Question

The Candy Shoppe assembles gift boxes that contain 8 chocolate truffles and two handmade caramel nougats. The truffles have a mean weight of 2 ounces with a standard deviation of 0.5 ounce, and the nougats have a mean weight of 4 ounces with a standard deviation of 1 ounce. The empty boxes weigh 3 ounces with a standard deviation of 0.2 ounce.

(a) Assuming that the weights of the truffles, nougats, and boxes are independent, what are the mean and standard deviation of the weight of a box of candy?

(b) Assuming that the weights of the truffles, nougats, and boxes are approximately Normally distributed, what is the probability that a randomly selected box of candy will weigh more than 30 ounces?

(c) If five gift boxes are randomly selected, what is the probability that at least one of them will weigh more than 30 ounces?

(d) If five gift boxes are randomly selected, what is the probability that the mean weight of the five boxes will be more than 30 ounces?

Short answer

(a). Mean of the weight of the box of a candy =$27$

Standard deviation of the weight of the box of a candy =$2.00998$ounces.

(b). Probability that a candy box selected will weigh more than $30$ounces is $0.06772$.

(c). Probability that at least one of the boxes selected will weigh more than 30 ounces is$0.2957$.

(d). Probability that the mean weight of five boxes is more than 30 ounces is$0.000423.$

Step-by-step solution

Part (a) Step 1. Given information

It is given that

Number of chocolate truffles in a gift box= $8$

Number of handmade caramel nougats = $2$

Mean of truffle = $2$ounces

Standard deviation of a truffle= $0.5$ounces

Mean of nougat =$4$

Standard deviation of nougat =$1$ounces

Mean of empty box=$3$ounces

Standard deviation of an empty box = $0.2$ounces

Part (a) Step 2. simplify

Mean of $8$truffles=$8\times E$(truffle) $=8\times 2=16$ounces

Variance of $8$truffles $8\times$Var (a truffle) role="math" localid="1650472885729" $=8\times 0.5^2=2$ounces

Mean of $2$nougats $=2\times E$(nougat ) $=2\times 4=8$ounces

Variance of $2$nougats $2\times$Var (nougat )$=2\times 1^2=2$ounces

As the truffle, nougat and empty box are independent, so

the mean of a candy box is $16+8+3=27$ounces.

Variance of a candy box

$2+2+0.2^2=4.04$ounces

Standard deviation of a candy box=

$\sqrt{Varianceofacandybox}=\sqrt{4.04}=2.009$

Part (b). Step 1. Given information

The weight of truffles, nougats and empty box are normally distributed.

Part (b) Step 2. simplify

Let X be the weight of a candy box.

The distribution will be approximately normal due to the additive property of normal distributions.

So,$$\begin{gathered} X~N(27,4.04) \\ P(X>30)=P(\frac{x-\mu }{\sigma }>\frac{30-27}{2.00998}) \\ P(z>1.493)=1-P(z\le 1.493) \end{gathered}$$

From z tables,

$$\begin{gathered} P(z<1.49)=0.93189 \\ P(z<1.50)=0.93319 \end{gathered}$$

Using linear interpolation

$P(z<1.493)=0.93189+(0.93319-0.9318)$

Hence,$P(X>30)=1-0.93228=0.06772$

Part (c ) Step 1. Given information

we have to find out that probability that at least one of the $5$boxes selected will weigh more than 30 ounces.

Part (c) Step. 2 Simplify

As each gift box has same probability of weighing more than$30$ounces and are independent ,so it can be modelled by a binomial distribution.

Let Y be the number of gift boxes weighing more than $30$ounces

so,$$\begin{gathered} Y~Bin(5,0.06772) \\ P(Y\ge 1)=1-P(Y=0) \\ SO,P(Y=0)={}^{5}C_0\times 0.{06772}^0\times {(1-0.06772)}^{5-0}0.704258 \end{gathered}$$

Hence the required probability is

$1-0.704258=0.2957$

Part (d) Step 1.  Given information

We have to find that the probability that the mean weight of five boxes is more than $30$ounces.

Part (d) Step 2. Simplify

we know that by central limit theorem

$\overline{X}~N({\mu }_x,\frac{{\sigma }_x^2}{n})$

Here $\overline{X}$= Mean

$\sigma$=Standard deviation

N= $30$

Now ,after putting all the value in the above formula we get,

$$\begin{gathered} P(\overline{X}>30)=P(\frac{\overline{x}-{\sigma }_{\overline{x}}}{{\sigma }_{\overline{x}}}>\frac{30-27}{{\displaystyle \frac{2.00998}{\sqrt{5}}}}) \\ P(z>3.337)=1-P(z<3.337) \end{gathered}$$

From z tables,

$$\begin{gathered} P(z<3.33)=0.99957 \\ P(z<3.34)=0.99958 \end{gathered}$$

Now, by using linear interpolation we get

$P(z<3.337)=0.99957+(0.99958-0.9957)\times 0.7=0.999577$