# In a distribution of fingernail length where n = 196, population standard deviation = 7 and x-bar...

## Question:

In a distribution of fingernail length where n = 196, population standard deviation = 7 and {eq}\bar{x} {/eq}= 0.75, compute a 92% Interval Estimate and determine if a nail is 1.87 long, will it be within the interval?

## Confidence Interval for a Mean:

The confidence interval is a type of interval estimation that gives a range of all values likely to be the true population mean at a given level of confidence. The interval is determined by the size of the margin of error.

Given that;

{eq}n=196\\\sigma=7\\\bar x=0.75 {/eq}

Use the equation below to construct a 92% confidence interval:

{eq}\displaystyle \left(\bar X\pm Z_{\frac{\alpha}{2}}\frac{\sigma}{\sqrt{n}}\right) {/eq}

Find the critical value z that correspond to a 95% level of confidence:

{eq}\displaystyle \frac{\alpha}{2}=\frac{1-0.92}{2}=0.04\\Z_{0.04}=\pm 1.75 {/eq}

Plug in the values into the formula and solve for the upper and lower bounds of a 92% confidence interval:

{eq}\displaystyle \left(0.75\pm 1.75\times \frac{7}{\sqrt{196}}\right)\\(0.75\pm 0.875)\\(-0.125, 1.625) {/eq}

1.87 is ABOVE and NOT WITHIN the 92% confidence interval. 