Question

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?

Answer 1

Complete Question

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Answer:

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

So

    $P(x \ge 60) \approx P(z > 1.34)$

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

                    $=1-0.9099$

                  $=0.0901$