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Brrunno [24]
3 years ago
11

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?

Mathematics
1 answer:
Dovator [93]3 years ago
3 0

Complete Question

The complete question is shown on the first uploaded image

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

             

   

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