Question

A major U.S. automaker has determined that the city mileage for its new hybrid gas/electric model is normally distributed with a mean equal to 52 miles per gallon (μ = 52) and a standard deviation of 3.7 mpg (σ = 3.7).

A. What is the probability a vehicle would get less than 50 miles per gallon, i.e., P(x < 50)?
Enter your response rounded to 3 decimals, in the format 0.123 with no other text or symbols.

B. What is the probability a vehicle would get more than 60 miles per gallon, i.e., P(x > 60)?
Enter your response rounded to 3 decimals, in the format 0.123 with no other text or symbols.

C. What vehicle mileage is at the 90th percentile?
Enter your response rounded to 1 decimal, in the format 1.2 with no other text or symbols.

Answer 1

Solution :

Given that ,

mean = = 52

standard deviation = = 3.7

A)

P(x < 50) = P((x - ) / < (50 - 52) / 3.7)

= P(z < -0.54)

Using standard normal table,

P(x < 50) = 0.2946 = 0.295

Probability = 0.295

B)

P(x > 60) = 1 - P(x < 60)

= 1 - P((x - ) / < (60 - 52) / 3.7)

= 1 - P(z < 2.162)

= 1 - 0.9847

= 0.0153

P(x > 60) = 0.0153

Probability = 0.0153

C)

P(Z < z) = 90%

P(Z < 1.28) = 0.90

z = 1.28

Using z-score formula,

x = z * +

x = 1.28 * 3.7 + 52 = 56.7

90th percentile = 56.7