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:The probability that the baby will weigh more than 7.5 lbs is [tex]\sim[/tex] 0.28Step-by-step explanation:Let's call [tex]x[/tex] 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 ([tex]\mu = 7\ \rm{lb}[/tex]) with an standard deviation [tex]\sigma = 14\ \rm{oz} = 0.875\ \rm{lb}[/tex]. Thus, for this hospital we can say [tex]x \sim N(\mu,\sigma^2)[/tex]. Note that [tex]\sigma^2[/tex] is the variance of the distribution and not the standard deviation, which is the square root of the variance. For our problem we have:[tex]x\sim N(7, 0.766)[/tex]. The probability of an event is then given by:[tex]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 }[/tex]Now the probability that the baby will weigh more than 7.5 lbs is given by:[tex]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 }[/tex]Calculating the integral we obtain:[tex]\boxed{P(x\geq 7.5) = 0.28}[/tex]