91影视

Assuming that the variation in the ages of the population of female students is lesser than the general population, let $蟽=17.7\text{years}$.

A 95% level of confidence is required that the sample mean lies within one-half year of the true mean.

The sample size n can be determined by using the following formula.

$n={\left[\frac{z_{\frac{伪}{2}}脳蟽}{E}\right]}^2鈥�鈥�鈥�鈥�鈥�鈥�鈥�鈥�鈥�鈥�鈥�鈥�鈥�鈥�鈥�鈥�鈥�鈥�鈥�鈥�鈥�鈥�鈥�鈥�鈥�鈥�...\left(1\right)$

Here, E is the margin of error.

The$z_{\frac{伪}{2}}$ is a z-score that separates an area of $\frac{伪}{2}$in the right tail of the standard normal distribution.

The confidence level of 95% corresponds to $伪=0.05鈥�鈥�鈥�鈥�\text{and}鈥�鈥�鈥�鈥�\frac{伪}{2}=0.025$.

The value$z_{\frac{伪}{2}}$has cumulative area $1-\frac{伪}{2}$to its left .

Mathematically,

$$\begin{gathered} P\left(z<z_{\frac{伪}{2}}\right)=1-\frac{伪}{2} \\ =0.975 \end{gathered}$$

From the standard normal table, the area of 0.975 is observed corresponding intersection of the row value 1.9 and column value 0.06, which implies $z_{\frac{伪}{2}}$is 1.96.

The sample size is calculated by substituting the values of $z_{\frac{伪}{2}}$,$蟽$, and E in equation (1).

$$\begin{gathered} \text{n}={\left[\frac{z_{\frac{伪}{2}}脳蟽}{\text{E}}\right]}^2 \\ ={\left[\frac{1.96脳17.7}{0.5}\right]}^2 \\ =4814.14 \\ =4815鈥�鈥�鈥�鈥�\left(\text{rounded}鈥�鈥�\text{off}\right) \end{gathered}$$

So, with 4815 samples values, you can be 95% confident that your sample mean lies within a one-half year of the true mean.

The sample size required to estimate the Mean Age of Female Statistics Studentsis 4815.

The assumption seems reasonable as the female Statistics students are expected to show less variation than the general population of females.