Answer to Question #160361 in Statistics and Probability for Andrew

Question #160361

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.


1
Expert's answer
2021-02-03T01:16:13-0500

X ~ N(3380, 475^2)

a.

"P(X<2100) = P(\\frac{x-\u03bc}{\u03c3} < \\frac{2100-3380}{475}) \\\\\n\n= P(Z < -2.6947) \\\\\n\n= 0.00353 \\\\\n\nExpect = 500 \\times 0.00353 \\\\\n\n= 1.75 \\\\\n\n\u2248 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\u2264C) = 0.03 \\\\\n\nP(\\frac{X-\u03bc}{\u03c3} \u2264 \\frac{c-3380}{475}) = 0.03 \\\\\n\nP(Z \u2264 -1.88) = 0.03 \\\\\n\n\\frac{c-3380}{475} = -1.88 \\\\\n\nc = 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}}) \\\\\n\n= P( -1.69 <Z<1.13 ) \\\\\n\n= P(Z<1.13) - P(Z< -1.69) \\\\\n\n= 0.8707-0.045 \\\\\n\n= 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.


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