Question

What is the probability that out of 250 babies born, 110 or fewer will be boys?

Aaaume that boys and girls are equally probable, and round your answer to
the nearest tenth of a percent,
A. 3.3%
B. 75.7%
C. 28.5%
O D. 97.5%

Answer 1

A. 3.3%

Explanation:

We use the normal approximation to the binomial to solve this question.

Binomial probability distribution

Probability of exactly x sucesses on n repeated trials, with p probability.

Can be approximated to a normal distribution, using the expected value and the standard deviation.

The expected value of the binomial distribution is:

`E(X) = np`

The standard deviation of the binomial distribution is:

`√(V(X)) = √(np(1-p))`

Normal probability distribution

Problems of normally distributed distributions can be solved using the z-score formula.

In a set with mean
`\mu` and standard deviation
`\sigma`, the zscore of a measure X is given by:

`Z = (X - \mu)/(\sigma)`

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

When we are approximating a binomial distribution to a normal one, we have that
`\mu = E(X)`,
`\sigma = √(V(X))`.

250 babies, boys and girls equally as likely:

This means that
`n = 250, p = 0.5`.

Mean and standard deviation:

`\mu = E(X) = np = 250*0.5 = 125`

`\sigma = √(V(X)) = √(np(1-p)) = √(250*0.5*0.5)`

Probability that out of 250 babies born, 110 or fewer will be boys?

Using continuity correction, this is P(X < 110 + 0.5) = P(X < 110.5), which is the p-value of Z when X = 110.5. So

`Z = (X - \mu)/(\sigma)`

`Z = (110.5 - 125)/(√(250*0.5*0.5))`

`Z = -1.83`

`Z = -1.83` has a p-value of 0.033

0.033*100% = 3.3%, so option A.