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 = 92

The 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 0.9

P(Z <z) = 0.9

Using excel function to find z score corresponding to probability 0.9 is

z = NORM.S.INV(0.9) = 1.28

z =1.28

Now convert z score into x value using the formula

x = z *σ + μ

x = 1.28 * 92 + 1028

x = 1145.76

The 90th percentile score value is 1145.76

The probability that randomly selected score exceeds 1200 is

P(X > 1200)

Z score corresponding to x=1200 is

z =

z =

z = 1.8695 ~ 1.87

P(Z > 1.87 ) = 1 - P(Z < 1.87)

Using z-score table to find probability z < 1.87

P(Z < 1.87) = 0.9693

P(Z > 1.87) = 1 - 0.9693

P(Z > 1.87) = 0.0307

The probability that a randomly selected score exceeds 1200 is 0.0307

90th percentile score is 1145.76.

The probability that a randomly selected score exceeds 1200 is 0.0307.

Step-by-step explanation:

Given :

Mean,

Standard Deviation,

Calculation :

To find 90th percentile we have to find z score such that probability below z is 0.9.

P(Z < z) = 0.9

Using excel function to find z score corresponding to probability 0.9 is,

z = 1.28  

Now, x value is

90th percentile score is 1145.76.

Now, Z score corresponding to x = 1200 is,

The probability that a randomly selected score exceeds 1200 is 0.0307.

For more information, refer the link given below

link

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step-by-step explanation:

its wrong

this is a tricky question, lol,   but i think the answer is 100,000.