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 (X76) ?

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 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 value. Assuming the distribution is symmetrical, then the value of > + 2 SD alone is 5%/2 = 2.5%