Assume that the red blood cell counts of women are normally distributed with a mean of 4.577 million cells per microliter and a standard deviation of 0.382 million cells per microliter. Find the value closest to the probability that a randomly selected woman has a red blood cell count above the normal range of 4.2 to 5.4 million cells per microliter. Round to four decimal places.

Answer: 0.8231Step-by-step explanation:Given: Mean : [tex]\mu = 4.577\text{ million cells per microliter }[/tex]Standard deviation : [tex]\sigma = 0.382 \text{ million cells per microliter}[/tex]The formula to calculate z is given by :-[tex]z=\dfrac{x-\mu}{\sigma}[/tex]For x=  4.2[tex]z=\dfrac{4.2-4.577}{0.382}=-0.986910994764\approx-0.99[/tex]The P Value =[tex]P(z<-0.99)=0.1610871[/tex]For x= 5.4[tex]z=\dfrac{5.4-4.577}{0.382}=2.15445026178\approx2.15[/tex]The P Value =[tex]P(z<2.15)= 0.9842223[/tex][tex]\text{Now, }P(4.2<X<5.4)=P(X<5.4)-P(X<4.2)\\\\=P(z<2.15)-P(z<-0.99)\\\\=0.9842223-0.1610871=0.8231352\approx 0.8231[/tex]Hence, the probability that a randomly selected woman has a red blood cell count above the normal range of 4.2 to 5.4 million cells per microliter=0.8231