Respuesta :
Answer:
Probability that the test will lead the consumer group to "accept" the claimed mileage for this car is 0.00043.
Step-by-step explanation:
We are given that a consumer group plans to test whether a new passenger car that is advertised to have a mean highway miles per gallon of at least 33 actually meets this level.
It is believed that the population standard deviation is 3 mpg. Based upon this information, the "true" population mean is 32.0 mpg.
Let [tex]\bar X[/tex] = sample mean highway miles per gallon
The z-score probability distribution for sample mean is given by;
Z = [tex]\frac{\bar X -\mu}{\frac{\sigma}{\sqrt{n} } }[/tex] ~ N(0,1)
where, [tex]\mu[/tex] = true population mean = 32.0 mpg
[tex]\sigma[/tex] = population standard deviation = 3 mpg
n = sample of cars = 100
The Z-score measures how many standard deviations the measure is away from the mean. After finding the Z-score, we look at the z-score table and find the p-value (area) associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X.
Now, Probability that a mean highway miles per gallon of at least 33 actually meets this level is given by = P([tex]\bar X[/tex] [tex]\geq[/tex] 33)
P([tex]\bar X[/tex] [tex]\geq[/tex] 33) = P( [tex]\frac{\bar X -\mu}{\frac{\sigma}{\sqrt{n} } }[/tex] [tex]\geq[/tex] [tex]\frac{33-32}{\frac{3}{\sqrt{100} } }[/tex] ) = P(Z [tex]\geq[/tex] 3.33) = 1 - P(Z < 3.33)
= 1 - 0.99957 = 0.00043
The above probability is calculated by looking at the value of x = 3.33 in the z table which has an area of 0.7673.
Therefore, the probability that the test will lead the consumer group to "accept" the claimed mileage for this car is 0.00043.