Question

A telephone exchange operator assumes that 7% of the phone calls are wrong numbers. If the operator is accurate, what is the probability that the proportion of wrong numbers in a sample of 459 phone calls would differ from the population proportion by more than 3%

Answer 1

Answer:

0.0118 = 1.18% probability that the proportion of wrong numbers in a sample of 459 phone calls would differ from the population proportion by more than 3%

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 telephone exchange operator assumes that 7% of the phone calls are wrong numbers.

This means that $p = 0.07$

Sample of 459 phone calls:

This means that $n = 459$

Mean and standard deviation:

$\mu = p = 0.07$

$s = \sqrt{\frac{p(1-p)}{n}} = \sqrt{\sqrt{\frac{0.07*0.93}{459}}} = 0.0119$

What is the probability that the proportion of wrong numbers in a sample of 459 phone calls would differ from the population proportion by more than 3%?

Proportion below 0.07 - 0.03 = 0.04 or above 0.07 + 0.03 = 0.1. Since the normal distribution is symmetric, these probabilities are the same, which means that we find one of them and multiply by 2.

Probability the proportion is below 0.04.

p-value of Z when X = 0.04. So

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

By the Central Limit Theorem

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

$Z = \frac{0.04 - 0.07}{0.0119}$

$Z = -2.52$

$Z = -2.52$ has a p-value of 0.0059

2*0.0059 = 0.0118

0.0118 = 1.18% probability that the proportion of wrong numbers in a sample of 459 phone calls would differ from the population proportion by more than 3%