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 lbs with a standard deviation of 14 oz (1lb = 16 oz). What is the probability that the average weight of the four babies will be more than 7.5 lbs?

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()

The standard normal z distribution is given by;

              Z = ~ 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( >  ) = 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 .

y=mx+b

step-by-step explanation:

m slope, b y-intercept

answer: at first, in 1635, the evening meetings had been just for women . .

what relationship between ideas is shown in the text above?

question 1 options:

location of events

similarity of events

timing of events

examples of an idea

step-by-step explanation: