Answer:
Probability that a randomly selected newborn is low-weight is 0.1038.
Step-by-step explanation:
We are given that the birth weights of full-term babies in a certain area are normally distributed with mean 7.13 pounds and standard deviation 1.29 pounds.
Also, a newborn weighing 5.5 pounds or less is a low-weight baby.
Let X = weights of full-term babies
So, X ~ N([tex]\mu=7.13,\sigma^{2} = 1.29^{2}[/tex])
The z score probability distribution is given by;
Z = [tex]\frac{X-\mu}{\sigma}[/tex] ~ N(0,1)
where, [tex]\mu[/tex] = mean weight = 7.13 pounds
[tex]\sigma[/tex] = standard deviation = 1.29 pounds
So, probability that a randomly selected newborn is low-weight is given by = P(X [tex]\leq[/tex] 5.5 pounds)
P(X [tex]\leq[/tex] 5.5 pounds) = P( [tex]\frac{X-\mu}{\sigma}[/tex] [tex]\leq[/tex] [tex]\frac{5.5-7.13}{1.29}[/tex] ) = P(Z [tex]\leq[/tex] -1.26) = 1 - P(Z < 1.26)
= 1 - 0.8962 = 0.1038
Hence, probability that a randomly selected newborn is low-weight is 0.1038.
