Question

A federal government agency that is responsible for setting vehicle fuel economy standards for automobile manufacturers is conducting research in order to update its fuel economy standards for the year 2030. Automobile​ manufacturers, and​ consumers, are highly interested in what the​ agency's findings and determinations will be as this will affect every vehicle in the United States. The federal government agency is very interested in the relationship between the weight of a vehicle and the​ vehicle's fuel economy​ (average miles per gallon​ (MPG)). Specifically, the agency is concerned that if the current trend of automobile manufacturers producing heavier new vehicles continues that its fuel economy targets will not be met. The​ agency's research department recently collected data for analysis in order to support the​ agency's upcoming discussion with industry regarding its proposed 2030 fuel economy standards. The average MPG from a random sample of 750 vehicles was recently calculated by the agency. The research division also collected the vehicle weight of these 750 randomly sampled vehicles. The Vehicle​ Number, Type, Vehicle​ Weight, Average​ MPG, Fuel Tank Size​ (Gallons), Engine Size​ (Liters), and Meet or Not Meet Current Standards data were collected for these 750 vehicles. StatCrunch Data Set

Answer 1

Using the normal distribution, it is found that there is a 0.0307 = 3.07% probability of randomly selecting a vehicle with an engine size less than 2.7 L.

Normal Probability Distribution

In a normal distribution with mean $\mu$ and standard deviation $\sigma$, the z-score of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

• It measures how many standard deviations the measure is from the mean.

• After finding the z-score, we look at the z-score table and find the p-value associated with this z-score, which is the percentile of X.

Researching the problem on the internet, it is found that:

• The mean engine size in L is of $\mu = 3.59804$.

• The standard deviation in L is of $\sigma = 0.47986$.

The probability of randomly selecting a vehicle with an engine size less than 2.7 L is the p-value of Z when X = 2.7, hence:

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{2.7 - 3.59804}{0.47986}$

$Z = -1.87$

$Z = -1.87$ has a p-value of 0.0307.

0.0307 = 3.07% probability of randomly selecting a vehicle with an engine size less than 2.7 L.

More can be learned about the normal distribution at https://brainly.com/question/24663213