It is believed that 4% of children have a gene that may be linked to juvenile di

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 24 children with the gene for their study, the researchers test 680 newborns for the presence of the gene linked to diabetes.

a) Assuming the 4% figure is correct, what is the probability that they find enough subjects for their study?

b) Assuming the 4% figure is correct, what is the probability they find less than 22 but more than 16 subjects for their study?

c) What is the expected value and standard deviation for the number of children in the sample with the gene linked to diabetes?

Explanation / Answer

Here we use sampling distribution of normal to binomial distribution.

Here random variable X ~ Binomial(n = 24, p = 4% = 0.04)

Here we have to find P(p^ > 24/680) = P(p^ > 0.035)

Here distribution of p^ is Normal with mean = p and sd = sqrt[(p*q)/n]

where q=1-p

q = 1 - 0.04 = 0.96

mean = p = 0.04

sd = sqrt[(0.04*0.96)/24] = 0.04

Now we have to find z-score for p^ = 0.035

z-score is defined as,

z = (p^ - mean) / sd

z = (0.035 - 0.04) / 0.04= -0.12

Now we have to find P(Z > -0.12).

This probability we can find in EXCEL.

syntax :

=1 - NORMSDIST(z)

where z is z-score

P(Z >-0.12) = 0.5468

Now in the next part we have to find P(16/680 <p^ < 22/680).

That is we have to find P(0.024 < p^ < 0.032).

Now convert p^ = 0.024 and p^ = 0.032 into z-score.

z = (0.024 - 0.04) / 0.04 = -0.41

z = (0.032 - 0.04) / 0.04 = -0.19

That is we have to find P(-0.41 < Z < -0.19)

P(-0.41 < Z < -0.19) = P(Z < -0.19) - P(Z < -0.41)

This probability we can find in EXCEL.

syntax :

=NORMSDIST(z)

where z is z-score.

P(-0.41 < Z < -0.19) = 0.4242 - 0.3403 = 0.0839

Expected value = n*p = 24 * 0.04 = 0.96

variance = npq = 24*0.04 *0.96 = 0.9216

sd = sqrt(0.9216) = 0.96

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