Answer to Question #220493 in Statistics and Probability for Vickie

Question

the distribution of wages in a manufacturing company is approximately normal company has 5,000 workers the main wage per month is 800standard deviation is 200
The number of workers getting salary between 700 and 800

the percentage of workers getting salary above 900
The percentage of workers getting a salary before below 400

Answer 1

$N=5000 \\ \mu=800 \\ \sigma=200 \\ P(700<X<800) = P(X<800) -P(X<700) \\ = P(Z< \frac{800-800}{200}) - P(Z< \frac{700-800}{200}) \\ = P(Z<0) -P(Z< -0.5) \\ = 0.5 -0.3085 \\ = 0.1915$

The number of workers getting salary between 700 and 800 $= 5000 \times 0.1915 = 957.5 ≈958$

$P(X>900) = 1 -P(X<900) \\ = 1 -P(Z< \frac{900-800}{200}) \\ = 1 -P(Z< 0.5) \\ = 1 -0.6914 \\ = 0.3086 \\ = 30.86 \; \%$

The percentage of workers getting salary above 900 is 30.86 %

$P(X<400) = P(Z< \frac{400-800}{200}) \\ = P(Z< -2) \\ = 0.0227 \\ = 2.27 \; \%$

The percentage of workers getting a salary before below 400 is 2.27 %