Mathematics, 11.06.2020 11:57, justiceemani12

The mean annual salary for intermediate level executives is about $74000 per year with a standard deviation of $2500. A random sample of 36 intermediate level executives is selected. What is the probability that the mean annual salary of the sample is between $71000 and $73500?

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answered: chaeyoung22

11.51% probability that the mean annual salary of the sample is between $71000 and $73500

Step-by-step explanation:

To solve this question, we need 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 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 $s = \frac{\sigma}{\sqrt{n}}$ .

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:

$\mu = 74000, \sigma = 2500, n = 36, s = \frac{2500}{\sqrt{36}} = 416.67$

What is the probability that the mean annual salary of the sample is between $71000 and $73500?

This is the pvalue of Z when X = 73500 subtracted by the pvalue of Z when X = 71000. So

X = 73500

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

By the Central Limit Theorem

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

$Z = \frac{73500 - 74000}{416.67}$

Z = -1.2

Z = -1.2 has a pvalue of 0.1151

X = 71000

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

$Z = \frac{71000 - 74000}{416.67}$

Z = -7.2

Z = -7.2 has a pvalue of 0.

0.1151 - 0 = 0.1151

11.51% probability that the mean annual salary of the sample is between $71000 and $73500

answered: Guest

ca = ec

cpctc makes it so that corresponding parts of the triangle you proved to be equal are equal, the other options aren't part of the part of the triangles you proved to be equal.

In this figure,afc=efc . which statement is true by cpctc?

answered: Guest

d.) yes, because the volume of the cone is greater than the volume of the sphere

step-by-step explanation:

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The mean annual salary for intermediate level executives is about $74000 per year with a standard de...