Birth weights of babies born to full-term pregnancies follow roughly a Normal distribution. At Meadowbrook Hospital, the mean weight of babies born to full-term pregnancies is 7 pounds with a standard deviation of 14 ounces (1 pound = 16 ounces). Dr. Watts (who works at Meadowbrook Hospital) has four deliveries (all for full-term pregnancies) coming up during the night. Assume that the birth weights of these four babies can be viewed as a simple random sample. What is the probability that all four babies will weigh more than 7.5 pounds?

Answer: 0.006496Step-by-step explanation:Given : [tex]\mu=7\ pounds[/tex][tex]\sigma=14\ ounces = \dfrac{14}{16}=0.875\ pounds[/tex]   [∵ 1 pound = 16 ounces]Using [tex]z=\dfrac{x-\mu}{\sigma}[/tex] , for x= 7.5 , we have[tex]z=\dfrac{7.5-7}{0.875}=0.571428571429\approx0.5714[/tex] Using standard normal z-value table,P-value [tex]= P(z> 0.5714)=1-P(z>0.5714)=1-0.7161357[/tex] [tex]0.2838643\approx0.2839[/tex]Since, birth weights of these four babies can be viewed as a simple random sample. i.e. probability that bay has weigh more than 7.5 pounds is equal for all babies.Also, the weight of babies are independent .Then, the  the probability that all four babies will weigh more than 7.5 pounds =[tex](0.2839)^4=0.00649623265262\approx0.006496[/tex] [Rounded to nearest 4 decimal places]