standard deviation

Question

Assume that the salaries of elementary school teachers in the United States are normally distributed with a mean of $32,000 and a standard deviation of $3000. If a teacher is selected at random, find the probability that he or she makes more than $36,000.

Answer 1

You will need to use the z-score formula. 

Let  X = 36000
Z = (X - μ ) / σ

 

Step 1: Insert given information and calculate.

Z = (36000 - 32000) / 3000 = 1.33

 

 

simplify the problem to 

 

Z=(36-32)/3   by reducing by a factor of 1,000

 

z=1.33

 

Step 2: look at a standardized normal distribution table to find the probability for P(Z<1.33),

P(Z<1.33) =0.9082 for those that make less than $36,000

 

To find the probability for those that make more than $36000 we have to subtract those that make less than $36000 from the whole sample (1).

 

P ( Z>1.33 )=1−P ( Z<1.33 )=1−0.9082=0.0918

 

 

 

Answer 2

The above explanation is great, however, since you are looking for the probability that x > 36,000, you need to subtract that value from 1. This is because the z table tells you the probability of scoring below a certain z score, but we are looking for the probability that we score above it. 

 

1-.9082=.0917 so the probability of a teacher making above $36,000 is about 9.17%

Answer 3

So, I'm not sure if this is completely necessary, but it might be easier to look at the standard normal distribution.  We use a translation to get from normal to standard normal.  It condenses our probability function now to a normal distribution centered on 0 with SD of 1.  So, we are going to go from P(X>36,000) to the translation P(X>(36,000-mean)/3000), because we often refer to a standard normal chart that displays the probabilities for given values.  I'm sure there is one in the book you're using.  Now, P(X>1.33333).  Which, if you look that up, is 90.82%.  This makes sense since 36,000 is almost two standard deviations out from the mean.