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 23 children with the gene for their study, the researchers test 732 newborns for the presence of the gene linked to diabetes. What is the probability that they find enough subjects for their study?

0.882

Step-by-step explanation:

Given that:

Positive gene (p) = 4% = 0.04

Not positive (1 - p) = r = 1 - 0.04 = 0.96

Number of samples (n) = 732

Standard deviation = (s) :

√[(p)(1 - p)] / n

s = √(0.04)(0.96) / 732

s = √ 0.0384 / 732

s = 0.0072428

x = 23/732 = 0.0314207

Hence,

Z = (x - p) / s

Z = (0.0314207 - 0.04) / 0.0072428

Z = −1.184528

P(Z ≥ −1.184528)

1 - p(Z ≤ - 1.18) = 1 - 0.118

= 0.882

answer: hi

step-by-step explanation:

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