Mathematics, 05.03.2020 15:29 sierram298

To estimate the mean age for a population of 4000 employees, a simple random sample of 40 employees is selected. If the population standard deviation is 8.2 years, computer the standard error of the mean. (Round to one decimal place) What is the probability that the sample mean age of the employees will be within 2 years of the population mean age

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Answer from: sneakersolequeen

The standard error of the mean is 1.3.

87.64% probability that the sample mean age of the employees will be within 2 years of the population mean age

Step-by-step explanation:

To solve this question, we have to understand the normal probability distribution and the central limit theorem.

Normal probability distribution:

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the zscore 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 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 random variable X, with mean $\mu$ and standard deviation $\sigma$ , a large sample size can be approximated to a normal distribution with mean $\mu$ and standard deviation, which is also called standard error $s = \frac{\sigma}{\sqrt{n}}$

In this problem, we have that:

$\sigma = 8.2, n = 40$

Computer the standard error of the mean

$s = \frac{8.2}{\sqrt{40}} = 1.3$

The standard error of the mean is 1.3.

What is the probability that the sample mean age of the employees will be within 2 years of the population mean age

This is the pvalue of Z when $X = \mu + 2$ subtracted by the pvalue of Z when $X = \mu - 2$ . So

$X = \mu + 2$

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

By the Central Limit Theorem

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

$Z = \frac{\mu + 2 - \mu}{1.3}$

Z = 1.54

Z = 1.54 has a pvalue of 0.9382

-----

$X = \mu - 2$

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

$Z = \frac{\mu - 2 - \mu}{1.3}$

Z = -1.54

Z = -1.54 has a pvalue of 0.0618

0.9382 - 0.0618 = 0.8764

87.64% probability that the sample mean age of the employees will be within 2 years of the population mean age

Answer from: Quest

250

step-by-step explanation:

Answer from: Quest

sinn is the total amount of students now and m were the students before, you subtract them both (n-m) and divide that by m. then, since to get a percentage from a fraction you have to multiply by 100, your answer should look like this: n-m/m*100

step-by-step explanation:

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