Question

# Normal mu 544 sigma 110 xi P(X<=xi) 380 0.0680 420 0.1298 500 0.3446 610 0.7257 770...

 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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