Let ¯¯¯ x be the mean life expectancy in a sample of 100 non‑Hispanic white males. Determine the interval centered at the population mean μ such that 95 % of sample means ¯¯¯ x will fall in the interval.

Assuming that: "According to the Center for Disease Control and Prevention (CDC), the mean life expectancy in 2015 for non-Hispanic white males was 76.3 years. Assume that the standard deviation was 15 years, as suggested by the Bureau of Economic Research"

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

Assuming the X follows a normal distribution

[tex]X \sim N(\mu, \sigma)[/tex]

The distribution for the sample mean is given by:

[tex]\bar X \sim N(\mu, \frac{\sigma}{\sqrt{n}})[/tex]

[tex]\bar x =76.3[/tex] represent the mean

[tex]\sigma =15[/tex] represent the population standard deviation

n= 100 sample size selected.

The confidence interval is given by this formula:

[tex]\bar X \pm z_{\alpha/2} \frac{\sigma}{\sqrt{n}}[/tex] (1)

And for a 95% of confidence the significance is given by [tex]\alpha=1-0.95=0.05[/tex], and [tex]\frac{\alpha}{2}=0.025[/tex]. Since we know the population standard deviation we can calculate the critical value [tex]z_{0.025}= \pm 1.96[/tex]

[tex]n=100,\bar X=76.3,\sigma=15[/tex]

If we use the formula (1) and we replace the values we got:

[tex]76.3 - 1.96 \frac{15}{\sqrt{100}}=73.36[/tex]

[tex]76.3 + 1.96 \frac{15}{\sqrt{100}}=79.24[/tex]

The 95% confidence interval is given by (73.36;79.24)