Assume that females have pulse rates that are normally distributed with a mean of mu equals 73.0 beats per minute and a standard deviation of sigma equals 12.5 beats per minute. Complete parts​ (a) through​ (c) below. a. If 1 adult female is randomly​ selected, find the probability that her pulse rate is between 69 beats per minute and 77 beats per minute. The probability is nothing. ​(Round to four decimal places as​ needed.)

Answer: 0.2510Step-by-step explanation:Given : Mean =[tex]\mu=\text{73.0 beats per minute}[/tex]Standard deviation : [tex]\sigma=\text{12.5 beats per minute}[/tex]Sample size : n=1We assume that females have pulse rates that are normally distributed.Then , the formula to calculate the z-score is given by :-[tex]z=\dfrac{x-\mu}{\dfrac{\sigma}{\sqrt{n}}}[/tex]For x =69[tex]z=\dfrac{69-73}{\dfrac{12.5}{\sqrt{1}}}=-0.32[/tex]For x= 77[tex]z=\dfrac{77-73}{\dfrac{12.5}{\sqrt{1}}}=0.32[/tex]The p-value =[tex]P(-0.32<z<0.32)[/tex][tex]=P(z<0.32)-P(z<-0.32)\\= 0.6255158-0.3744842\\=0.2510316\approx0.2510[/tex]Hence, the  probability that her pulse rate is between 69 beats per minute and 77 beats per minute =0.2510