P(z> 1.17) = normal cdf (1.17, 100) = 0.4152

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JeanaShupp JeanaShupp · Answer 2 · 2019-08-13T16:28:16+02:00

Answer: 0.121

Step-by-step explanation:

Given : The manufacturer of a new compact car claims the miles per gallon (mpg) for the gasoline consumption is mound-shaped and symmetric i.e. a normal distribution with

[tex]\mu=24.6\text{ mpg}[/tex]

[tex]\sigma=11.2\text{ mpg}[/tex]

Sample size : n= 30

Let x be a random variable that represents the gasoline consumption.

Z-score : [tex]z=\dfrac{x-\mu}{\dfrac{\sigma}{\sqrt{n}}}[/tex]

For x= 27

[tex]z=\dfrac{27-24.6}{\dfrac{11.2}{\sqrt{30}}}\approx1.17[/tex]

Now, the probability the average mpg achieved by these 30 cars will be greater than 27 will be :-

[tex]P(X>27)=P(z>1.17)=1-P(\leq1.17)\\\\=1-0.8789995\\\\=0.1210005\approx0.121[/tex]

Hence, the probability the average mpg achieved by these 30 cars will be greater than 27 = 0.121