Question

The Candy Shoppe assembles gift boxes that contain $8$chocolate truffles and $2$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 have mean weight $3$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

Part a. The mean is $27$ounces.

The standard deviation of the weight of a box of candy is $2.01$ounces

Part b. The probability that a randomly selected box of candy will weigh more than $30$ounces is $0.0681$

Part c. The probability that at least one of them will weigh more than $30$ounces is $0.2972$

Part d. The probability that the mean weight of five boxes will weigh more than$30$ ounces is$0.0004$

Step-by-step solution

Part a. Step 1. Calculation

It is given that the candy shop assembles gift boxes that contains eight chocolate truffles and two handmade caramel nougats. Thus, now:

Let us define $X_i(i=1,2,...8)$represents the weight of a truffle$Y_i(i=1,2)$and represents the weight of the handmade caramel nougat and represents the weight of the empty box.

Now, the total weight will be:

$U=\sum _{i-1}^8{X}_i+Y_1+Y_2+Z$

And it is given in the question that:

$$\begin{gathered} {\mu }_{x_i}=2 \\ {\sigma }_{x_i}=0.5 \\ {\mu }_{y_i}=4 \\ {\sigma }_{y_i}=1 \\ {\mu }_{z_i}=3 \\ {\sigma }_{z_i}=0.2 \end{gathered}$$

As we know that the mean of the sum of random variables is equal to the sum of their means, then total mean will be calculated as:

$$\begin{gathered} {\mu }_u=8{\mu }_{x_i}+2{\mu }_{y_i}+{\mu }_z \\ =8\left(2\right)+2\left(4\right)+3\left(1\right) \\ =16+8+3 \\ =27 \end{gathered}$$

Thus the mean is $27$ounces.

The variance of the sum of random variables is equal to the sum of their variances, when the random variables are independent. So, the variance is:

$$\begin{gathered} {{\sigma }^2}_u=8{{\sigma }^2}_{x_i}+2{{\sigma }^2}_{y_i}+{{\sigma }^2}_z \\ =8{(0.5)}^2+2{\left(1\right)}^2+0.2^2 \\ =4.04 \end{gathered}$$

Now, the standard deviation will be calculated by square root of variance:

$$\begin{gathered} {\sigma }_u=\sqrt{{{\sigma }^2}_u} \\ =\sqrt{4.04} \\ =2.01 \end{gathered}$$

Thus, the standard deviation of the weight of a box of candy is $2.01$ounces.

Part b. Step 1. Calculation

From part (a), we know that,

$$\begin{gathered} \mu =27 \\ \sigma =2.01 \\ x=30 \end{gathered}$$

Now, to calculate the probability we have to find the z-score:

$$\begin{gathered} z=\frac{x-\mu }{\sigma } \\ =\frac{30-27}{2.01} \\ =1.49 \end{gathered}$$

Now, we will find the corresponding probability using the normal probability table. As,

$$\begin{gathered} P(X>30)=P(Z>1.49) \\ =1-P(Z>1.49) \\ =1-0.9319 \\ =0.0681 \\ =6.81\% \end{gathered}$$

Thus, the probability that a randomly selected box of candy will weigh more than $30$ounces is$0.0681$.

Part c. Step 1. Calculation

It is given that:

$$\begin{gathered} n=5 \\ p=P(X>30) \\ =0.0681 \end{gathered}$$ from part (c)

The number of successes among a fixed number of independent trials with a constant probability of success follows the binomial distribution.

Thus, now evaluate probability at $K=0$

role="math" localid="1663667060763" $$\begin{gathered} P(Y=0){}_{5}C_0\times {(0.0681)}^0\times {(1-0.0681)}^{5-0} \\ =\frac{5!}{0!(5-0)!}\times {(0.0681)}^0\times {(0.9319)}^5 \\ =1\times {(0.0681)}^0\times {(0.9319)}^5 \\ =0.7028 \end{gathered}$$

Now, use the complement rule:

$$\begin{gathered} P(Y\ge 1)=1-P(Y=0) \\ =1-0.7028 \\ =0.2972 \\ =29.72\% \end{gathered}$$

Thus, the probability that at least one of them will weigh more than $30$ounces is$0.2972$

Part d. Step 1. Calculation

Now, it is given and from part (a), we have,

$$\begin{gathered} \mu =27 \\ \sigma =2.01 \\ \stackrel{-}{x}=30 \\ n=5 \end{gathered}$$

Now, since we know the population is normal, the sampling distribution of the sample mean $\stackrel{-}{x}$is also normal.

Thus, the sampling distribution of the sample mean $\stackrel{-}{x}$has mean $\mu$and standard deviation $\frac{\sigma }{\sqrt{n}}$

So, we have to first find the z-score:

$$\begin{gathered} z=\frac{{\displaystyle \stackrel{-}{x}}-\mu }{{\sigma }_x} \\ =\frac{{\displaystyle \stackrel{-}{x}}-\mu }{\frac{\sigma }{\sqrt{n}}} \\ =\frac{30-27}{{\displaystyle \frac{2.01}{\sqrt{5}}}} \\ =3.34 \end{gathered}$$

And now we will find the corresponding probability using the normal probability table. As,

$$\begin{gathered} P(\overline{X}>30)=P(Z>3.34) \\ =1-P(Z<3.34) \\ =1-0.9996 \\ =0.0004 \\ =0.04\% \end{gathered}$$

Thus, the probability that the mean weight of five boxes will weigh more than $30$ounces is$0.0004$.