Answer to Question #172191 in Statistics and Probability for kmy

Question

Weights of newborn babies in a particular city are normally distributed with a mean of  3380 g and a standard deviation of 475 g. 

a. A newborn weighing less than 2100 g is considered to be at risk, because the  mortality rate for this group is very low. If a hospital in the city has 500 births in a  year, how many of those babies are in the “at-risk” category?

b. If we redefine a baby to be at risk if his or her birth weight is in the lowest 3%,  find the weight that becomes the cutoff separating at-risk babies from those who  are not at risk. 

c. If 20 newborn babies are randomly selected as a sample in a study, find the  probability that their mean weight is between 3200 g and 3500 g.

Answer 1

a.

$P(X<2100) = P(\frac{x-μ}{σ} < \frac{2100-3380}{475}) \\ = P(Z < -2.6947) \\ = 0.00353 \\ Expect = 500 \times 0.00353 \\ = 1.75 \\ ≈ 2$

If a hospital in the city has 500 births in a year, 2 of those babies are in the “at-risk” category.

b.

$P(X≤C) = 0.03 \\ P(\frac{X-μ}{σ} ≤ \frac{c-3380}{475}) = 0.03 \\ P(Z ≤ -1.88) = 0.03 \\ \frac{c-3380}{475} = -1.88 \\ c = 2487$

If we redefine a baby to be at risk if his or her birth weight is in the lowest 3%, the weight that becomes the cutoff separating at-risk babies from those who are not at risk will be 2487 g.

c.

$P(3200 < \bar{X}<3500) = P( \frac{3200-3380}{475/ \sqrt{20}} <Z< \frac{3500-3380}{475/ \sqrt{20}}) \\ = P( -1.69 <Z<1.13 ) \\ = P(Z<1.13) - P(Z< -1.69) \\ = 0.8707-0.045 \\ = 0.8257$

If 20 newborn babies are randomly selected as a sample in a study, the probability that their mean weight is between 3200 g and 3500 g will be 0.8257.