Normal | |
mu | 544 |
sigma | 110 |
xi | P(X<=xi) |
380 | 0.0680 |
420 | 0.1298 |
500 | 0.3446 |
610 | 0.7257 |
770 | 0.9800 |
P(X<=xi) | xi |
0.1 | 403.0293 |
0.2 | 451.4217 |
0.3 | 486.3159 |
0.4 | 516.1318 |
0.7 | 601.6841 |
0.9 | 684.9707 |
Use the excel output. The probability is 0.20 that a student's test score will be between two values equidistant from the mean. Find the upper endpoint.
Please be as detailed as possible with your answer. I need to learn the steps to do it.
Let x be the required z score corresponding to upper end
According to question, we have to find x such that P(-x < z < x) = 0.20
P(z < x) - P(z < -x) = 0.20
P(z < x) - [1-P(z < x)] = 0.20 ..................... by property P(z<-A) =1 - P(z<A)
P(z < x) - 1 + P(z < x) = 0.20
2*P(z < x) = 1+0.20
P(z < x) = 1.20/2 = 0.60
Using excel function NORMSINV(0.60) to get 0.253347
Now, we have the z score for upper end value
Using z score formula with mean and standard deviation, we get
Upper end value = mean +z*standard deviation = 544+0.2533*110 = 571.8682
[your excel data does not have P(X < = xi) = 0.60 value, thats why I need to use NORMSINV(0.60)]
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