5#Consider the following random sample of diameter measurements (in inches) of 12 softballs.4.72, 4.74, 4.83, 4.75, 4.73, 4.87, 4.69, 4.7, 4.76, 4.7, 4.79, 4.76Send data to calculatorIf we assume that the diameter measurements are normally distributed, find a 90% confidence interval for the mean diameter of a softball. Give the lower limit and upper limit of the 90% confidence interval. Carry your intermediate computations to at least three decimal places. Round your answers to two decimal places. (If necessary, consult a list of formulas.)Lower limit:??Upper limit:??

[tex]\begin{gathered} \mu=\frac{4.72+4.74+4.83+4.75+4.73+4.87+4.69+4.7+4.76+4.7+4.79+4.76}{12}=\frac{57.04}{12}=4.753 \\ \\ \sigma=\sqrt{\frac{\lparen4.72-4.753^2+\left(4.74-4.753\right)^2+\lparen4.83-4.753)^2+\left(4.75-4.753\right)^2+\left(4.73-4.753\right)^2+\lparen4.87-4.753)^2+\left(4.69-4.753\right)^2+\lparen4.7-4.753)^2+\left(4.76-4.753\right)^2+\left(4.7-4.753\right)^2+\left(4.79-4.753\right)^2+\left(4.76-4.753\right)^2}{12-1}} \\ \\ \sigma=0.054 \end{gathered}[/tex]

The critical limit of 90% confidence interval is 1.645

We can determine the limits as follows:

[tex]\begin{gathered} \text{ Lower limit: }\mu-Z\cdot\frac{\sigma}{\sqrt{n}}=4.753-1.645\cdot\frac{0.054}{\sqrt{12}}=4.727\cong4.73 \\ \\ \text{ Upper limit:: }\mu+Z\cdot\frac{\sigma}{\sqrt{n}}=4.753+1.645\cdot\frac{0.054}{\sqrt{12}}=4.778\cong4.78 \end{gathered}[/tex]