A researcher wishes to estimate within 2 points the average systolic blood pressure of female college students. If she wishes to be 90% confidence, how large a sample should she select if the population standard deviation of female systolic blood pressure is 4.8? Please show steps, please and thank you.

Answer:n=16Step-by-step explanation:A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval". 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". [tex]\bar X[/tex] represent the sample mean for the sample  [tex]\mu[/tex] population mean (variable of interest) [tex]\sigma[/tex] represent the population standard deviation n represent the sample size  (variable of interest)The margin of error is given by this formula: [tex] ME=z_{\alpha/2}\frac{\sigma}{\sqrt{n}}[/tex]    (1) And on this case we have that ME =+2 and we are interested in order to find the value of n, if we solve n from equation (1) we got: [tex]n=(\frac{z_{\alpha/2} \sigma}{ME})^2[/tex]   (2) The critical value for 90% of confidence interval now can be founded using the normal distribution. And in excel we can use this formla to find it:"=-NORM.INV(0.05,0,1)", and we got [tex]z_{\alpha/2}=1.640[/tex], replacing into formula (2) we got: [tex]n=(\frac{1.64(4.8}{2})^2 =15.492 \approx 16[/tex] So the answer for this case would be n=16 rounded up to the nearest integer