SOLUTION: plz solve my qustion Given a normal distribution with u=40 and variance=6 find the value of x that has (i) 38% of the area below it and (ii) 5% of the area above it. chandumer

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Question 266971: plz solve my qustion
Given a normal distribution with u=40 and variance=6 find the value of x that has (i) 38% of the area below it and (ii) 5% of the area above it.
chandumer932@hotmail.com
The definition of Z is number of S ( Standard Deviation) Z= (Xbar-Xi)/S. Z table tells relationship between percent under graph and how many S.
In this problem given percentage, S square ( Variance =6) and X bar ( u=40)
So find Z number first, then times square root 6 then minus u.
Answer is Z*6^0.5-40
Found 2 solutions by stanbon, WYWT:
Answer by stanbon(75887) (Show Source):
You can put this solution on YOUR website!
plz solve my question
Given a normal distribution with u=40 and variance=6 find the value of x that has
(i) 38% of the area below it
1st: Find the z-value that has 38% to its left.
invNorm(0.38) = -0.3055
2nd: use x = zs+u to find your answer:
x = -0.3055*6 + 40
x = 38.17
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(ii) 5% of the area above it.
1st: Find the z-value that has 5% above it
invNorm(0.95) = 1.645
Then x = 1.645*6+40
x = 49.87
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Cheers,
Stan H.
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Answer by WYWT(1) (Show Source):
You can put this solution on YOUR website!