Question

The mean salary offered to students who are graduating from Coastal State University this year is , with a standard deviation of . A random sample of Coastal State students graduating this year has been selected. What is the probability that the mean salary offer for these students is or less

Answer 1

Answer:

The probability that the mean salary offer is of X or less is the p-value of $Z = \frac{X - \mu}{\frac{\sigma}{\sqrt{n}}}$, in which $\mu$ is the mean salary for the population, $\sigma$ is the standard deviation for the population and n is the size of the sample.

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 normal distributions can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$, the z-score 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 p-value, we get the probability that the value of the measure is greater than X.

Central Limit Theorem

The Central Limit Theorem establishes 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.

Z-score with the Central Limit Theorem:

Z-is given by:

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

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

What is the probability that the mean salary offer for these students is X or less?

The probability that the mean salary offer is of X or less is the p-value of $Z = \frac{X - \mu}{\frac{\sigma}{\sqrt{n}}}$, in which $\mu$ is the mean salary for the population, $\sigma$ is the standard deviation for the population and n is the size of the sample.