The weight of a fully harvested apple has a mean of 200 grams and a standard dev

Question

The weight of a fully harvested apple has a mean of 200 grams and a standard deviation of 12 grams and is normally distributed. Suppose

X1,X2,...,X50 are the weights of 50 randomly selected apples. Let Xbar be the mean of the weights and let Xsum be the sum of the weights. Then

a) What is the probability that 197 < Xbar < 203?

b) What is the probability that 10000< Xsum < 10100?

c) what is the 99th percentile of the distribution pof Xsum?

d) What is the standard deviation of the distribution of Xsum?

e) What is the probability that the maximum apple weight (of the 50) is more than 236 grams?

Explanation / Answer

(a)
P(197 < Xbar < 203) = P((197 - 200)/(12/sqrt(50)) < z < (203 - 200)/(12/sqrt(50)) = (-1.7678 < z < 1.7678) = 0.9229

(b)
Xsum = 50*200 = 10000
std. dev. (Xsum) = sqrt(50)*12 = 84.8528

P(10000 < Xbar < 10100) = P((10000 - 10000)/84.8528 < z < (10100 - 10000)/84.8528) = (0 < z < 1.1785) = 0.3807

(c)
For 99th percentile, z-value =2.33
Xsum(99) = Xsum + 2.33*(std.dev.(Xsum)) = 10000 + 2.33*84.8528 = 10197.71

(d)
std. dev. (Xsum) = sqrt(50)*12 = 84.8528

(e)
P(X>236) = P(z > (236 - 200)/(12/sqrt(50))) = P(z > 21.2)= 0

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