It is well documented that the birth weight of babies from women who receive no

Question

It is well documented that the birth weight of babies from women who receive no prenatal care is significantly lower than for women who do receive prenatal care. Recent studies estimate that 85% of all premature babies born to women who receive no pre-natal care have birth weights ranging from 4.5 lbs to 7.5 lbs.

(a) Assuming that the birth weight distribution for this segment of the population can be adequately approximated by a Normal distribution, and assuming that 4.5 lbs and 7.5 lbs are equi-distant from the mean weight, :, calculate the standard deviation of the distribution.

(b) A random sample of N = 16 babies are taken from this group. What is the probability that the sample mean weight is less than 5.4 lbs?

Explanation / Answer

(a) mean=(4.5+7.5)/2 = 6

a=1-0.85= 0.15 , Z(0.15/2) = Z(0.075) =0.27 (from standard normal table)

mean- Z*s = 4.5

--> 6 - 0.27*s = 4.5

--> 0.27*s = 6-4.5 =1.5

So standard deviation = 1.5/0.27 =5.555556

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(b) So the probability that the sample mean weight is less than 5.4 lbs is

P(xbar<5.4) = P((xbar-mean)/(s/vn) <(5.4-6)/(5.555556/sqrt(16)))

=P(Z<-0.43) = 0.3336 (from standard normal table)

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