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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?

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Answer: 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

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