Western Pygmy-Possum. Refer to Problem $12$

a. Find the percentage of all samples of four pygmy possums that have mean weights within $0.225\mathrm{g}$the population mean weight of $8.5\mathrm{g}$

b. Obtain the probability that the mean weight of four randomly selected pygmy possums will be within $0.225\mathrm{g}$the population mean weight of $8.5\mathrm{g}$

c. Interpret the probability you obtained in part (b) in terms of sampling error.

d. Repeat parts (a) -(c) for samples of size $9$

The weight of adult male pygmy-possums $\left(x\right)$ is assumed to be regularly distributed, with a mean of $\left(\mu \right)8.5\mathrm{g}$ and a standard deviation of $\left(\sigma \right)0.3\mathrm{g}$

The formula used:${\sigma }_x=\frac{\sigma }{\sqrt{n}}$

Here $n=4,{\mu }_n=8.5$and

$$\begin{gathered} {\sigma }_x=\frac{\sigma }{\sqrt{n}} \\ =\frac{0.3}{\sqrt{4}} \\ =\frac{0.3}{2} \\ =0.15 \end{gathered}$$

That is, to find $P(8.275\le \overline{x}\le 8.725)$

The $z-$score $8.275$is,

$$\begin{gathered} z=\frac{8.275-8.5}{0.15} \\ =\frac{-0.225}{0.15} \\ =-1.5 \end{gathered}$$

The $z-$score $8.725$is,

$$\begin{gathered} z=\frac{8.725-8.5}{0.15} \\ =\frac{0.225}{0.15} \\ =1.5 \end{gathered}$$

To find the area between the z-scores, use Table II: Areas under the standard normal curve.

To the left of the entrance, $z-$score $1.5$is $0.0668$

To the left of the entrance, $z-$score $1.5$is $0.9332$

Thus, the area between the $z-$scores is,

The area between $z-$scores $=(Areatotheleftof1.5)-(Areatotheleftof-1.5)$

$$\begin{gathered} =0.9332-0.0668 \\ =0.8664 \end{gathered}$$

Thus, the mean weights of all four pygmy-possum samples are within $0.225\mathrm{g}$ of the population mean weight of $8.5\mathrm{g}$ in $86.64\%$ of the cases.

The probability that the mean weight of four randomly chosen pygmy-possums will be within $0.225\mathrm{g}$ of the population mean weight of $8.5\mathrm{g}$ is $0.8664$ as shown in part (a).

The sampling error made in determining the mean weight by a sample of four possums is likely to be within $0.225\mathrm{g}$

Here $n=4,{\mu }_x=8.5$and

$$\begin{gathered} \backslash \mathrm{beginaligned}{\sigma }_{\overline{x}}=\frac{\sigma }{\sqrt{n}} \\ =\frac{0.3}{\sqrt{9}} \\ =\frac{0.3}{3} \\ =0.1 \end{gathered}$$

That is, to find $P(8.275\le \overline{x}\le 8.725)$

The $z-$score for $8.275$is,

$$\begin{gathered} z=\frac{8.275-8.5}{0.1} \\ =\frac{-0.225}{0.1} \\ =-2.25 \end{gathered}$$

The $z-$score for $8.725$is,

$$\begin{gathered} z=\frac{8.725-8.5}{0.1} \\ =\frac{0.225}{0.1} \\ =2.25 \end{gathered}$$

To find the area between the $z-$scores, use Table II: Areas under the standard normal curve.

To the left of the entrance,$z-$score $2.25$ is $0.0122$

To the left of the entrance,$z-$score $2.25$ is $0.9878$

Thus, the area between the $z-$scores is,

$Areabetweenz-scores=(Areatotheleftof2.25)-(Areatotheleftof-2.25)$

$$\begin{gathered} =0.9878-0.0122 \\ =0.9756 \end{gathered}$$

As a result, the mean weights of all nine pygmy-possum samples are within $0.225\mathrm{g}$of the population mean weight of $8.5\mathrm{g}$

The probability that the mean weight of nine randomly chosen pygmy-possums will be within $0.225\mathrm{g}$ of the population mean weight of $8.5\mathrm{g}$ is $0.9756$ as shown in part (a).

There is a $97.56\%$ chance that the sampling error caused in predicting the mean weight from a sample of four possums is less than $0.225\mathrm{g}$