Question 933491
For a normal distribution with µ = 500 and σ = 100, find the following values:
 a. What X value separates the highest 10% of the distribution from the rest of the scores?
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Find the z-value with a 10% right tail:: invNorm(0.10) = 1.2816
Find the corresponding X-value:: X = 1.2816*100+500 = 628.16
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 b. What X values form the boundaries for the middle 60% of the distribution?
Find the z-value with a left tail of 20%:: invNorm(0.20) = -0.8416
Lower boundary:: X = -0.8416*100+500 = 500-84.16 = 415.84
Upper boundary:: X = +0.8416*100+500 = 500+84.16 = 584.16
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 c. What is the probability of randomly selecting a score greater than X = 475?
z(475) = (475-500)/100 = -25/100 = -0.25 
P(X > 475) = P(z > -0.25) = normalcdf(-0.25,100) = 0.5987
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Cheers,
Stan H.
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