91影视

Given data is as follows

Let n denote the total number of parts in stock that is

The sample size is

The mean is calculated using the following formula,

$\stackrel{炉}{x}=\frac{1}{n}\underset{i=1}{\overset{n}{鈭�}}{f}_i{X}_i$

Therefore,

$$\begin{gathered} \stackrel{炉}{x}=\frac{1}{100}\left[3\left(108\right)+2\left(55\right)+...+19\left(100\right)\right] \\ =\frac{15646}{100} \\ =156.46 \end{gathered}$$

The standard deviation is calculated using the following formula,

$s=\sqrt{\frac{1}{n-1}\underset{i=1}{\overset{n}{鈭�}}{f}_i{\left(X_i-渭\right)}^2}$

Therefore,

$$\begin{gathered} s=\sqrt{\frac{1}{99}\left[3{\left(108-156.45\right)}^2+2{\left(55-156.46\right)}^2+...+19{\left(100-156.46\right)}^2\right]} \\ =\sqrt{\frac{1}{99}\left[4328363\right]} \\ =209.0953 \end{gathered}$$

Consider,

$$\begin{gathered} \frac{n}{N}=\frac{100}{500} \\ =0.2 \end{gathered}$$

Here,$\frac{n}{N}=0.2>0.05$the finite population correction factor should be included in the standard error calculation

The standard error is obtained by using the following formula

$\frac{s}{\sqrt{n}}\sqrt{\frac{N-n}{N}}$

Therefore,

$$\begin{gathered} \frac{s}{\sqrt{n}}\sqrt{\frac{N-n}{N}}=\left(\frac{209.0953}{\sqrt{100}}脳\sqrt{\frac{500-100}{500}}\right) \\ =18.7020 \end{gathered}$$

The 95% confidence intervals for the population mean by using the following formula,

$\stackrel{炉}{x}卤2\frac{s}{\sqrt{n}}\sqrt{\frac{N-n}{N}}$

Therefore,

$$\begin{gathered} \stackrel{炉}{x}卤2\frac{s}{\sqrt{n}}\sqrt{\frac{N-n}{N}}=156.46卤2\left(\frac{209.0953}{\sqrt{100}}脳\sqrt{\frac{500-100}{500}}\right) \\ =156.46卤37.4042 \\ =\left(119.0558,193.8642\right) \end{gathered}$$

Hence, the 95% confidence interval is (119.0558, 193.8642).

The 95% confidence interval is (119.0558, 193.8642). Thus, it is observed that the mean parts inventory value of $300 is not included in the confidence interval. So the mean parts inventory value of $300 is not reasonable.