Question

Suppose that past records indicate that the probability that a new car will need a warranty repair in the first 90 days of use is 0.04. If a random sample of 400 new cars is selected. what is the probability that the proportion of new cars needing a warranty repair in the first 90 days will be: a. between 0.05 and 0.06? i.e. P(0.05 SpS 0.06) = (round your answer to 4 decimal places). b. above 0.07? i.e. P(p > 0.07) = (round your answer to 5 decimal places) c. below 0.03?ie. P(p < 0.03) = (round your answer to 4 decimal places)

Answer 1

Answer:

a) P(0.05 < p < 0.06) = 0.1332

b) P(p > 0.07) = 0.0011.

c) P(p < 0.03) = 0.1539

Step-by-step explanation:

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

Normal Probability Distribution:

Problems of normal distributions can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$, the z-score 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 p-value, we get the probability that the value of the measure is greater than X.

Central Limit Theorem

The Central Limit Theorem establishes 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.

For a proportion p in a sample of size n, the sampling distribution of the sample proportion will be approximately normal with mean $\mu = p$ and standard deviation $s = \sqrt{\frac{p(1-p)}{n}}$

Suppose that past records indicate that the probability that a new car will need a warranty repair in the first 90 days of use is 0.04.

This means that $p = 0.04$

Sample of 400.

This means that $n = 400, s = \sqrt{\frac{0.04*0.96}{400}} = 0.0098$

a. between 0.05 and 0.06?

This is the p-value of Z when X = 0.05 subtracted by the p-value of Z when X = 0.05. So

X = 0.06

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

By the Central Limit Theorem

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

$Z = \frac{0.06 - 0.04}{0.0098}$

$Z = 2.04$

$Z = 2.04$ has a p-value of 0.9793.

X = 0.05

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

$Z = \frac{0.05 - 0.04}{0.0098}$

$Z = 1.02$

$Z = 1.02$ has a p-value of 0.8461.

0.9793 - 0.8461 = 0.1332

So

P(0.05 < p < 0.06) = 0.1332

b. above 0.07?

This is 1 subtracted by the p-value of Z when X = 0.07. So

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

$Z = \frac{0.07 - 0.04}{0.0098}$

$Z = 3.06$

$Z = 3.06$ has a p-value of 0.9989.

1 - 0.9989 = 0.0011. So

P(p > 0.07) = 0.0011.

c. below 0.03?

This is the p-value of Z when X = 0.03. So

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

$Z = \frac{0.03 - 0.04}{0.0098}$

$Z = -1.02$

$Z = -1.02$ has a p-value of 0.1539. So

P(p < 0.03) = 0.1539