Question

we choose a sample of size 100 from a population of monthly cable bills having standard deviation $20 If we assume the population mean bill is $65, what is the probability mean of our sample is greater than $70. .

Answer 1

0.62% probability that the mean of our sample is greater than $70.

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 = (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 = (\sigma)/(√(n))`.

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

In this question, we have that:

`\mu = 65, \sigma = 20, n = 100, s = (20)/(√(100)) = 2`

What is the probability mean of our sample is greater than $70.

This is 1 subtracted by the pvalue of Z when X = 70. So

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

By the Central Limit Theorem

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

`Z = (70 - 65)/(2)`

`Z = 2.5`

`Z = 2.5` has a pvalue of 0.9938

1 - 0.9938 = 0.0062

0.62% probability that the mean of our sample is greater than $70.