Question

It is believed that 4​% of children have a gene that may be linked to juvenile diabetes. Researchers at a firm would like to test new monitoring equipment for diabetes. Hoping to have 21 children with the gene for their​ study, the researchers test 733 newborns for the presence of the gene linked to diabetes. What is the probability that they find enough subjects for their​ study?

Answer 1

Answer:

the probability that they find enough subjects for their​ study is 0.9515

Step-by-step explanation:

From the given information:

Let X be the random variable that follows a normal distribution.

Therefore:

X $\sim$Binomial(n=733,p=0.04)

$\mu = np$

$\mu = 733*0.04$

$\mu = 29.32$

$\sigma = \sqrt{np(1-p)}$

$\sigma = \sqrt{29.32(1-0.04)}$

$\sigma = \sqrt{29.32(0.96)}$

$\sigma = \sqrt{28.1472}$

$\sigma = 5.305$

The probability of P(X ≥ 21) lies in the region between 20.5 and 21.5 by considering a discrete contribution of a continuous normal distribution. Eventually, X > 20.5

∴

P(X >20.5)= P(Z > z)

Using  standard normal z formula:

$z = \dfrac{x-\mu}{\sigma}$

$z = \dfrac{20.5-29.32}{5.305}$

$z = \dfrac{-8.82}{5.305}$

$z = -1.662582$

z = -1.66

P(X >20.5)= P(Z > -1.66)

From the standard z tables ;

P(X >20.5)= 1 - 0.0485

P(X >20.5)= 0.9515