Your firm has a contract to make staff uniforms for a fast-food retailer. The heights of the staff are normally distributed with a mean of 70 inches and a standard deviation of 3 inches. a) What percentage of uniforms will have to fit staff shorter than 67 inches, i. e., P(X<67)? b) What percentage will have to be suitable for staff taller than 76 inches, i. e., P(X>76)?

a) 16%

b) 2.5%

Step-by-step explanation:

a)

The mean is 70 with standard deviation(SD) of 3 and you are asked to find out the percentage of staff that have <67(70-3 inch= mean - 1 SD) inch size, which means 1 SD below the mean (<-1 SD). Using 68-95-99.7 rule, you can know that 68% of the population is inside 1 SD range from the mean ( -1 SD to + 1 SD).

To put it on another perspective, there are 32% of the population that have < -1 SD and > +1 SD value. Assuming the distribution is symmetrical, then the value of < - 1 SD alone is 32%/2= 16%

b)

The question asks how many populations have size >76 inches, or mean + 2 SD (70+3*2 inch).

You can also solve this using 68-95-99.7 rule, but you take 95% value as the question asking for 2 SD instead.  Since 95% of population is inside 2 SD range from the mean ( -2 SD to + 2 SD), so there are 5% of population that have < -2 SD and > +2 SD value. Assuming the distribution is symmetrical, then the value of > +2 SD alone is 5%/2= 2.5%

1. x> -5

2. x< 8

step-by-step explanation:

1. 2x> -10

divide by 2 from both sides of equation.

2x/2> -10/2

simplify, to find the answer.

-10/2=-5

x> -5 is the correct answer.

2. 3x< 24

divide by 3 from both sides of equation.

3x/3< 24/3

simplify, to find the answer.

24/3=8

x< 8 is the correct answer.

i hope this you, and have a wonderful day!