Question

In a random sample of 45 newborn babies, what is the probability that 40% or fewer of the sample are boys. (Use 3 decimal places). Hint: you are asked to find P( < .40 ). To calculate this first find the z-score and then calculate the probability.

Answer 1

The probability that 40% or fewer of the sample are boys P(x ≤ 0.4) = 0.090

Explanation:

Normally, the mean number of boys in the population should be given, but in the absence, we take 50% of the population to be boys (since, there are only two possibilities, with close probabilities).

Mean = 0.5

Standard deviation of a sample = √[p(1-p)/n]

p = 0.5

1-p = 0.5

n = sample size = 45

Standard deviation = √[(0.5×0.5)/45] = 0.0745

So, we can now standardize 0.4.

The standardized score is the value minus the mean then divided by the standard deviation.

z = (x - μ)/σ = (0.4 - 0.5)/0.0745 = - 1.34

To determine the probability that 40% or fewer of the sample are boys P(x ≤ 0.4) = P(z ≤ -1.34)

We'll use data from the normal probability table for these probabilities

P(x ≤ 0.4) = P(z ≤ -1.34) = 0.090

Answer 2

`z= (p- \mu_p)/(\sigma_p)`

And the z score for 0.4 is

`z = (0.4-0.4)/(\sigma_p) = 0`

And then the probability desired would be:

`P(p<0.4) = p(z<0) =0.5`

Explanation:

The normal approximation for this case is satisfied since the value for p is near to 0.5 and the sample size is large enough, and we have:

`np = 45*0.4= 18 >10`

`n(1-p) = 45*0.6= 27 >10`

For this case we can assume that the population proportion have the following distribution

`p \sim N(p,\sqrt{(p(1-p))/(n)})`

Where:

`\mu_(p)= \hat p = 0.4`

`\sigma_p = \sqrt{\frac{p(1-p)}(n} =\sqrt{\frac{0.4(1-0.4)}(45}= 0.0703`

And we want to find this probability:

`P(p <0.4)`

And we can use the z score formula given by:

`z= (p- \mu_p)/(\sigma_p)`

And the z score for 0.4 is

`z = (0.4-0.4)/(\sigma_p) = 0`

And then the probability desired would be:

`P(p<0.4) = p(z<0) =0.5`