Question

The population mean annual salary for environmental compliance specialists is about $63,500. A random sample of 35 specialists is drawn from this population. What is the probability that the mean salary of the sample is less than $60,000? Assume σ = $6100

Answer 1

Answer:

0.003

Step-by-step explanation:

By formula we know that:

z (x) = (x - m) / [sd / sqrt (n)]

where x is the value we want to know (60,000), m is the mean (63500), sd is the standard deviation (6100) and n is the sample size (35).

Replacing we have:

z (60000) = (60000 - 63500) / [6100 / sqrt (35)]

z = -3.39

If we look in the normal distribution table (attached), we have that the probability is 0.0003.

Answer 2

Using the normal distribution and the central limit theorem, it is found that there is a 0.0003 = 0.03% probability that the mean salary of the sample is less than $60,000.

In a normal distribution with mean $\mu$ and standard deviation $\sigma$, the z-score of a measure X is given by:

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

• It 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, which is the percentile of X.

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

In this problem:

• Mean of $63,000, thus $\mu = 63000$.
• Standard deviation of $\sigma = 6100$.
• Sample of 35, thus $n = 35, s = \frac{6100}{\sqrt{35}}$.

The probability that the mean salary of the sample is less than $60,000 is the p-value of Z when X = 60000, thus:

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

By the Central Limit Theorem

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

$Z = \frac{60000 - 63500}{\frac{6100}{\sqrt{35}}}$

$Z = -3.39$

$Z = -3.39$ has a p-value of 0.0003.

0.0003 = 0.03% probability that the mean salary of the sample is less than $60,000.

A similar problem is given at https://brainly.com/question/24663213