SOLUTION: Can someone please help me understand this! Suppose we have a population of scores with a mean (μ) of 975 and a standard deviation (σ) of 15. Assume that the distribu

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Question 669832: Can someone please help me understand this!
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?
b. What percentage of the population will lie below 975?
c. What percentage of the population will lie below 990?
d. What two values of X (the counts) would encompass the middle 50 percent of scores?
Answer by stanbon(75887) About Me  (Show Source):
You can put this solution on YOUR website!
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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