Question

Weights of newborn babies in a particular city are normally distributed with a mean of

3360 g and a standard deviation of 450 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 900 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 1%,

find the weight that becomes the cutoff separating at-risk babies from those who

are not at risk.

c.

If 30 newborn babies are randomly selected as a sample in a study, find the probability that their mean weight is between 3000 g and 3500 g.

Answer 1

X ~ $N(3360, 450^2)$

a.

$P(X<2100) = P(\frac{x-μ}{σ} < \frac{2100-3360}{450}) \\ = P(Z < -2.8) \\ = 0.002555 \\ Expect = 900 \times 0.002555 \\ = 2.299 \\ ≈ 2$

If a hospital in the city has 900 births in a year, 2 of those babies are in the “at-risk” category.

b.

$P(X≤C) = 0.01 \\ P(\frac{X-μ}{σ} ≤ \frac{c-3360}{450}) = 0.01 \\ P(Z ≤ -1.88) = 0.01 \\ frac{c-3360}{450} = -2.32 \\ c = 3360 + 450 \times (-2.32) \\ c = 2316$

If we redefine a baby to be at risk if his or her birth weight is in the lowest 1%, the weight that becomes the cutoff separating at-risk babies from those who are not at risk will be 2316 g.

c.

$P(3000 < \bar{X}<3500) = P( \frac{3000-3360}{450/ \sqrt{30}} <Z< \frac{3500-3360}{450/ \sqrt{30}}) \\ = P( -4.38 <Z<1.70 ) \\ = P(Z<1.70) - P(Z< -4.38) \\ = 0.955434 -0.000031\\ = 0.955403$

If 30 newborn babies are randomly selected as a sample in a study, the probability that their mean weight is between 3000 g and 3500 g will be 0.955403.