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

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 = (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 = √(npq)`

Substituting values

`= √(50000 * 0.001 * 0.999)`

`= √(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 = (x-\mu )/(\sigma )`

`= (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`