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The average fuel efficiency of U.S. light vehicles (cars, SUVs, minivans, vans, and light trucks) for 2005 was 21 mpg. If the standard deviation of the population was 2.9 and the gas ratings were normally distributed, what is the probability that the mean mpg for a random sample of 25 light vehicles is Under 20

Answer :

Answer:

0.042341

Step-by-step explanation:

We solve using z score formula

z = (x-μ)/σ/n, where

x is the raw score = 20 mpg

μ is the population mean = 21 mpg

σ is the population standard deviation = 2.9

n = random number of samples

Hence:

For x < 20

= z = 20 - 21/2.9/√25

= z = -1/2.9/5

= z = -1.72414

P-value from Z-Table:

P(x<20) = 0.042341

The probability that the mean mpg for a random sample of 25 light vehicles is Under 20 mpg is 0.042341

The probability that the mean mpg for a random sample of 25 light vehicles is 0.042341

Calculation:

Here we used z score formula

  • z = (x-μ)/σ/n, where
  • x is the raw score = 20 mpg
  • μ is the population mean = 21 mpg
  • σ is the population standard deviation = 2.9
  • n = random number of samples

So,

For x < 20

= z = 20 - 21/2.9/√25

= z = -1/2.9/5

= z = -1.72414

Now

P-value from Z-Table:

P(x<20) = 0.042341

Learn more about the probability here: https://brainly.com/question/795909?referrer=searchResults

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