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Answer:
Step-by-step explanation:
Hello!
The monitoring system warn the driver when the tire pressure of the vehicle is 28% below target pressure.
Be X: target tire pressure of a certain car (pounds per square inch)
a)
X= 28 psi
If the monitoring system will warn the driver when the pressure is 28% below the target pressure: X-0.28X
First step, you have to calculate the 28% of 28psi
28*0.28= 7.84
Second step, is to subtract the calculated 28% to the target pressure:
28 - 7.84= 20.16
The TPMS will trigger a warning at 20.16 psi.
b)
If X~N(μ;σ²)
μ= 28psi (since the average is on target, then the target pressure for the car will be the average value of the distribution)
σ= 3psi
P(X≤20.16)
The standard normal distribution is tabulated. Any value of any random variable X with normal distribution can be "converted" by subtracting the variable from its mean and dividing it by its standard deviation.
So to calculate each of the asked probabilities, you have to first, "transform" the value of the variable to a value of the standard normal distribution Z, then you use the standard normal tables to reach the corresponding probability.
Z= (X-μ)/σ= (20.16-28)/3= -2.61
Now you have to look for the corresponding value of probability using the Z-table. Since the value is negative you have to the use the left entry of the Z-table, in the first column you'll find the integer and first decimal of the value -2.6- and in the first row you'll find the second decimal value -.-1
The value of probability that corresponds to -2.61 is:
P(Z≤-2.61)= 0.005
c)
You have to calculate the probability of inspecting a tire at random and it being inflated within recommended range, symbolically this is:
P(30≤X≤26)= P(X≤30)-P(X≤26)
Calculate 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
The probability of the tire being inflated within recommended inflation range is 0.498.
I hope this helps!
Using the normal distribution, it is found that:
a) The warnings triggers at 20.16 psi.
b) 0.0045 = 0.45% probability that the TPMS will trigger a warning.
c) 0.4972 = 49.72% probability that the tire’s inflation is within the recommended range.
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Normal Probability Distribution
We use the z-score formula, given by:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
- [tex]\mu[/tex] is the mean.
- [tex]\sigma[/tex] is the standard deviation.
- It measures how many standard deviations the measure is from the mean.
- Each z-score has a p-value, which is the probability that the value of the measure is smaller than X, that is, the percentile of X.
- Subtracting 1 by the p-value, we get the probability that the value of the measure is greater than X.
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Item a:
- 28% below 28 psi is 100 - 28 = 72% of 28 psi, that is:
[tex]0.72(28) = 20.16[/tex]
The warnings triggers at 20.16 psi.
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Item b:
- Mean of 28 psi, thus, [tex]\mu = 28[/tex]
- Standard deviation of 3 psi, which means that [tex]\sigma = 3[/tex]
- The probability of a warning is the p-value of Z when X = 20.16.
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{20.16 - 28}{3}[/tex]
[tex]Z = -2.61[/tex]
[tex]Z = -2.61[/tex] has a p-value of 0.0045.
0.0045 = 0.45% probability that the TPMS will trigger a warning.
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Item c:
Between 26 and 30 psi, which is the p-value of Z when X = 30 subtracted by the p-value of Z when X = 26. Thus:
X = 30
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{30 - 28}{3}[/tex]
[tex]Z = 0.67[/tex]
[tex]Z = 0.67[/tex] has a p-value of 0.7486.
X = 26
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{26 - 28}{3}[/tex]
[tex]Z = -0.67[/tex]
[tex]Z = -0.67[/tex] has a p-value of 0.2514.
0.7486 - 0.2514 = 0.4972.
0.4972 = 49.72% probability that the tire’s inflation is within the recommended range.
A similar problem is given at https://brainly.com/question/16866369
