Question

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 (1 lb = 16 oz). Dr. Watts (who works at Meadowbrook Hospital) has one delivery (for a full-term pregnancy) coming up tonight. What is the probability that the baby will weigh more than 7.5 lbs?

Answer 1

The probability that the baby will weigh more than 7.5 lbs is
`\sim` 0.28

Explanation:

Let's call
`x` the weights of babies born to full-term pregnancies. As the problems states that this follows roughly a Normal distribution, and that in Meadowbrook Hospital this distribution has a mean value (
`\mu = 7\ \rm{lb}`) with an standard deviation
`\sigma = 14\ \rm{oz} = 0.875\ \rm{lb}`.

Thus, for this hospital we can say
`x \sim N(\mu,\sigma^2)`. Note that
`\sigma^2` is the variance of the distribution and not the standard deviation, which is the square root of the variance. For our problem we have:

`x\sim N(7, 0.766)`.

The probability of an event is then given by:

`P(x\leq x_0) = {\displaystyle { \int\limits^(\x_0)_(-\infty) {\frac {1}{\sqrt {2\pi \sigma ^(2)}}}e^{-{\frac {(x-\mu )^(2)}{2\sigma ^(2)}}}} \, dx }`

Now the probability that the baby will weigh more than 7.5 lbs is given by:

`P(x\geq 7.5) = {\displaystyle { \int\limits^(\infty)_(7.5) {\frac {1}{\sqrt {2\pi\cdot 0.875 ^(2)}}}e^{-{\frac {(x-7 )^(2)}{2\cdot0.875^(2)}}}} \, dx }`

Calculating the integral we obtain:

`\boxed{P(x\geq 7.5) = 0.28}`