Answer to Question #210029 in Statistics and Probability for Tefi

Question

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.

Answer 1

Whole problem:

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

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

$X \sim N(\mu_X, \sigma_X^2) \\ Y\sim N(\mu_Y, \sigma_Y^2) \\ W=X+Y \\ X∼N(\mu_X + \mu_Y, \sigma^2_X+ \sigma^2_Y)$

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

$\mu_W=\mu_X+\mu_{Y_1}+\mu_{Y_2}+...+\mu_{Y_{25}} \\ =150+25(40)=1150 (g)=150+25(40)=1150\; g$

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:

$V\sim N(\mu_V, \sigma_V^2) \\ P(V<800)=P(Z<\dfrac{800-\mu_V}{\sigma_V})=0.19 \\ \dfrac{800-\mu_V}{\sigma_V} \approx -0.877896 \\ P(V>1200)=1-P(Z\leq\dfrac{1200-\mu_V}{\sigma_V})=0.28 \\ \dfrac{1200-\mu_V}{\sigma_V} \approx 0.582842 \\ \mu_V=800+0.877896\sigma_V \\ 1200-800-0.8777896 \sigma_V=0.582842 \sigma_V \\ \sigma_V \approx 273.8541\; g \\ \mu_V \approx 1040.4138 \; g$