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Maru [420]
3 years ago
12

In the general population, what is the probability that an individual will have the birth defect, assuming that maternal and pat

ernal genes
Mathematics
1 answer:
vfiekz [6]3 years ago
7 0

Answer:

0.0625

Step-by-step explanation:

The prevalence of gene a = 25 %, P (a) = 0.25

birth defect occurs when both parents have prevalence of gene a.

P (Defect) = P ( Both parents have gene a)

If both parents inherit the gene a independently, the the individual will have a birth defect when both parents have gene a.

P ( Father having gene a) = 0.25

P ( Mother having gene a) = 0.25

Hence,

P (Birth Defect) = P ( Both parents have gene a) = 0.25 * 0.25 = 0.0625

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Step-by-step explanation:

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Laboratory tests show that the lives of
SIZIF [17.4K]
<h3>Answer:  34%</h3>

This result is approximate.

=========================================================

Explanation:

mu = 750 = mean

sigma = 75 = standard deviation

The raw scores or x values are x = 750 and x = 825

Let's compute the z score for each x value

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and

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z = (825 - 750)/75

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Therefore P(750 ≤ x ≤ 825) is equivalent to P(0 ≤ z ≤ 1) in this context.

Use a z score table to determine that

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So,

P(a ≤ z ≤ b) = P(z ≤ b) - P(z ≤ a)

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The value 0.34314 then converts to 34.314% which rounds to <u>34%</u>

Or you could use the empirical rule as shown below. The pink section on the right is marked <u>34%</u> which is approximate. This pink section is between z = 0 and z = 1.

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2 years ago
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Step-by-step explanation:

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