Respuesta :
Answer:
Required Probability = 0.1283 .
Step-by-step explanation:
We are given that at Meadow brook Hospital, the mean weight of babies born to full-term pregnancies is 7 lbs with a standard deviation of 14 oz.
Firstly, standard deviation in lbs = 14 ÷ 16 = 0.875 lbs.
Also, Birth weights of babies born to full-term pregnancies follow roughly a Normal distribution.
Let X = mean weight of the babies, so X ~ N([tex]\mu = 7 lbs , \sigma^{2} = 0.875^{2} lbs[/tex])
The standard normal z distribution is given by;
Z = [tex]\frac{Xbar-\mu}{\frac{\sigma}{\sqrt{n} } }[/tex] ~ N(0,1)
where, X bar = sample mean weight
n = sample size = 4
Now, probability that the average weight of the four babies will be more than 7.5 lbs = P(X bar > 7.5 lbs)
P(X bar > 7.5) = P( [tex]\frac{Xbar-\mu}{\frac{\sigma}{\sqrt{n} } }[/tex] > [tex]\frac{7.5-7}{\frac{0.875}{\sqrt{4} } }[/tex] ) = P(Z > 1.1428) = 0.1283 (using z% table)
Therefore, the probability that the average weight of the four babies will be more than 7.5 lbs is 0.1283 .
