Respuesta :
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 = [tex] \frac{x - mean}{standard deviation} [/tex]
z = [tex] \frac{1200-1028}{92} [/tex]
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, [tex]\mu = 1028[/tex]
Standard Deviation, [tex]\sigma = 92[/tex]
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
[tex]x = z\times \sigma +\mu[/tex]
[tex]x = 1.28\times92+1028[/tex]
[tex]x=1145.76[/tex]
90th percentile score is 1145.76.
Now, Z score corresponding to x = 1200 is,
[tex]\rm z = \dfrac{x - mean}{standard \; deviation}[/tex]
[tex]\rm z = \dfrac{1200-1028}{92}[/tex]
[tex]\rm z=1.87 \;\;\;\;(Approx)[/tex]
[tex]\rm P(Z>1.87)= 1-P(Z<1.87)[/tex]
[tex]\rm P(Z<1.87) = 0.9693[/tex]
[tex]\rm P(Z>1.87)=1-0.9693[/tex]
[tex]\rm P(Z>1.87)=0.0307[/tex]
The probability that a randomly selected score exceeds 1200 is 0.0307.
For more information, refer the link given below
https://brainly.com/question/23017717?referrer=searchResults
