Question

$16.05$A machine is designed to fill 16-ounce bottles of shampoo. When the machine is working properly, the amount poured into the bottles follows a Normal distribution with mean 16.05 ounces and standard deviation $0.1$ounce. Assume that the machine is working properly. If 4 bottles are randomly selected and the number of ounces in each bottle is measured, then there is about a $95\%$probability that the sample mean will fall in which of the following intervals?

a. $16.05$to $16.15$ounces

b. $16.00$to $16.10$ounces

c. $15.95$to role="math" localid="1654391534650" $16.15$ounces

d. $15.90$to $16.20$ounces

e. $15.85$to $16.25$ounces

Short answer

The correct option is (c) $15.95$to $16.15$ounces

Step-by-step solution

Given Information

Given,

$$\begin{gathered} \beta =165 \\ a-0.1 \\ n-4 \end{gathered}$$

Formula used:

$z=\frac{va}{n_i}$

Explanation for correct option

The sampling distribution of the sample mean $\overline{x}$is also normal because the population distribution is normal.

The Z-score is

$z=\frac{x-{\mu }_{\overline{x}}}{{\sigma }_{\overline{x}}}=\frac{\overline{x}-\mu }{\sigma /\sqrt{n}}=\frac{15.85-16.05}{0.1/\sqrt{4}}=-4.00$

$$\begin{gathered} z=\frac{x-{\mu }_{\overline{x}}}{{\sigma }_{\overline{x}}}=\frac{\overline{x}-\mu }{\sigma /\sqrt{n}}=\frac{15.90-16.05}{0.1/\sqrt{4}}=-3.00 \\ z=\frac{x-{\mu }_{\overline{x}}}{{\sigma }_{\overline{x}}}=\frac{\overline{x}-\mu }{\sigma /\sqrt{n}}=\frac{15.95-16.05}{0.1/\sqrt{4}}=-2.00 \\ z=\frac{x-{\mu }_{\overline{x}}}{{\sigma }_{\overline{x}}}=\frac{\overline{x}-\mu }{\sigma /\sqrt{n}}=\frac{16.00-16.05}{0.1/\sqrt{4}}=-1.00 \\ z=\frac{x-{\mu }_{\overline{x}}}{{\sigma }_{\overline{x}}}=\frac{\overline{x}-\mu }{\sigma /\sqrt{n}}=\frac{16.05-16.05}{0.1/\sqrt{4}}=0.00 \\ z=\frac{x-{\mu }_{\overline{x}}}{{\sigma }_{\overline{x}}}=\frac{\overline{x}-\mu }{\sigma /\sqrt{n}}=\frac{16.10-16.05}{0.1/\sqrt{4}}=1.00 \\ z=\frac{x-{\mu }_{\overline{x}}}{{\sigma }_{\overline{x}}}=\frac{\overline{x}-\mu }{\sigma /\sqrt{n}}=\frac{16.15-16.05}{0.1/\sqrt{4}}=2.00 \\ z=\frac{x-{\mu }_{\overline{x}}}{{\sigma }_{\overline{x}}}=\frac{\overline{x}-\mu }{\sigma /\sqrt{n}}=\frac{16.20-16.05}{0.1/\sqrt{4}}=3.00 \\ z=\frac{x-{\mu }_{\overline{x}}}{{\sigma }_{\overline{x}}}=\frac{\overline{x}-\mu }{\sigma /\sqrt{n}}=\frac{16.25-16.05}{0.1/\sqrt{4}}=4.00 \end{gathered}$$

The normal probability table is used to calculate the associated probability.

$P(Z<-3.00)$is given in the row starting with $-3.0$and in the column starting with .00 of the appendix's standard normal probability table The same can be said for the other probability.

localid="1654392139846" $$\begin{gathered} P(15.95<\overline{X}\ge 16.15)=P(-2.00<Z<2.00) \\ =P(Z<2.00)-P(Z<-2.00) \\ =0.9772-0.0228 \\ =0.9544 \\ =95.44\% \end{gathered}$$

We know that the probability is nearest to $95\%$is $P(15.95<\overline{X}<16.15)$

Hence, the correct option is (c)

Explanation for incorrect option

a)

$$\begin{gathered} P(16.05<\overline{X}\ge 16.15)=P(0.00<Z<2.00) \\ =P(Z<2.00)-P(Z<0.00) \\ =0.9772-0.05000 \\ =0.4772 \\ =47.72\% \end{gathered}$$

b)

$$\begin{gathered} P(16.00<\overline{X}\ge 16.10)=P(-1.00<Z<1.00) \\ =P(Z<1.00)-P(Z<-1.00) \\ =0.8413-0.1587 \\ =0.6826 \\ =68.26\% \end{gathered}$$

d)

$$\begin{gathered} P(15.95<\overline{X}\ge 16.15)=P(-2.00<Z<2.00) \\ =P(Z<2.00)-P(Z<-2.00) \\ =0.9772-0.0228 \\ =0.9544 \\ =95.44\% \end{gathered}$$

e)

$$\begin{gathered} P(15.90<\overline{X}\ge 16.20)=P(-3.00<Z<3.00) \\ =P(Z<3.00)-P(Z<-3.00) \\ =1-0 \\ =1 \\ =100\% \end{gathered}$$