91Ó°ÊÓ

The mean $\mathrm{μ}$is given below,

$$\begin{gathered} \mathrm{μ}=\frac{\underset{x_i}{∑}}{N} \\ =\frac{2+3+5+5+7+8}{6} \\ =\frac{30}{6} \\ =5 \end{gathered}$$

For each of the possible sample sizes, we construct a table.

If the sample size taken $n=1$,

If the sample size taken $n=2$,

If the sample size taken $n=3$,

If the sample size taken $n=4$,

If the sample size taken $n=5$,

If the sample size taken $n=6$,

We will construct the dot plot for the sampling distribution of the sample mean.

To construct dot plot for the sampling distribution of the sample mean,

We can observe that from the dot plot there is two dot corresponding to $\mathrm{μ}=5$ when nis $1$.

Hence, the probability that sample mean will be equal to population mean$$\begin{gathered} =\frac{2}{6} \\ =\frac{1}{3} \end{gathered}$$

Similarly, the probability that sample mean will be equal to population mean when n is $2$is $$\begin{gathered} =\frac{3}{15} \\ =\frac{1}{5} \end{gathered}$$(As there are data-custom-editor="chemistry" $3$dots corresponding $\mathrm{μ}=5$)

The probability that sample mean will be equal to population mean when n is $3$is $$\begin{gathered} =\frac{4}{20} \\ =\frac{1}{5} \end{gathered}$$(As there are no dots corresponding $\mathrm{μ}=5$)

We can observe that from the dot plot there is one dot corresponding to $\mathrm{μ}=5$ when n is data-custom-editor="chemistry" $4$.

The probability that sample mean will be equal to population mean when n is $4$is $$\begin{gathered} =\frac{3}{15} \\ =\frac{1}{5} \end{gathered}$$(As there are $3$dots corresponding $\mathrm{μ}=5$)

The probability that sample mean will be equal to population mean when n is $5$is $\frac{2}{6}=\frac{1}{3}$.

The probability that sample mean will be equal to population mean for$n=6$is$1$.

Number of dots within $0.5$or less of $\mathrm{μ}=5$is $2$out of $6$ when nis $1$.

Hence, the probability that $\overline{x}$will be within $0.5$or less of $\mathrm{μ}$is $\frac{2}{6}=\frac{1}{3}$.

Number of dots within $0.5$or less of $\mathrm{μ}$is $5$out of $15$ when nis $2$.

Hence, the probability that $\overline{x}$will be within $0.5$or less of $\mathrm{μ}$is $\frac{5}{15}=\frac{1}{3}$.

Number of dots within $0.5$or less of $\mathrm{μ}$is role="math" localid="1652598843137" $8$out of $20$ when n is $3$.

Hence, the probability that $\overline{x}$will be within $0.5$or less of $\mathrm{μ}$is $\frac{8}{20}=\frac{2}{5}$,

Number of dots within $0.5$or less of $\mathrm{μ}$is $9$out of $15$ when nis $4$.

Hence, the probability that $\overline{x}$will be within $0.5$or less of $\mathrm{μ}$is $\frac{9}{15}=\frac{3}{5}$.

Number of dots within $0.5$or less of $\mathrm{μ}$is $4$out of $6$ when n is $5$.

Hence, the probability that $\overline{x}$will be within $0.5$or less of $\mathrm{μ}$is $\frac{4}{6}=\frac{2}{3}$.

Number of dots within data-custom-editor="chemistry" $0.5$or less of $\mathrm{μ}$is $1$out of $1$ when n is $6$.

Hence, the probability that $\overline{x}$will be within $0.5$or less of$\mathrm{μ}$is$\frac{1}{1}=1$.