91Ó°ÊÓ

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

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

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

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

Variance of $2$nougats $2×$Var (nougat )$=2×1^2=2$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=

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

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

Let X be the weight of a candy box.

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

So,$$\begin{gathered} X~N(27,4.04) \\ P(X>30)=P(\frac{x-\mathrm{μ}}{\mathrm{σ}}>\frac{30-27}{2.00998}) \\ P(z>1.493)=1-P(z≤1.493) \end{gathered}$$

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

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

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,$$\begin{gathered} Y~Bin(5,0.06772) \\ P(Yâ‰\yen 1)=1-P(Y=0) \\ SO,P(Y=0)={}^{5}C_0×0.{06772}^0×{(1-0.06772)}^{5-0}0.704258 \end{gathered}$$

Hence the required probability is

$1-0.704258=0.2957$

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

we know that by central limit theorem

$\overline{X}~N({\mathrm{μ}}_x,\frac{{\mathrm{σ}}_x^2}{n})$

Here $\overline{X}$= Mean

$\mathrm{σ}$=Standard deviation

N= $30$

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

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

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

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