Question

The distribution of the number of words in text messages between employees at a large company is skewed right with a mean of 8.6 words and a standard deviation of 4.3 words. If a random sample of 39 messages is selected, what is the probability the sample mean is more than 10 words?

0.0210
0.2454
0.3724
0.9790

Answer 1

Using the normal distribution and the central limit theorem, the probability the sample mean is more than 10 words is of 0.0210.

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 $s = \frac{\sigma}{\sqrt{n}}$.

In this problem, the parameters are given as follows:

$\mu = 8.6, \sigma = 4.3, n = 39, s = \frac{4.3}{\sqrt{39}} = 0.6885$

The probability that the sample mean is more than 10 words is one subtracted by the p-value of Z when X = 10, hence:

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

By the Central Limit Theorem:

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

$Z = \frac{10 - 8.6}{0.6885}$

Z = 2.03.

Z = 2.03 has a p-value of 0.979.

1 - 0.979 = 0.021.

The probability the sample mean is more than 10 words is of 0.0210.

More can be learned about the normal distribution and the central limit theorem at https://brainly.com/question/24663213

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