Question

QUESTION 17 EPA studies of fuel consumption indicate compact car mileage to be normally distributed with a mean of 25.5 miles per gallon and a standard deviation of 5.0 mpg. What is the probability of compact cars gets between 20 and 40 mpg? (Please answer in terms of a decimal, eg 0.12) QUESTION 18 Suppose the amount of heating oil used annually by households in lowa is normally distributed with mean 200 gallons per household per year and standard deviation 40 gallons of heating oil per household per year. If the members of a particular household were scared into using fuel conservation measures by newspaper accounts of the probable price of heating oil next year, and they decided they wanted to use less oil than 97.5% of all other lowa households currently using heating oil, what is the maximum amount of oil they can use and still accomplish their conservation objective?

Answer 1

Solution :

Given that ,

Question 17) mean = $\mu$ = 25.5

standard deviation = $\sigma$ = 5.0

P(20 < x < 40) = P[(20 - 25.5)/ 5.0) < (x - $\mu$ ) /$\sigma$  < (40 - 25.5) / 5.0) ]

= P(-1.1 < z < 2.9)

= P(z < 2.9) - P(z < -1.1)

= 0.9981 - 0.1357

0.86

Probability = 0.86

Question 18) mean = $\mu$ = 200

standard deviation = $\sigma$ = 40

Using standard normal table ,

P(Z < z) = 97.5%

P(Z < z) = 0.975

P(Z < 1.96) = 0.975

z = 1.96

Using z-score formula,

x = z * $\sigma$ + $\mu$

x = 1.96 * 40 + 200 = 278.40

Maximum  amount of oil is 278.40 gallons