The national average sat score (for verbal and math) is 1028. if we assume a normal distribution with standard deviation 92, what is the 90th percentile score? what is the probability that a randomly selected score exceeds 1200?

Let X be the national sat score. X follows normal distribution with mean μ =1028, standard deviation σ = 92The 90th percentile score is nothing but the x value for which area below x is 90%.To find 90th percentile we will find find z score such that probability below z is 0.9P(Z <z) = 0.9Using excel function to find z score corresponding to probability 0.9 isz = NORM.S.INV(0.9) = 1.28z =1.28Now convert z score into x value using the formulax = z *σ + μx = 1.28 * 92 + 1028x = 1145.76The 90th percentile score value is 1145.76The probability that randomly selected score exceeds 1200 isP(X > 1200) Z score corresponding to x=1200 isz = [tex] \frac{x - mean}{standard deviation} [/tex]z = [tex] \frac{1200-1028}{92} [/tex]z = 1.8695 ~ 1.87P(Z > 1.87 ) = 1 - P(Z < 1.87)Using z-score table to find probability z < 1.87P(Z < 1.87) = 0.9693P(Z > 1.87) = 1 - 0.9693P(Z > 1.87) = 0.0307The probability that a randomly selected score exceeds 1200 is 0.0307