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oee
3 months ago
11

Tire pressure monitoring systems (TPMS) warn the driver when the tire pressure of the vehicle is 28% 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
below

20.16
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



(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.)
Mathematics
1 answer:
Zina [12.3K]3 months ago
5 0

Answer:

Step-by-step explanation:

Hello!

The monitoring system alerts the driver when the vehicle's tire pressure drops to 28% below the set target pressure.

Let X be: the target tire pressure for a particular vehicle (measured in pounds per square inch)

a)

X= 28 psi

When the monitoring device alerts at a pressure of 28% below the designated target: X-0.28X

Initially, calculate 28% of 28 psi.

28*0.28= 7.84

Next, subtract this 28% calculation from the target pressure:

28 - 7.84= 20.16

The TPMS will activate its warning at 20.16 psi.

b)

Assuming X~N(μ;σ²)

μ= 28 psi (as the average indicates accurate targeting, this represents the expected tire pressure)

σ= 3 psi

P(X≤20.16)

The standard normal distribution is available in tables. To convert any random variable X with a normal distribution, one subtracts its mean and divides by the standard deviation.

To compute the desired probabilities, the variable value undergoes transformation to fit the standard normal distribution Z, after which standard normal tables are referenced to find probabilities.

Z= (X-μ)/σ= (20.16-28)/3= -2.61

Now locate the probability corresponding to the Z value using the Z-table. Given the negative result, refer to the left entry; in the first column, find the integer and first decimal for -2.6- and in the first row locate the second decimal for -.-1

The probability link for -2.61 is:

P(Z≤-2.61)= 0.005

c)

The task is to determine the likelihood of encountering a tire at random within the recommended inflation range, expressed as:

P(30≤X≤26)= P(X≤30)-P(X≤26)

Determine both Z values:

Z= (30-28)/3= 0.67

Z= (26-28)/3= -0.67

P(Z≤0.67)-P(Z≤-0.67)= 0.749 - 0.251= 0.498

<pThe calculated probability of a tire being inflated within the suggested range is 0.498.

I hope this information is useful!

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Part 2)

The chance that one gets selected while the other does not =

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These two scenarios are precisely equal, so it suffices to compute one.


Let's analyze the scenario: husband chosen, wife not selected.

Assuming the husband is selected, we need to determine the possible formations of 3 from the 11 total excluding the wife=10 individuals.

This results in:

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P(husband selected, wife not selected)=120/495=0.24


Thus, the overall probability that one is picked while the other is not =

P(husband selected, wife not selected) + P(wife selected, husband not selected) =

0.24+0.24=0.48



Result:


A) 0.09


B) 0.48

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