Part (a)

The table containing the mean, variance is shown below.

$\begin{array}{|ccc|}\hline X& X-\overline{X}& (X-\overline{X}{)}^{2}\\ 74& 9& 81\\ 59& -6& 36\\ 62& -3& 9\\ 61& -4& 16\\ 70& 5& 25\\ 62& -3& 9\\ 66& 1& 1\\ 62& -3& 9\\ 75& 10& 100\\ 59& -6& 36\\ Total& & 322\\ \hline\end{array}$

The mean is 65.

Use the equation $V\left(X\right)=\frac{\sum {(X-\stackrel{\u2015}{X})}^{2}}{n-1}$ to find the variance.

$V\left(X\right)=\frac{322}{9}$

$=35.78$

Use the equation $S.D=\sqrt{V\left(X\right)}$ to find the standard deviation.

$S.D=\sqrt{35.78}$

$=5.98$

The t value corresponding to $95\mathrm{\%}$ confidence interval is 2.26.

Use the equation $C.I=\stackrel{\u2015}{X}\pm \text{}t\frac{{\sigma}^{2}}{n}$ to construct the confidence interval.

$C.I=65\pm 2.26\frac{{5.98}^{2}}{10}$

$=65\pm 4.28$

Therefore, the upper and lower limits are $65+4.28$ and $65-4.28$

Part (b)

Given that, the variance of the distribution is 25.

$V\left(X\right)=25$

$S.D=\sqrt{25}$

$=5$

Use the equation $C.I=\stackrel{\u2015}{X}\pm \text{}t\frac{{\sigma}^{2}}{n}$ to construct the interval.

$C.I=65\pm 2.26\times \frac{{5}^{2}}{10}$

$C.I=65\pm 5.65$

The upper and lower limits of the confidence interval is $65+5.65$ and $65-5.65.$

Part (c)

The standard deviation is defined as the degree of variance of value from mean value.

In part (b) the standard deviation is higher compared to part (a).

Therefore, the deviation is higher. So, the confidence interval got widened.