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Normal distribution, probabilities (3 problems)

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Problem 1

We know the IQ of people is normally distributed with mean 120 and standard deviation (SD) of 25. Calculate the probability that an adult will have

a ) IQ less than 100

b) IQ greater than 150

c) IQ between 100 and 150

d) Any person with a score over 200 is considered a genius. In a population of 10000 how many will score above 200.

e) MENSA is a set up to recruit the top 1% of the population. What should the IQ score be set at to qualify for MENSA membership?

Problem 2

A chemical company produces a product which has an impurity that is expensive to remove. The mean level is 20 parts per million with SD of 6 parts.

a) What level of impurity could the chemical company quote to its customers and be 99% confident that it will not exceed the stated impurity level?

A customer, however, is only prepared to accept a maximum impurity of 30 parts.

b) How much product will meet this specification?

c) Suppose the product deteriorates so that the mean drifts out to 25 parts per million, but with the same SD of 6 parts. How much of production is now unsuitable?

d) Finally, if the mean is brought back to 20 parts but the SD is the found to be 8 parts, how much product will not meet the specification?

Problem 3

During tests on a group of workers it is found that only 15% of them can hear sound above a frequency of 200 units and only 6% can detect sound above 210 units. Calculate the mean and SD of the highest frequency that can be heard for the population of workers.

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Solution Summary

Three problems involving the normal distribution are solved. The solutions show how to calculate probabilities of specific outcomes when sampling from a normal distribution with a known mean and standard deviation.

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Problem 1

We know the IQ of people is normally distributed with mean 120 and standard deviation (SD) of 25. Calculate the probability that an adult will have

a ) IQ less than 100

X has a normal distribution (mean=120, SD=25)

Please note that all the problems on this sheet involve using one simple equation. That equation is:

X = mean + (SD)*Z

Z = (X - mean) / (SD) (the same equation rearranged)

Where X is a value in the distribution and Z is the Z-score corresponding to the X-value. The value of Z tells us how many standard deviations the corresponding X value is from the mean. For example, if a distribution has a mean of 10 and SD of 1, an X value of 12 has a Z value of 2, since it's 2 standard deviations above the mean. An X value of 9 has a Z-score of -1, since it's 1 standard deviation below the mean.

Now back to the problem at hand.

Calculate the z score for X=100 by subtracting the mean and dividing by the standard deviation.

P {X < 100} = P {Z < [(100-120)/25]} = P {Z < -0.8} = .2119

The value 0.2119 was found in a table of standard normal probabilities.

b) IQ greater than 150

P (X > 150) = P {Z > [(150-120)/25]} = P {Z > ...

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