Question

A bottle maker believes that 23% of his bottles are defective. If the bottle maker is accurate, what is the probability that the proportion of defective bottles in a sample of 602 bottles would differ from the population proportion by less than 4%

Answer 1

Answer:

0.9802 = 98.02% probability that the proportion of defective bottles in a sample of 602 bottles would differ from the population proportion by less than 4%

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 $s = \sqrt{\frac{p(1-p)}{n}}$

A bottle maker believes that 23% of his bottles are defective.

This means that $p = 0.23$

Sample of 602 bottles

This means that $n = 602$

Mean and standard deviation:

$\mu = p = 0.23$

$s = \sqrt{\frac{p(1-p)}{n}} = \sqrt{\frac{0.23*0.77}{602}} = 0.0172$

What is the probability that the proportion of defective bottles in a sample of 602 bottles would differ from the population proportion by less than 4%?

p-value of Z when X = 0.23 + 0.04 = 0.27 subtracted by the p-value of Z when X = 0.23 - 0.04 = 0.19.

X = 0.27

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

By the Central Limit Theorem

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

$Z = \frac{0.27 - 0.23}{0.0172}$

$Z = 2.33$

$Z = 2.33$ has a p-value of 0.9901

X = 0.19

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

$Z = \frac{0.19 - 0.23}{0.0172}$

$Z = -2.33$

$Z = -2.33$ has a p-value of 0.0099

0.9901 - 0.0099 = 0.9802

0.9802 = 98.02% probability that the proportion of defective bottles in a sample of 602 bottles would differ from the population proportion by less than 4%