A random sample of n=400 students is selected from a population of N=4000 students to...

Step 1
The ${\displaystyle (1-\alpha )100\mathrm{\%}}$ confidence interval formula for the population mean when population standard deviation is not known, is defined as follows:
${\displaystyle CI=\bar{x}\pm t_{\frac{\alpha }{2},n-1}\left(\frac{s}{\sqrt{n}}\right).}$
${\displaystyle t_{\frac{\alpha }{2},n-1}}$ is the critical value of the t-distribution with degrees of freedom of ${\displaystyle n-1}$ above which, ${\displaystyle 100\left(\frac{\alpha }{2}\right)\mathrm{\%}\text{or}\frac{\alpha }{2}}$ proportion of the observations lie, and below which, ${\displaystyle 100(1-\alpha +\frac{\alpha }{2}=100(1-\frac{\alpha }{2})\mathrm{\%}\text{or}(1-\frac{\alpha }{2})}$ proportion of the observation lie, ${\displaystyle \bar{x}}$ is the sample mean, s is the sample standard deiation, and n is the sample size.
Step 2
The sample mean is ${\displaystyle \bar{x}=140lb}$, and the sample standard deviation is ${\displaystyle s=15lb(=\surd 225)}$.
The sample size is ${\displaystyle n=400}$. The degrees of freedom is ${\displaystyle 399(=400–1).}$
The confidence level is 0.95. Hence, the level of significance is ${\displaystyle 1–0.95=0.05.}$
Using Excel formula: ${\displaystyle =T.INV.2T(0.05,399)}$, the critical value is, ${\displaystyle t_{\frac{\alpha }{2}}=t_{0.025}\approx 1.96.}$
Thus,
${\displaystyle CI=\bar{x}\pm t_{\frac{\alpha }{2}},n-1\left(\frac{s}{\sqrt{n}}\right)}$
${\displaystyle =140\pm \left(1.96\right)\left(\frac{15}{\sqrt{400}}\right)}$
${\displaystyle =140\pm \left(1.96\right)\left(0.75\right)}$
${\displaystyle =140\pm 1.47}$
${\displaystyle =(138.53,141.47)}$
Thus, the required $95\mathrm{\%}$ confidence interval is (138.53 lb, 141.47 lb).