Chapter 10, Problem 34PT3

(a)

To determine

To compute: Mean and standard deviation of the weight of a box of candy.

Answer to Problem 34PT3

Mean $=27\text{ ounces}$

Standard deviation $=2.00998\text{ ounces}$

Explanation of Solution

Given information:

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

Number of handmade caramel nougats $=2$

Mean of a truffle $=2\text{ ounces}$

Standard deviation of a truffle $=0.5\text{ ounces}$

Mean of nougat $=4\text{ ounces}$

Standard deviation of nougat $\text{1 ounces}$

Mean of an empty box $=3\text{ ounces}$

Standard deviation of an empty box $=0.2\text{ ounces}$

Formula used:

If independent

  $\begin{array}{l}E(X+Y)=E(X)+E(Y)\\ Var(X+Y)=Var(X)+Var(Y)\\ {\displaystyle \sum E(X)=n\times E(X)}\\ {\displaystyle \sum Var(X)=n\times Var(X)}\end{array}$

Calculation:

Mean of 8 truffles

  $=8\times \text{E (truffle) = 8 }\times \text{ 2 = 16 ounces}$

Variance of 8 truffles

  $=8\times \text{ Var (a truffle) = 8 }\times {{\displaystyle 0.5}}^2=2\text{ ounces}$

Mean of 2 nougats

  $=2\times \text{E (nougat) = 2 }\times \text{ 4 = 8 ounces}$

Variance of 2 nougats

  $=2\times \text{ Var (nougat) = 2 }\times {{\displaystyle 1}}^2=2\text{ ounces}$

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

The mean of a candy box is $16+8+3=27\text{ ounces}$

Variance of a candy box $=2+2+{{\displaystyle 0.2}}^2=4.04\text{ ounces}$

Standard deviation of a candy box = $\sqrt{Variance\text{ of a candy box}}=\sqrt{4.04}=2.00998\text{ ounces}$

(b)

To determine

To compute: Probability that a candy box selected will weigh more than 30 ounces.

Answer to Problem 34PT3

Probability that a candy box selected will weigh more than 30 ounces is 0.06772.

Explanation of Solution

Given information:

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

Formula used:

Z score: $z=\frac{x-\mu }{\sigma }$

Calculation:

Let X the weight of a candy box.

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

So, $X~N(27,4.04)$

  $\begin{array}{l}P(X>30)=P\left(\frac{x-\mu }{\sigma }>\frac{30-27}{2.00998}\right)\\ P(z>1.493)=1-P(z\le 1.493)\end{array}$

From z tables,

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

Using linear interpolation

  $P(z<1.493)=0.93189+(0.93319-0.93189)\times 0.3=0.93228$

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

(c)

To determine

To compute: Probability that at least one of the 5 boxes selected will weigh more than 30 ounces.

Answer to Problem 34PT3

Probability that at least one of the 5 boxes selected will weigh more than 30 ounces is 0.2957.

Explanation of Solution

Formula used:

  $\text{ P (Y = y) = }{{\displaystyle C}}_y\times {{\displaystyle p}}^y\times {{\displaystyle (1-p)}}^{5-y}$

Calculation:

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, $Y~Bin(5,0.06772)$

  $\begin{array}{l}P(Y\ge 1)=1-P(Y=0)\\ So,\text{ P (Y = 0) = }{{\displaystyle C}}_0\times {{\displaystyle 0.06772}}^0\times {{\displaystyle (1-0.06772)}}^{5-0}=0.704258\end{array}$

Hence, the required probability is

  $1-0.704258=0.2957$

(d)

To determine

To compute: Probability that the mean weight of five boxes is more than 30 ounces.

Answer to Problem 34PT3

Probability that the mean weight of five boxes is more than 30 ounces is 0.000423.

Explanation of Solution

Concept used:

Central limit theorem

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

Calculation:

  $\begin{array}{l}P\left(\overline{X}>30\right)=P\left(\frac{\overline{x}-{\mu }_{\overline{X}}}{{\sigma }_{\overline{X}}}>\frac{30-27}{\frac{2.00998}{\sqrt{5}}}\right)\\ P(z>3.337)=1-P(z<3.337)\end{array}$

From z tables,

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

Using linear interpolation

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

Hence, the required probability is $1-0.999577=0.000423$