The sick days of employees every two years in a company are normally distributed with a population standard deviation of 7 days and an unknown population mean. If a random sample of 20 employees is taken and results in a sample mean of 21 days, find a 95% confidence interval for the population mean.

Step-by-step explanation:

Let [tex]$X_1, X_2, X_3,$[/tex] . . . ,Xn be the random sample of n employee's sick days. It is given that the random samples follows the  Normal distribution along with standard deviation of 7 days. Let

[tex]X_i\sim N(\mu ,7)[/tex]

[tex]\bar{X} =\frac{1}{n}\Sigma X_i\sim N(\mu,\frac{7}{\sqrt n})[/tex]

or   [tex]Z=\frac{\bar{X}-\mu}{7 /\sqrt n} \sim N(0,1)[/tex]

So,

[tex]P(-Z_{\alpha /2} \leq Z\leq Z_{\alpha /2} ) = 1- \alpha[/tex]

[tex]P(-Z_{\alpha /2} \leq \frac{\bar{X}- \mu}{7 / \sqrt n}\leq Z_{\alpha /2} ) = 1- \alpha[/tex]

[tex]P(\bar{X}-\frac{7}{\sqrt n} Z_{\alpha / 2} \leq \mu \leq \bar{X}+\frac{7}{\sqrt n} Z_{\alpha / 2} ) = 1- \alpha[/tex]

Therefore, the confidence interval of the population mean for α = 0.05 is

= [tex]P(\bar{X}-\frac{7}{\sqrt n} Z_{\alpha / 2} , \bar{X}+\frac{7}{\sqrt n} Z_{\alpha / 2} )[/tex]

= [tex]P(21-\frac{7}{\sqrt 20} Z_{0.05 / 2} , 21+\frac{7}{\sqrt 20} Z_{0.05 / 2} )[/tex]

= (17.93, 24.07)

Simplify: 6 (4x - 5)

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