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Assume that the speed of automobiles on an expressway during rush hour is normally distributed with a mean of 68 mph and a standard deviation of 55 mph. What percent of cars are traveling faster than 77 ​mph?



Answer :

reinhard10158

Step-by-step explanation:

let's call X = speed of the cars on the expressway during rush hour.

the probability function is P(X).

mean = 68

SD = 55

P(X > 77) = P(X - mean > 77 - 68) =

= P((X - mean)/SD > (77 - 68)/55)

Z = (x - mean)/SD

(77 - 68)/55 = 9/55 = 0.163636364...

so,

P(X > 77) = P(Z > 0.163636364...) =

= 1 - P(Z <= 0.163636364...) ≈

≈ 1 - 0.565 (between 0.56356 and 0.56749) ≈

≈ 0.435

therefore, about 43.5% of the cars are traveling faster than 77 mph.

remember, while the probabilty distribution area is normed to 1, percentages are normed to 100.

so, % = probabilty × 100

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