Chapter 10, Problem AP3.34CPT

(a)

To determine

To find out what are the mean and standard deviation of the weight of a box of candy.

Answer to Problem AP3.34CPT

The mean is $27$ ounces.

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

Explanation of Solution

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 $X_i(i=1,2,\mathrm{....},8)$ represents the weight of a truffle and $Y_i(i=1,2)$ represents the weight of the handmade caramel nougat and $Z$ represents the weight of the empty box.

Now, the total weight will be:

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

And it is given in the question that:

  $\begin{array}{l}{\mu }_{X_i}=2\\ {\sigma }_{X_i}=0.5\\ {\mu }_{Y_i}=4\\ {\sigma }_{Y_i}=1\\ {\mu }_Z=3\\ {\sigma }_Z=0.2\end{array}$

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:

  $\begin{array}{l}{\mu }_U=8{\mu }_{X_i}+2{\mu }_{Y_i}+{\mu }_Z\\ =8(2)+2(4)+3(1)\\ =16+8+3\\ =27\end{array}$

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{array}{l}{\sigma }^2{}_U=8{\sigma }^2{}_{X_i}+2{\sigma }^2{}_{Y_i}+{\sigma }^2{}_Z\\ $ =8{(0.5)}^2+2{(1)}^2+{0.2}^2$=4.04\end{array}$

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

  $\begin{array}{l}{\sigma }_U=\sqrt{{\sigma }^2{}_U}\\ =\sqrt{4.04}\\ =2.01\end{array}$

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

(b)

To determine

To calculate what is the probability that a randomly selected box of candy will weigh more than $30$ ounces.

Answer to Problem AP3.34CPT

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

Explanation of Solution

From part (a), we know that,

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

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

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

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

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

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

(c)

To determine

To calculate what is the probability that at least one of them will weigh more than $30$ ounces.

Answer to Problem AP3.34CPT

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

Explanation of Solution

It is given that:

  $\begin{array}{l}n=5\\ p=P(X>30)\\ =0.0681\end{array}$ (From part (c))

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

Thus, now evaluate probability at $K=0$ :

  $\begin{array}{l}$ P(Y=0)=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$=0.7028\end{array}$

Now, use the complement rule:

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

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

(d)

To determine

To calculate what is the probability that the mean weight of five boxes will weigh more than $30$ ounces.

Answer to Problem AP3.34CPT

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

Explanation of Solution

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

  $\begin{array}{l}\mu =27\\ \sigma =2.01\\ \overline{x}=30\\ n=5\end{array}$

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

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

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

  $\begin{array}{l}z=\frac{\overline{x}-{\mu }_{\overline{x}}}{{\sigma }_{\overline{x}}}\\ =\frac{\overline{x}-{\mu }_{\overline{x}}}{\frac{\sigma }{\sqrt{n}}}\\ =\frac{30-27}{\frac{2.01}{\sqrt{5}}}\\ =3.34\end{array}$

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

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

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