Obstetrics Assume that birthweights are normally distributed with a mean of 3400

Question

Obstetrics Assume that birthweights are normally distributed with a mean of 3400 g and a standard deviation of 700 g. Find the probability of a low-birthweight child, where low birthweight is defined as lessthanorequalto 2500 g. Find the probability of a very low birthweight child, where very low birthweight is defined as lessthanorequalto 2000 g. Assuming that successive deliveries by the same woman have the same probability of being low birthweight, what is the probability that a woman with exactly 3 deliveries will have 2 or more low-birthweight deliveries?

Explanation / Answer

Solution:

We are given that the birth weights are normally distributed.

Mean = 3400 and SD = 700

Question 5.16

Here, we have to find P(X2500)

The Z-score formula is given as below:

Z = (X – mean) / SD

Z = (2500 – 3400) / 700

Z = -1.28571

P(X2500) = P(Z< -1.28571) = 0.099271

(By using normal or z-table or excel)

Required probability = 0.099271

Question 5.17

Here, we have to find P(X2000)

The Z-score formula is given as below:

Z = (X – mean) / SD

Z = (2000 – 3400) / 700

Z = -2

P(X2000) = P(Z<-2) = 0.02275

(By using normal or z-table or excel)

Required probability = 0.02275

Question 5.18

For the given scenario, we are given

n = 3, p = 0.099271

We have to find P(X2)

Here, we have to use the binomial distribution.

P(X=x) = nCx*p^x*q^(n – x)

Where, q = 1 – p = 1 – 0.099271 = 0.900729

P(X2) = 1 – P(X<2) = 1 – P(X1)

P(X1) = P(X=0) + P(X=1)

P(X=0) = 3C0*0.099271^0*0.900729^(3 – 0)

P(X=0) = 0.730772

P(X=1) = 3C1*0.099271^1*0.900729^(3 – 1)

P(X=1) = 0.24162

P(X1) = P(X=0) + P(X=1) = 0.730772 + 0.24162 = 0.972392

P(X2) = 1 – P(X<2) = 1 – P(X1) = 1 – 0.972392 = 0.027608

Required probability = 0.027608

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