Question 615228: I have problems figuring out 3 questions
1) Assume that a procedure yields a binomial distribution with a trial repeated n=5 times. Find the probability of x= 2 successes given that the probability p=0.75 of successes on a single trial.
2) Assume that adults have IQ scores that are normally distributed with a mean of 102 and a standard deviation of 12. Find the IQ score that separates the bottom 25% from the top 75%
3) Heights of adult men are normally distributed with a mean of 69 inches and standard deviation of 2.8 inches. If an adult man is randomly selected, what is the probability that he is taller than 73.2 inches
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! 1) Assume that a procedure yields a binomial distribution with a trial repeated n=5 times. Find the probability of x= 2 successes given that the probability p=0.75 of successes on a single trial.
Ans: P(x = 2) = 5C2*0.75^2*0.25*3 = binompdf(5,0.75,2) = 0.0879
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2) Assume that adults have IQ scores that are normally distributed with a mean of 102 and a standard deviation of 12. Find the IQ score that separates the bottom 25% from the top 75%
Find the z-score with a left tail of 0.25
invNorm(0.25) = -0.6745
Find the corresponding "IQ" value using IQ = z*s + u
IQ = -0.6745*12+102 = 93.9
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3) Heights of adult men are normally distributed with a mean of 69 inches and standard deviation of 2.8 inches. If an adult man is randomly selected, what is the probability that he is taller than 73.2 inches
z(73.2) = (73.2-69)/2 = 4.2/2 = 2.1
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P(x > 73.2) = P(z > 2.1) = normalcdf(2.1,100) = 0.0179
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Cheers,
Stan H.
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