# 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

Algebra ->  Algebra  -> Probability-and-statistics -> 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      Log On

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 Click here to see ALL problems on Probability-and-statistics 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*6^0.5-40Found 2 solutions by stanbon, WYWT:Answer by stanbon(57222)   (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 =========================== (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 ================= Cheers, Stan H. ------------------ Answer by WYWT(1)   (Show Source):