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%

x^6

step-by-step explanation:

this is correct because:

x-6=0

0*x=0

so

x^6