Question

answer all questions please!!!

The mean gas mileage for a hybrid car is

56

miles per gallon. Suppose that the gasoline mileage is approximately normally distributed with a standard deviation of

3.2

miles per gallon.​ (a) What proportion of hybrids gets over

62

miles per​ gallon? (b) What proportion of hybrids gets

52

miles per gallon or​ less?

left parenthesis c right parenthesis What(c) What

proportion of hybrids gets between

58

and

62

miles per​ gallon? (d) What is the probability that a randomly selected hybrid gets less than

45

miles per​ gallon?

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​(a) The proportion of hybrids that gets over

62

miles per gallon is

nothing.

​(Round to four decimal places as​ needed.)

Answer 1

Answer a) The proportion of hybrids gets over 62 is 0.0304

The following information has been provided: μ 56, σ 3.2 We need to compute Pr(X 62). The corresponding z-value needed to be computed is: z-X-62 56 62-56 1.875 Therefore, we get that Pr (z 62-56 %3.2 Pr(Z 1.875) Pr(X > 62) 1 0.9696 0.0304

Answer b) The proportion of hybrids gets 52 miles per gallon or​ less is 0.1056

The following information has been provided μ= 56, σ= 3.2 ec he comspondig ed o X-μ 52-56 T2-=-1.25 z- 3.21.25 - Therefore Pr(X < 52) = Pr ( Z < 023.200 ) = Pr(Z <-1.25) = 0.1056

Answer c) The proportion of hybrids gets between 58 and 62 miles per​ gallon is 0.2356

The following information has been provided: μ 56, σ = 3.2 We need to compute Pr(58 X 62). The corresponding z-values needed to be computed are: 58-56 0625 3.2 z -, Xo_μ 62-56 -3.2-= 1.875 Therefore, we get: 58 -56 62 56 Pr(58 < X < 62) = Pr (リ(3.2 < z < 3.2Pr 0.625 s Z s 1.875) Pr(Z < 1.875) Pr(Z < 0.625) = 0.9696 0.734 = 0.2356

Answer d) The probability that a randomly selected hybrid gets less than 45 miles per​ gallon is 0.0003

The following information has been provided: μ = 56, σ = 3.2 We need to compute Pr(X 45). The corresponding z-value needed to be computed: Χ μ-45 56_-3.4375 Therefore Pr(X < 45) = Pr ( Z < 45 -56 3.2 -Pr(Z 3.4375) 0.0003