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](https://tex.z-dn.net/?f=P%28x%20%5Cge%2060%29%3D0.0901)
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](https://tex.z-dn.net/?f=n%3D%2050%2C000)
The probability of observing the defect is mathematically evaluated as
![p = \frac{1}{1000}](https://tex.z-dn.net/?f=p%20%3D%20%5Cfrac%7B1%7D%7B1000%7D)
![= 0.001](https://tex.z-dn.net/?f=%3D%200.001)
The probability of not observing the defect is mathematically evaluated as
![q = 1-p](https://tex.z-dn.net/?f=q%20%3D%201-p)
![= 1-0.001](https://tex.z-dn.net/?f=%3D%201-0.001)
![= 0.999](https://tex.z-dn.net/?f=%3D%200.999)
The mean of this probability is mathematically represented as
![\mu = np](https://tex.z-dn.net/?f=%5Cmu%20%3D%20np)
Substituting values
![\mu = 50000*0.001](https://tex.z-dn.net/?f=%5Cmu%20%3D%2050000%2A0.001)
![= 50](https://tex.z-dn.net/?f=%3D%2050)
The standard deviation of this probability is mathematically represented as
![\sigma = \sqrt{npq}](https://tex.z-dn.net/?f=%5Csigma%20%3D%20%5Csqrt%7Bnpq%7D)
Substituting values
![= \sqrt{50000 * 0.001 * 0.999}](https://tex.z-dn.net/?f=%3D%20%5Csqrt%7B50000%20%2A%200.001%20%2A%200.999%7D)
![= \sqrt{49.95}](https://tex.z-dn.net/?f=%3D%20%5Csqrt%7B49.95%7D)
![= 7.07](https://tex.z-dn.net/?f=%3D%207.07)
the probability of detecting
defects can be represented in as normal distribution like
![P(x \ge 60)](https://tex.z-dn.net/?f=P%28x%20%5Cge%2060%29)
in standardizing the normal distribution the normal area used to approximate
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 }](https://tex.z-dn.net/?f=z%20%3D%20%5Cfrac%7Bx-%5Cmu%20%7D%7B%5Csigma%20%7D)
![= \frac{59.5 - 50 }{7.07}](https://tex.z-dn.net/?f=%3D%20%5Cfrac%7B59.5%20-%2050%20%7D%7B7.07%7D)
![=1.34](https://tex.z-dn.net/?f=%3D1.34)
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)](https://tex.z-dn.net/?f=P%28x%20%5Cge%2060%29%20%5Capprox%20P%28z%20%3E%201.34%29)
![= 1-P(z \le 1.34)](https://tex.z-dn.net/?f=%3D%201-P%28z%20%5Cle%201.34%29)
![=1-0.9099](https://tex.z-dn.net/?f=%3D1-0.9099)
![=0.0901](https://tex.z-dn.net/?f=%3D0.0901)