Question

Supposed the salaries for recent graduates of one University have a mean of $24,800 with a standard deviation of $1100. Using Chebyshevs theorem what is the minimum percentage of recent graduates who have salaries between $21,500 and $28,100 Round your answer to one decimal place

Answer 1

The Solution:

Given:

$\begin{gathered} \mu=\text{ \ }24800 \\ \\ \sigma=\text{ \ }1100 \\ \\ Lower\text{ limit}=\text{ \ }21500 \\ \\ Upper\text{ limit}=\text{ \ }28100 \end{gathered}$

Required:

To find the minimum percentage of recent graduates who have salaries between $21,500 and $28,100.

Step 1:

We need to compute:

$Pr\left(21500≤X≤28100\right)$

Notice that in this case:

$\begin{gathered} \frac{Upper−μ}{\sigma}=\frac{28100-24800}{1100}=3 \\ \\ \frac{μ−Lower}{\sigma}=\frac{24800-21500}{1100}=3 \end{gathered}$

Since these two values coincide, it means that the event (21500, 28100) is centered around the mean = $24800 and, also, the event 21500≤X≤28100) is equivalent to having the sample mean to be within 3 standard deviations of the mean.

Apply the Chebychev theorem:

The probability that X is within k standard deviations of the mean can be estimated as follows:

[tex]Pr\left(∣X−μ∣In this case, since k=3, the probability that X is within 3 standard deviations from the mean is at least:[tex]\begin{gathered} Pr(X-\mu|Thus, the minimum percentage of the specified recent graduates is 88.9%