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14 + 28 + 42 + 56 + … 280
= 14 (1 + 2 + 3 + 4 + … + 20)
= 14 × 20 × 21 / 2
= 2940
where we use the well-known identity,
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()
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 .