Question

The average number of miles (in thousands) that a car's tire will function before needing replacement is 70 and the standard deviation is 12. Suppose that 50 randomly selected tires are tested. Round all answers to 4 decimal places where possible and assume a normal distribution.
a. What is the distribution of X ? X ~ N( , )
b. What is the distribution of ¯ x ? ¯ x ~ N( , )
c. f a randomly selected individual tire is tested, find the probability that the number of miles (in thousands) before it will need replacement is between 72.1 and 73.8.
For the 50 tires tested, find the probability that the average miles (in thousands) before need of replacement is between 72.1 and 73.8.
d. Is the assumption that the distribution is normal necessary? No Yes

Answer 1

Answer:

a) X ~ N(70,12)

b) ¯ x ~ N(70,1.6971)

c) For a single tire, 0.0541 = 5.41% probability that the number of miles (in thousands) before it will need replacement is between 72.1 and 73.8. For the 50 tires tested, 0.0950 = 9.50% probability that the average miles (in thousands) before need of replacement is between 72.1 and 73.8.

d) No

Step-by-step explanation:

To solve this question, we need to understand the normal probability distribution and the central limit theorem.

Normal probability distribution

When the distribution is normal, we use the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$, the zscore of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

Central Limit Theorem

The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean $\mu$ and standard deviation $\sigma$, the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean $\mu$ and standard deviation $s = \frac{\sigma}{\sqrt{n}}$.

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

a. What is the distribution of X ?

Single tire, so normal with mean $\mu = 70$, standard deviation $\sigma = 12$

So X ~ N(70,12).

b. What is the distribution of ¯ x ?

Sample of 50.

By the Cental Limit Theorem, the mean stays the same.

The standard deviation will be $s = \frac{12}{\sqrt{50}} = 1.6971$

So

¯ x ~ N(70,1.6971)

c. f a randomly selected individual tire is tested, find the probability that the number of miles (in thousands) before it will need replacement is between 72.1 and 73.8.

This is the pvalue of Z when X = 73.8 subtracted by the pvalue of Z when X = 72.1. So

X = 73.8

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{73.8 - 70}{12}$

$Z = 0.32$

$Z = 0.32$ has a pvalue of 0.6255

X = 72.1

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{72.1 - 70}{12}$

$Z = 0.18$

$Z = 0.18$ has a pvalue of 0.5714

0.6255 - 0.5714 = 0.0541

For a single tire, 0.0541 = 5.41% probability that the number of miles (in thousands) before it will need replacement is between 72.1 and 73.8.

For the 50 tires tested, find the probability that the average miles (in thousands) before need of replacement is between 72.1 and 73.8

50 tires, so by the Central Limit Theorem, we use s in the z-score formula.

X = 73.8

$Z = \frac{X - \mu}{s}$

$Z = \frac{73.8 - 70}{1.6971}$

$Z = 2.24$

$Z = 2.24$ has a pvalue of 0.9875

X = 72.1

$Z = \frac{X - \mu}{s}$

$Z = \frac{72.1 - 70}{1.6971}$

$Z = 1.24$

$Z = 1.24$ has a pvalue of 0.8925

0.9875 - 0.8925 = 0.0950

For the 50 tires tested, 0.0950 = 9.50% probability that the average miles (in thousands) before need of replacement is between 72.1 and 73.8.

d. Is the assumption that the distribution is normal necessary?

Sample size of 30 or larger(in this case, 50), so the assumption is not necessary.