Mathematics, 07.06.2020 05:00 mcguirefam7071p2mbzz

Of all the companies on the New York Stock Exchange, profits are normally distributed with a mean of $6.54 million and a standard deviation of $10.45 million. In a random sample of 73 companies from the NYSE, what is the probability that the mean profit for the sample was between 0 million and 5.1 million?

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Answer from: webbhlharryteach

11.90% probability that the mean profit for the sample was between 0 million and 5.1 million

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:

$\mu = 6.54, \sigma = 10.45, n = 73, s = \frac{10.45}{\sqrt{73}} = 1.2231$

In a random sample of 73 companies from the NYSE, what is the probability that the mean profit for the sample was between 0 million and 5.1 million?

This is the pvalue of Z when X = 5.1 subtracted by the pvalue of Z when X = 0. So

X = 5.1

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

By the Central Limit Theorem

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

$Z = \frac{5.1 - 6.54}{1.2231}$

Z = -1.18

Z = -1.18 has a pvalue of 0.1190

X = 0

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

$Z = \frac{0 - 6.54}{1.2231}$

Z = -5.35

Z = -5.35 has a pvalue of 0

0.1190 - 0 = 0.1190

11.90% probability that the mean profit for the sample was between 0 million and 5.1 million

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