Question

At a certain hospital, the probability of a child being born female is 0.51. If 80 babies are born on the a certain day, the probability that more than half of these babies are female is closest to? a. 0.61, b. 0.70, c. 0.39, d. 0.47, 0.53 (I will award you the brainiest)

Answer 1

Answer:

The probability that more than half of these babies are female is closest to 0.53.

Step-by-step explanation:

We are given that at a certain hospital, the probability of a child being born female is 0.5. 80 babies are born on a certain day.

Let X = Number of babies that are female

The above situation can be represented through binomial distribution which means X ~ Binom(n = 80, p = 0.51)

Here, n = number of samples taken and p = probability of success

For finding the probability we will use the normal approximation because the sample size is greater than 30 (n > 30), i.e;

Mean of X, $\mu$ = $n \times p$

                    = 80 $\times$ 0.51 = 40.8

Standard deviation of X, $\sigma$ = $\sqrt{n \times p\times (1-p)}$

                                         = $\sqrt{80 \times 0.51 \times (1-0.51)}$  = 4.5

So, X ~ Normal($\mu=40.8, \sigma^{2} = 4.5^{2}$)

Now, the probability that more than half of these babies are female is given by = P(X > 40) = P(X > 40.5)  {using continuity correction}

      P(X > 40.5) = P( $\frac{X-\mu}{{\sigma}} }$ > $\frac{40.5-40.8}{{4.5} } }$ ) = P(Z > -0.07) = P(Z < 0.07)

                                                          = 0.5279 ≈ 0.53

The above probability is calculated by looking at the value of x = 0.07 in the z table which has an area of 0.5279.

Hence, the probability that more than half of these babies are female is closest to 0.53.