Question

The average fuel efficiency of U.S. light vehicles for 2005 was 21 mpg. If the standard deviation of the population was 2.9 and the gas ratings were normally distributed,
1) What is the probability that the mean mpg for a random sample of 25 light vehicles is under 20?

2) What is the probability that the mean mpg for a random sample of 25 light vehicles is between 20 and 25?

Answer 1

Answer:

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

2) between 20 and 25 --> 21-25/2.9 = -1.38

Step-by-step explanation:

Problem #1:

• Using the 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.

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

Problem #2:

21-25/2.9 = -1.38