Answer: 5.37
Step-by-step explanation:
Let x = ACT scores.
Given: ACT scores have a mean of 20.8 and 9 percent of the scores are above 28. The scores have a distribution that is approximately normal.
i.e. P(X>28)=0.09 (i)
Now,
(ii)
One -tailed z value for p-value of 0.09 =1.3408 [By z-table]
From (i) and (ii)
Hence, the standard deviation = 5.37
Complete Question
The complete question is shown on the first uploaded image
Answer:
The probability that there exist 60 or more defected children is
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
The probability of observing the defect is mathematically evaluated as
The probability of not observing the defect is mathematically evaluated as
The mean of this probability is mathematically represented as
Substituting values
The standard deviation of this probability is mathematically represented as
Substituting values
the probability of detecting defects can be represented in as normal distribution like
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
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
