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pogonyaev
2 years ago
13

Last year, the numbers of skateboards produced per day at a certain factory were normally distributed with a mean of 20,500 skat

eboards 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?
Mathematics
1 answer:
BartSMP [9]2 years ago
5 0
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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a) Failed to reject the null hypothesis (P-value=0.09).

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The z-value for this CI is 1.96.

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\sigma=\sqrt{\frac{p_1(1-p_1)}{n_1} +\frac{p_2(1-p_2)}{n_2}} =\sqrt{\frac{0.1116(1-0.1116)}{2142} +\frac{0.1353(1-0.1353)}{399}} =0.0184

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The upper limit is:

UL=(p_1-p_2)+z*\sigma=(0.1116-0.1353)+1.96*0.0184=-0.0238+0.0361\\\\UL=0.0124

The 95% CI for the difference in proportions is:

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In this case, we can conclude that the difference between the proportions, with 95% confidence, can still be equal or greater than zero, meaning that it is possible passenger car owners violate laws more than truck owners.

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|---|---|---|---|---|---|---|---|---|---|---|---|

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Description A describes that exact box-and-whisker plot.

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