Question

The average telephone bill in a locality is $70, with a standard deviation of $40. In a sample of 50 randomly selected phone connections, what is the probability that the sample average will exceed $75?

Answer 1

Using the normal distribution, there is a 0.1894 = 18.94% probability that the sample average will exceed $75.

Normal Probability Distribution

The z-score of a measure X of a normally distributed variable with mean $\mu$ and standard deviation $\sigma$ is given by:

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

• The z-score measures how many standard deviations the measure is above or below the mean.

• Looking at the z-score table, the p-value associated with this z-score is found, which is the percentile of X.

• By the Central Limit Theorem, the sampling distribution of sample means of size n has standard deviation $s = \frac{\sigma}{\sqrt{n}}$.

The parameters for this problem are given as follows:

$\mu = 70, \sigma = 40, n = 50, s = \frac{40}{\sqrt{50}} = 5.66$

The probability that the sample average will exceed $75 is one subtracted by the p-value of Z when X = 75, hence:

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

By the Central Limit Theorem

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

Z = (75 - 70)/5.66

Z = 0.88

Z = 0.88 has a p-value of 0.8106.

1 - 0.8106 = 0.1894.

0.1894 = 18.94% probability that the sample average will exceed $75.

More can be learned about the normal distribution at brainly.com/question/28096232

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