Respuesta :
Answer:
The guaranteed mileage should be 31800 miles.
Step-by-step explanation:
Consider the provided information.
It is given that the normally distributed with a mean of 40,000 miles and a standard deviation of 4,000 miles. Also the company desires that no more than 2 percent of the tires will fail to meet the guaranteed mileage.
That means the p < 0.02
Now with the help of the table the respective Z-score value of p < 0.02 is -2.05
From the provided information mean and standard deviation is given as [tex]\mu = 40,000\ \text{and}\ \sigma = 4000[/tex]
Now use the formula: [tex]z=\frac{x-\mu}{\sigma}[/tex]
Substitute the respective values in the above formula.
[tex]-2.05=\frac{x-40000}{4000}[/tex]
[tex]-8200=x-40000[/tex]
[tex]-8200+40000=x[/tex]
[tex]31800=x[/tex]
Hence, the guaranteed mileage should be 31800 miles.
Normal distribution can be converted to standard normal distribution for easy probability calculations.
The guaranteed mileage the company should provide which doesn't get more than 2 percent of the tires failed is 31,800 miles.
How to get the z scores?
If we've got a normal distribution, then we can convert it to standard normal distribution and its values will give us the z score.
If we have
[tex]X \sim N(\mu, \sigma)[/tex]
(X is following normal distribution with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex] )
then it can be converted to standard normal distribution as
[tex]Z = \dfrac{X - \mu}{\sigma}, \\\\Z \sim N(0,1)[/tex]
For this case, let the random variable X represent the mileage of tires produced by the considered company.
Then, by the given data, we have
[tex]X \sim N(40000, 4000)[/tex]
Let the mileage that the company would tell to guarantee that no more than 2 percent of the tires will fail to meet the guaranteed mileage be [tex]X = x[/tex]
Then,
Its z score is calculated as
[tex]z = \dfrac{x-\mu}{\sigma} = \dfrac{x - 40000}{4000}[/tex]
The p value of this z score will show [tex]P(Z \leq z)[/tex]
This probability is telling the probability that the mileage is at least [tex]x \: \rm miles[/tex] and we need it to be 98% at least (so that the probability of mileage being more than x get at max 2%)
Thus,
[tex]P(Z \leq z) \geq 98\%\\\\z \geq -2.05 \text{\:(From z tables)}\\\\\dfrac{x - 40000}{4000} \geq -2.05\\\\x \geq 40000 - 8200 = 31800[/tex]
Thus,
The guaranteed mileage the company should provide which doesn't get more than 2 percent of the tires failed is 31,800 miles.
Learn more about standard normal distribution here:
https://brainly.com/question/10984889
