Answer to Question #210015 in Statistics and Probability for Warrior

Question #210015

A machine fills boxes with chocolate balls. The weights of the empty boxes are normally

distributed with a mean of 150 g and a standard deviation of 10 g. The mean weight of each

chocolate ball is 40 g with a standard deviation of 0.8 g. A full box contains 25 chocolate balls.

a. Calculate the mean weight of a full box of chocolate balls

b. Calculate the standard deviation of the weight of a full box of chocolate balls.

c. Calculate the probability that a full box of chocolate balls weighs less than 1140 g.

d. The company also produces boxes of chocolate squares. It was found that 19% of the boxes

weigh less than 800 g and 28% weigh more than 1200 g. Calculate the mean and standard

deviation of the weight of these boxes


1
Expert's answer
2021-06-24T09:08:46-0400

Let "X"and "Y" be independent random variables that are normally distributed (and therefore also jointly so), then their sum is also normally distributed. i.e., if

"X\\sim N(\\mu_X, \\sigma_X^2), Y\\sim N(\\mu_Y, \\sigma_Y^2),"

"W=X+Y,"

Then "W\\sim N(\\mu_X+\\mu_Y, \\sigma_X^2+\\sigma_Y^2)."

a. Let "X=" the weight of the empty box, "Y_1, Y_2, ...Y_{25}" represent the weights of chocolate balls, "W=X+Y_1+Y_2+...+Y_{25}" represents the weight of of a full box of chocolate balls.

Given "X\\sim N(\\mu_X, \\sigma_X^2), \\mu_X=150\\ g, \\sigma_X=10\\ g"


"Y_i\\sim N(\\mu_{Y_i}, \\sigma_{Y_i}^2), \\mu_{Y_i}=40\\ g, \\sigma_{Y_i}=0.8\\ g, i=1,2,...,25"

Then

"W\\sim N(\\mu_W, \\sigma_W^2))"

The mean weight of a full box of chocolate balls is


"\\mu_W=\\mu_X+\\mu_{Y_1}+\\mu_{Y_2}+...+\\mu_{Y_{25}}"

"=150+25(40)=1150 (g)"

b.

The standard deviation of the weight of a full box of chocolate balls is


"\\sigma_W=\\sqrt{\\sigma_W^2}"

"=\\sqrt{\\sigma_X^2+\\sigma_{Y_1}^2+\\sigma_{Y_2}^2+...++\\sigma_{Y_{25}}^2}"

"=\\sqrt{10^2+25(0.8)^2}=2\\sqrt{29}\\approx10.77033"

c.


"P(W<1140)=P(Z<\\dfrac{1140-1150}{2\\sqrt{29}})"

"\\approx(Z<-0.9284767)\\approx0.17658"

d. Let "V" =the weight of the box of chocolate squares: "V\\sim N(\\mu_V, \\sigma_V^2)."


"P(V<800)=P(Z<\\dfrac{800-\\mu_V}{\\sigma_V})=0.19"

"\\dfrac{800-\\mu_V}{\\sigma_V}\\approx-0.877896"


"P(V>1200)=1-P(Z\\leq\\dfrac{1200-\\mu_V}{\\sigma_V})=0.28"

"\\dfrac{1200-\\mu_V}{\\sigma_V}\\approx0.582842"


"\\mu_V=800+0.877896\\sigma_V"

"1200-800-0.8777896\\sigma_V=0.582842\\sigma_V"

"\\sigma_V\\approx273.8541\\ g"


"\\mu_V\\approx1040.4138\\ g"


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