Complete Question
The complete question is shown on the first uploaded image
Answer:
The probability that there exist 60 or more defected children is [tex]P(x \ge 60)=0.0901[/tex]
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 [tex]n= 50,000[/tex]
The probability of observing the defect is mathematically evaluated as
[tex]p = \frac{1}{1000}[/tex]
[tex]= 0.001[/tex]
The probability of not observing the defect is mathematically evaluated as
[tex]q = 1-p[/tex]
[tex]= 1-0.001[/tex]
[tex]= 0.999[/tex]
The mean of this probability is mathematically represented as
[tex]\mu = np[/tex]
Substituting values
[tex]\mu = 50000*0.001[/tex]
[tex]= 50[/tex]
The standard deviation of this probability is mathematically represented as
[tex]\sigma = \sqrt{npq}[/tex]
Substituting values
[tex]= \sqrt{50000 * 0.001 * 0.999}[/tex]
[tex]= \sqrt{49.95}[/tex]
[tex]= 7.07[/tex]
the probability of detecting [tex]x \ge60[/tex] defects can be represented in as normal distribution like
[tex]P(x \ge 60)[/tex]
in standardizing the normal distribution the normal area used to approximate [tex]P(x \ge 60)[/tex] is the right of 59.5 instead of 60 because x= 60 is part of the observation
The z -score is obtained mathematically as
[tex]z = \frac{x-\mu }{\sigma }[/tex]
[tex]= \frac{59.5 - 50 }{7.07}[/tex]
[tex]=1.34[/tex]
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
[tex]P(x \ge 60) \approx P(z > 1.34)[/tex]
[tex]= 1-P(z \le 1.34)[/tex]
[tex]=1-0.9099[/tex]
[tex]=0.0901[/tex]
