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

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%

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 = (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 = (\sigma)/(√(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{(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{(p(1-p))/(n)} = \sqrt{\sqrt{(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 = (X - \mu)/(\sigma)`

By the Central Limit Theorem

`Z = (X - \mu)/(s)`

`Z = (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%