Question

The speeds of car traveling on Interstate Highway I-35 are normally distributed with a mean of 74
miles per hour and a standard deviation of 6 miles per hour.
(a) Find the percentage of the cars traveling on this highway with a speed
i. of more than 85,
ii. between 65 to 72.
(b) If a BMW is at the speed that is faster than 90 percentage of cars, what is the speed of the
BMW?

Answer 1

here the mean speed of car traveling on Interstate Highway I-35 = = 74 mph

standard deviation = = 6 mph

(a) Here if x is the spedd of a random car then

(i) Pr(x > 85) = 1 - Pr(x < 85) = 1 - NORMDIST(x < 85 ; 74 ; 6)

here we have to find the z value first

z = (x - )/ = (85 - 74)/6 = 1.83333

now we look for given z value in z table or calculator for cumulative probability value

Pr(x > 85) = 1 - Pr(x < 85) = 1 - NORMDIST(x < 85 ; 74 ; 6) = 1 - Pr(Z < 1.8333)

= 1 - 0.9666

= 0.0334

(b) Betwen 65 to 72

Pr(65 < X < 72) = Pr(X < 72 mph ; 74 mph ; 6 mph) - Pr(X < 65 mph ; 74 mph ; 6 mph)

Z₂ = (72 - 74)/6 = -0.3333

Z₁ = (65 - 74)/6 = -1.5

Pr(65 < X < 72) = Pr(X < 72 mph ; 74 mph ; 6 mph) - Pr(X < 65 mph ; 74 mph ; 6 mph)

= NORMSDIST(-0.3333) - NORMSDIST( -1.5)

= 0.3694 - 0.0668

= 0.3026

(b) here the speed of BMW is at the speed that is faster than 90 percentage of cars,

so let say the speed of BMW is x₀

that means writing mathematically,

Pr(x < x₀) = 0.90

so here now we have to look for the value of z for the given cumulative probability of 0.90

Z = normsinv(0.90) = 1.2816

so here

Z = (x₀ -)/

1.2816 = (x₀ - 74)/6

x₀ = 74 + 1.2816 * 6 = 81.69 mph