Question 669832
Suppose we have a population of scores with a mean (μ) of 975 and a standard deviation (σ) of 15. Assume that the distribution is normal. Provide answers to the following questions:
a. What percentage of the population will lie between 960 and 990?
z(960) = (960-975)/15 = -15/15 = -1
z(990) = (990-975)/15 = 15/15 = 1
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P(960<= x <= 990) = P(-1<= z <=1) = 0.6827
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b. What percentage of the population will lie below 975?
z(975) = (975-975)/15 = 0
P(x > 975) = P(z < 0) = 0.50
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c. What percentage of the population will lie below 990?
P(x < 990) = P(z < 1) = 0.8413
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d. What two values of X (the counts) would encompass the middle 50 percent of scores?
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Find the z-value with a left tail of 25%
invNorm(0.25) = -0.6745
By symmetry the z-value with a right tail of 25% is 0.6745
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Find the corresponding x-values using x = z*s + u
Lower limit: x = -0.6745*15+975 = 964.88
Upper limit: x = 0.6745*15+975 = 985.12
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Cheers,
Stan H.
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