The number of points TEDU basketball team scores against Stanford basketball team is normally distributed. Construct a 95\% two-sided confidence interval for the mean of the number of points scored by TEDU.

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 ${\displaystyle V\left(X\right)=\frac{\sum {(X-\bar{X})}^2}{n-1}}$ to find the variance.
${\displaystyle V\left(X\right)=\frac{322}{9}}$
${\displaystyle =35.78}$
Use the equation ${\displaystyle S.D=\sqrt{V\left(X\right)}}$ to find the standard deviation.
${\displaystyle S.D=\sqrt{35.78}}$
${\displaystyle =5.98}$
The t value corresponding to ${\displaystyle 95\mathrm{\%}}$ confidence interval is 2.26.
Use the equation ${\displaystyle C.I=\bar{X}\pm t\frac{{\sigma }^2}{n}}$ to construct the confidence interval.
${\displaystyle C.I=65\pm 2.26\frac{{5.98}^2}{10}}$
${\displaystyle =65\pm 4.28}$
Therefore, the upper and lower limits are ${\displaystyle 65+4.28}$ and ${\displaystyle 65-4.28}$
Part (b)
Given that, the variance of the distribution is 25.
${\displaystyle V\left(X\right)=25}$
${\displaystyle S.D=\sqrt{25}}$
${\displaystyle =5}$
Use the equation ${\displaystyle C.I=\bar{X}\pm t\frac{{\sigma }^2}{n}}$ to construct the interval.
${\displaystyle C.I=65\pm 2.26\times \frac{5^2}{10}}$
${\displaystyle C.I=65\pm 5.65}$
The upper and lower limits of the confidence interval is ${\displaystyle 65+5.65}$ and ${\displaystyle 65-5.65.}$
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.