3. Last year, the numbers of skateboards produced per day at a certain factory were normally distributed with a mean of 20,300 skateboards and a standard deviation of 52 skateboards.

(a) On what percent of the days last year did the factory produce 20,404 skateboards or fewer?

(b) On what percent of the days last year did the factory produce 20,248 skateboards or more?

(c) On what percent of the days last year did the factory produce 20,196 skateboards or fewer?

luckystar042003 Apr 25, 2021

#1**+2 **

3. Last year, the numbers of skateboards produced per day at a certain factory were normally distributed with a mean of 20,300 skateboards and a standard deviation of 52 skateboards.

(a) On what percent of the days last year did the factory produce 20,404 skateboards or fewer?

Calculate the z score = ( 20404 - 20300) / 52 = 2

The z score translates to .9772 = about 97.7 %

(b) On what percent of the days last year did the factory produce 20,248 skateboards or more?

Calculate the z score = [ 20248 - 20300 ] / 52 = -1

The z score translates to = .1587

So....the % days that they produced 20,248 skateboards or more = 1 - .1587 = .8413 = about 84.13%

(c) On what percent of the days last year did the factory produce 20,196 skateboards or fewer?

z = [ 20196 -20300 ] / 52 = -2

The z score translates to . 0228 = 2.28%

CPhill Apr 25, 2021