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. 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 

Answer 1

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: $V\sim N(\mu_V, \sigma_V^2).$

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

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

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

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

$$\mu_V=800+0.877896\sigma_V$$

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

$$\sigma_V\approx273.8541\ g$$

$$\mu_V\approx1040.4138\ g$$