+0

0
52
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+54

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?

Apr 25, 2021

#1
+118504
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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%

Apr 25, 2021