91Ó°ÊÓ

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:

$U=\underset{i-1}{\overset{8}{∑}}{X}_i+Y_1+Y_2+Z$

And it is given in the question that:

$$\begin{gathered} {\mathrm{μ}}_{x_i}=2 \\ {\mathrm{σ}}_{x_i}=0.5 \\ {\mathrm{μ}}_{y_i}=4 \\ {\mathrm{σ}}_{y_i}=1 \\ {\mathrm{μ}}_{z_i}=3 \\ {\mathrm{σ}}_{z_i}=0.2 \end{gathered}$$

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:

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

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:

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

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

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

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

From part (a), we know that,

$$\begin{gathered} \mathrm{μ}=27 \\ \mathrm{σ}=2.01 \\ x=30 \end{gathered}$$

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

$$\begin{gathered} z=\frac{x-\mathrm{μ}}{\mathrm{σ}} \\ =\frac{30-27}{2.01} \\ =1.49 \end{gathered}$$

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$0.0681$.

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

Now, use the complement rule:

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

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

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

$$\begin{gathered} \mathrm{μ}=27 \\ \mathrm{σ}=2.01 \\ \stackrel{-}{x}=30 \\ n=5 \end{gathered}$$

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 $\mathrm{μ}$and standard deviation $\frac{\mathrm{σ}}{\sqrt{n}}$

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

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

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

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

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