Mathematics, 17.08.2020 01:01 xeno777

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 29 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.) (b) Suppose tire pressure is a normally distributed random variable with a standard deviation equal to 2 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.) (c) The manufacturer’s recommended correct inflation range is 27 psi to 31 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.)

Answers: 2

Show answers

Answers

Answer from: cc14ty

At 21.46 psi, the TPMS trigger a warning for this car.

The probability that the TPMS will trigger a warning is 0.0001.

The probability that the tire’s inflation is within the recommended range is 0.6826.

Given that,

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 29 psi (pounds per square inch).

We have to determine,

At what psi will the TPMS trigger a warning for this car.

What is the probability that the TPMS will trigger a warning.

What is the probability that the tire’s inflation is within the recommended range.

According to the question,

It is stated that TPMS warns the driver when the tire pressure of the vehicle is 26% below the target pressure.

So, the TPMS trigger a warning for this car when;

Pressure = 29 psi - 26% of 29 psi

$Pressure = 29- (0.26 \times 29) = 21.46psi$

At 21.46 psi, the TPMS trigger a warning for this car.

Suppose tire pressure is a normally distributed random variable with a standard deviation equal to 2 psi.

Let X = The pressure at which TPMS will trigger a warning

So, X ~ Normal $\mu = 29, \sigma^2 = 2^2\\$

Now, the probability that the TPMS will trigger a warning is given by = P(X≤ 21.46).

$P(X\leq 21.46) = P (\dfrac{X-\mu}{\sigma}\leq \dfrac{21.46-29}{2}) = P(Z\leq -3.77) = 1-P(Z$

The above probability is calculated by looking at the value of x = 3.77 in the z table which has an area of 0.9999.

The manufacturer’s recommended correct inflation range is 27 psi to 31 psi.

So, the probability that the tire’s inflation is within the recommended range is given by = P(27 psi < X < 31 psi)

P(27psi

Therefore, , P(27 psi < X < 31 psi) = 0.8413 - 0.1587 = 0.6826.

To know more about Probability click the link given below.

link

Answer from: abbygreen

(a) At 21.46 psi, the TPMS trigger a warning for this car.

(b) The probability that the TPMS will trigger a warning is 0.0001.

Step-by-step explanation:

We are given that 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 29 psi (pounds per square inch).

(a) It is stated that TPMS warns the driver when the tire pressure of the vehicle is 26% below the target pressure.

So, the TPMS trigger a warning for this car when;

Pressure = 29 psi - 26% of 29 psi

= $29-(0.26 \times 29)$ = 21.46 psi

At 21.46 psi, the TPMS trigger a warning for this car.

(b) Suppose tire pressure is a normally distributed random variable with a standard deviation equal to 2 psi.

Let X = The pressure at which TPMS will trigger a warning

So, X ~ Normal( $\mu=29, \sigma^{2} =2^{2}$ )

Now, the probability that the TPMS will trigger a warning is given by = P(X $\leq$ 21.46)

P(X $\leq$ 21.46) = P( $\frac{X-\mu}{\sigma}$ $\leq$ $\frac{21.46-29}{2}$ ) = P(Z $\leq$ -3.77) = 1 - P(Z < 3.77)

= 1 - 0.9999 = 0.0001

The above probability is calculated by looking at the value of x = 3.77 in the z table which has an area of 0.9999.

So, the probability that the tire’s inflation is within the recommended range is given by = P(27 psi < X < 31 psi)

P(27 psi < X < 31 psi) = P(X < 31 psi) - P(X $\leq$ 27 psi)

P(X < 31 psi) = P( $\frac{X-\mu}{\sigma}$ < $\frac{31-29}{2}$ ) = P(Z < 1) = 0.8413

P(X $\leq$ 27 psi) = P( $\frac{X-\mu}{\sigma}$ $\leq$ $\frac{27-29}{2}$ ) = P(Z $\leq$ -1) = 1 - P(Z < 1)

= 1 - 0.8413 = 0.1587

Therefore, P(27 psi < X < 31 psi) = 0.8413 - 0.1587 = 0.6826.

Answer from: Quest

Ibelieve the answer is c i worked it out three times but if my answer is wrong someone correct me

Answer from: Quest

step-by-step explanation:

Another question on Mathematics

Mathematics, 21.06.2019 14:10

What is the perimeter, p, of a rectangle that has a length of x + 8 and a width of y − 1? p = 2x + 2y + 18 p = 2x + 2y + 14 p = x + y − 9 p = x + y + 7

Answers: 3

Answer

Mathematics, 21.06.2019 19:30

Finally, the arena decides to offer advertising space on the jerseys of the arena’s own amateur volley ball team. the arena wants the probability of being shortlisted to be 0.14. what is this as a percentage and a fraction? what is the probability of not being shortlisted? give your answer as a decimal. those shortlisted are entered into a final game of chance. there are six balls in a bag (2 blue balls, 2 green balls and 2 golden balls). to win, a company needs to take out two golden balls. the first ball is not replaced. what is the probability of any company winning advertising space on their volley ball team jerseys?

Answers: 3

Answer

Mathematics, 21.06.2019 20:10

Which ratios are equivalent to 10: 16? check all that apply. 30 to 48 25: 35 8: 32 5 to 15

Answers: 1

Answer

Mathematics, 21.06.2019 23:00

If 3 3/4 lb. of candy costs 20.25 how much would 1lb. of candy cost

Answers: 1

Answer

You know the right answer?

Tire pressure monitoring systems (TPMS) warn the driver when the tire pressure of the vehicle is 26%...