Question

proportion outside the interval: $31,000 to $67,000
the average starting salary for recent college graduates is $mu = $49,000$ and $sigma = $9,000$.
what proportion of working recent college graduates will earn a salary outside the interval of $31,000 to $67,000? what rule should be applied to solve this problem?
this question will ask you to do complete all steps to answers this question.
follow these steps to solve the problem:

1. select the correct rule for an outside proportion
2. compute both z - scores and shade the tails of the distribution
3. look up the proportions for both z - scores using the z - table
4. subtract the proportions for the two z - scores, subtract that proportion from 1.00, and report that final proportion

$z=\frac{x - mu}{sigma}$
now, select the correct final proportion from the answer options below:
rule 2: $1 - p(z)=.9550$ (shade to the right of 1 z - score)
rule 3: $1 - p(z_2)-p(z_1) =.9544$ (shade with the 2 z - scores)
rule 4: $1 - p(z_2)-p(z_1) =.0456$ (shade outside the 2 z - scores)
rule 1: $p(z)=.0250$ (shade to the left of 1 z - score)

Explanation:

Step1: Calculate the z - scores

For $X_1 = 31000$, $z_1=\frac{X_1-\mu}{\sigma}=\frac{31000 - 49000}{9000}=\frac{- 18000}{9000}=-2$.
For $X_2 = 67000$, $z_2=\frac{X_2-\mu}{\sigma}=\frac{67000 - 49000}{9000}=\frac{18000}{9000}=2$.

Step2: Look up the proportions in the z - table

The proportion corresponding to $z = - 2$ is $p(z_1)=0.0228$, and the proportion corresponding to $z = 2$ is $p(z_2)=0.9772$.

Step3: Calculate the proportion within the interval

The proportion within the interval $[31000,67000]$ is $p(z_2)-p(z_1)=0.9772 - 0.0228 = 0.9544$.

Step4: Calculate the proportion outside the interval

The proportion outside the interval is $1-(p(z_2)-p(z_1))=1 - 0.9544=0.0456$.

Answer:

Rule 4: $1 - [p(z_2)-p(z_1)] = 0.0456$ (Shade outside the 2 z - scores)