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

Answer:

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

Step-by-step 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:

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