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poohbearbaby299
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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.
AOn what percent of the days last year did the factory produce 20,555 skateboards or fewer?
BOn what percent of the days last year did the factory produce 20,610 skateboards or more?
COn what percent of the days last year did the factory produce 20,445 skateboards or fewer?

Verified Answer :

nobillionaireNobley

Let x be a random variable representing the number of skateboards produced
a.) P(x ≤ 20,555) = P(z ≤ (20,555 - 20,500)/55) = P(z ≤ 1) = 0.84134 = 84.1%

b.) P(x ≥ 20,610) = P(z ≥ (20,610 - 20,500)/55) = P(z ≥ 2) = 1 - P(z < 2) = 1 - 0.97725 = 0.02275 = 2.3%

c.) P(x ≤ 20,445) = P(z ≤ (20,445 - 20,500)/55) = P(z ≤ -1) = 1 - P(z ≤ 1) = 1 - 0.84134 = 0.15866 = 15.9%

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