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# PLEASE HELP ME!!!!

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1.     Last year, the numbers of skateboards produced per day at a certain factory were normally distributed with a mean of 20,500 skateboards and a standard deviation of 55 skateboards.

(a)   On what percent of the days last year did the factory produce 20,555 skateboards or fewer?

(b)   On what percent of the days last year did the factory produce 20,610 skateboards or more?

(c)   On what percent of the days last year did the factory produce 20,445 skateboards or fewer?

SO.... I keep getting this for (a)

(20555-20500)/55=

55/55=1

but I dont understand how to get the 84%

this goes for (b) =2 and (c)=-1

Please help me!

May 6, 2019

### 1+0 Answers

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$$\text{Let }\Phi(x) \text{ be the CDF of the standard normal distribution.}\\ \text{It's the table you probably have seen somewhere}\\ z = \dfrac{20555-20500}{55}=1\\ \Phi(1) \approx 0.8413 = \%84.13$$

https://www.math.arizona.edu/~rsims/ma464/standardnormaltable.pdf

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May 6, 2019
edited by Rom  May 6, 2019