Suppose the lengths of the pregnancies of a certain animal are approximately normally distributed with mean mu equals 192 daysμ=192 days and standard deviation sigma equals 12 daysσ=12 days.What is the probability that a randomly selected pregnancy lasts less than 188188 ​days? The probability that a randomly selected pregnancy lasts less than 188188 days is approximately

Answer: 0.5237Step-by-step explanation:Mean : [tex]\mu=192\text{ days}[/tex]Standard deviation : [tex]\sigma = 12\text{ days}[/tex]The formula to calculate the z-score is given by :-[tex]z=\dfrac{x-\mu}{\sigma}[/tex]For x = 188 ​days ,[tex]z=\dfrac{188-192}{12}\approx-0.33[/tex]For x = 107 miles per day ,[tex]z=\dfrac{107-92}{12}=1.25[/tex]The P-value =[tex]P(-0.33<z<1.25)=P(z<1.25)-P(z<-0.33)[/tex][tex]0.8943502-0.3707=0.5236502\approx0.5237[/tex]Hence, The probability that a randomly selected pregnancy lasts less than 188 days is approximately 0.5237.