Question

Suppose the speeds of people driving from Dallas to Houston normally distributed, with a mean average speed of 92 miles per hour. (a) If the probability that the speed is more than 108 mph is 0.8%, what is the standard deviation? (6 points) (b) Given the standard deviation in (a), what is the probability that a person drove less than 85 miles per hour? (5 points) (c) If Ben drove at a speed that exceeds 85% of all other people. What was her speed? (6 points)

Answer 1

X : speeds of people driving from Dallas to Houston

X is Normally distributed with mean = 92 miles per hour

(a) probability that the speed is more than 108 mph is 0.8%

i.e P(X > 108) = 0.8/100 = 0.008

P(X > 108) = 1- P(X108) =0.008

P(X108) = 1- 0.008 = 0.992

Z1 be the z-score for 108

Therefore,

P(ZZ1) = 0.992

Z1 = (108-mean)/Standard deviation ;

Standard deviation = (108-92)/Z1 = 16/Z1

From standard normal tables, Z1 = 2.41

Standard deviation = 16/Z1 = 16/2.41 = 6.639

Standard deviation = 6.639

(b)

probability that a person drove less than 85 miles per hour = P(X85)

Z-score for 85 = (85-92)/6.639= - 7/6.639 = -1.05

From standard normal tables,

P(Z<-1.05) = 0.1469

P(X85)=P(Z<-1.05) = 0.1469

probability that a person drove less than 85 miles per hour = 0.1469

(c) If Ben drove at a speed that exceeds 85% of all other people. What was her speed? (6 points)

Let XB be the Ben's speed

If Ben drove at a speed that exceeds 85% of all other people i.e 85% of people driving at a speed less than that of Ben

i.e P(X XB) = 85/100 = 0.85

ZB be the Z-score for XB

P(X XB) =P(ZZB) = 0.85

ZB = (XB - mean)/standard deviation

XB = Mean + ZB x Standard deviation

P(ZZB) = 0.85 ; From standard normal tables ; ZB = 1.04

XB = Mean + ZB x Standard deviation = 92 + 1.04 x 6.639 = 92+6.9046=98.9046

Ben's speed = 98.9046