Mathematics, 30.03.2020 22:16, destinysmithds7790

Genetic Defects Data indicate that a particular genetic defect occurs in of every children. The records of a medical clinic show children with the defect in a total of examined. If the children were a random sample from the population of children represented by past records, what is the probability of observing a value of equal to or more?

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answered: ayoismeisalex

Complete Question

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The probability that there exist 60 or more defected children is $P(x \ge 60)=0.0901$

Looking at the value for this probability we see that it is not so small to the point that the observation of this kind would be a rare occurrence

Step-by-step explanation:

From the question we are told that

in every 1000 children a particular genetic defect occurs to 1

The number of sample selected is n= 50,000

The probability of observing the defect is mathematically evaluated as

$p = \frac{1}{1000}$

= 0.001

The probability of not observing the defect is mathematically evaluated as

q = 1-p

= 1-0.001

= 0.999

The mean of this probability is mathematically represented as

$\mu = np$

Substituting values

$\mu = 50000*0.001$

= 50

The standard deviation of this probability is mathematically represented as

$\sigma = \sqrt{npq}$

Substituting values

$= \sqrt{50000 * 0.001 * 0.999}$

$= \sqrt{49.95}$

= 7.07

the probability of detecting $x \ge60$ defects can be represented in as normal distribution like

$P(x \ge 60)$

in standardizing the normal distribution the normal area used to approximate $P(x \ge 60)$ is the right of 59.5 instead of 60 because x= 60 is part of the observation

The z -score is obtained mathematically as

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

$= \frac{59.5 - 50 }{7.07}$

=1.34

The area to the left of z = 1.35 on the standardized normal distribution curve is 0.9099 obtained from the z-table shown z value to the left of the standardized normal curve

Note: We are looking for the area to the right i.e 60 or more

The total area under the curve is 1

$P(x \ge 60) \approx P(z 1.34)$

$= 1-P(z \le 1.34)$

=1-0.9099

=0.0901

Genetic Defects Data indicate that a particular genetic defect occurs in of every children. The reco

answered: Guest

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answered: Guest

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