Question

Tire pressure monitoring systems (TPMS) warn the driver when the tire pressure of the vehicle is 26% below the target pressure. Suppose the target tire pressure of a certain car is 28 psi (pounds per square inch.)

(a) At what psi will the TPMS trigger a warning for this car? (Round your answer to 2 decimal place.)

When the tire pressure is 20.72 psi.

(b) Suppose tire pressure is a normally distributed random variable with a standard deviation equal to 3 psi. If the car’s average tire pressure is on target, what is the probability that the TPMS will trigger a warning? (Round your answer to 4 decimal places.)

Probability 0.0075

(c) The manufacturer’s recommended correct inflation range is 26 psi to 30 psi. Assume the tires’ average psi is on target. If a tire on the car is inspected at random, what is the probability that the tire’s inflation is within the recommended range? (Round your intermediate calculations and final answer to 4 decimal places.)

Probability

Answer 1

X : Tire pressure of a certain car

Target of X = 28

TPMS warning : when the tire pressure of the vehicle is 26% below the target i.e

if tire pressure = 28-(26% of 28) psi then TPMS trigger a warning

26% of 28 = (28*26)/100 = 7.28

28-(26% of 28) = 28-7.28 = 20.72

When the tire pressure is 20.72 psi TPMS will trigger a warning

(b)

X : Tire pressure of a certain car

X is normally distributed with a standard deviation equal to 3 psi

If the car’s average tire pressure is on target,  probability that the TPMS will trigger a warning = P(X 20.72)

Z-score of 20.72 = (20.72-mean)/standard deviation = (20.72-28)/3 = -2.43

From standard normal tables, P(Z -2.43) =0.0075

P(X 20.72) = P(Z -2.43) =0.0075

If the car’s average tire pressure is on target,  probability that the TPMS will trigger a warning = 0.0075

Answer : Probability = 0.0075

(c)

The manufacturer’s recommended correct inflation range is 26 psi to 30 psi i.e 26 X 30

If a tire on the car is inspected at random, probability that the tire’s inflation is within the recommended range = P(26 X 30)

P(26 X 30) = P(X30) - P(X26)

Z-score for 30 = (30-mean)/standard deviation = (30-28)/3 = 0.67

Z-score for 26 = (30-mean)/standard deviation = (26-28)/3 = -0.67

From standard normal tables,

P(Z0.67 ) =0.7486 P(Z-0.67 ) = 0.2514

P(X30) = P(Z0.67 ) = 0.7486

P(X26) = P(Z-0.67 ) = 0.2514

P(26 X 30) = P(X30) - P(X26) = 0.7486-0.2514=0.4972

If a tire on the car is inspected at random, probability that the tire’s inflation is within the recommended range = 0.4972