Question

Find the indicated probability. The weekly salaries of teachers in one state are normally distributed with a mean of $490 and a standard deviation of $45. What is the probability that a randomly selected teacher earns more than $525 a week

Answer 1

Given Information:

Mean weekly salary = μ = $490

Standard deviation of weekly salary = σ = $45

Required Information:

P(X > $525) = ?

Answer:

P(X > $525) = 21.77%

Step-by-step explanation:

We want to find out the probability that a randomly selected teacher earns more than $525 a week.

$P(X > 525) = 1 - P(X < 525)\\\\P(X > 525) = 1 - P(Z < \frac{x - \mu}{\sigma} )\\\\P(X > 525) = 1 - P(Z < \frac{525 - 490}{45} )\\\\P(X > 525) = 1 - P(Z < \frac{35}{45} )\\\\P(X > 525) = 1 - P(Z < 0.78)$

The z-score corresponding to 0.78 from the z-table is 0.7823

$P(X > 525) = 1 - 0.7823\\\\P(X > 525) = 0.2177\\\\P(X > 525) = 21.77 \%$

Therefore, there is 21.77% probability that a randomly selected teacher earns more than $525 a week.

How to use z-table?

Step 1:

In the z-table, find the two-digit number on the left side corresponding to your z-score. (e.g 0.7, 2.2, 1.5 etc.)

Step 2:

Then look up at the top of z-table to find the remaining decimal point in the range of 0.00 to 0.09. (e.g. if you are looking for 0.78 then go for 0.08 column)

Step 3:

Finally, find the corresponding probability from the z-table at the intersection of step 1 and step 2.