Respuesta :
Answer: 0.006496
Step-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]
The weight of babies in pregnancies.
The questions stated that the weight of the babies born during the pregnancies to the full term are in a normal distribution and the babies born at the Meadowbrook Hospital is of the mean weight of full-term is seven pounds and with the SD of 14 ounces that is 1 pound will be equal to 16 ounces.
Hence the answer is 0.006496.
- The birth weight of these 4 babies can be viewed as a simple random sample. here the probability that a baby has a weight of more than 7.5 pounds is equal for all babies. As the weight of babies is independent.
- Dr. Watts has measured the time of four delivery of the babies as pr their weight viewed in simple random terms and assets the probability to be 7.5 pounds.
Learn more about the babies born to full-term.
brainly.com/question/14958871.