SOLUTION: Of randomly selected students with random variable x with normal distribution. 12.10% score below 60 and 6.3% score above 95. I have to find � = E[X] and σ^2 = Variance (X)

Question 687161: Of randomly selected students with random variable x with normal distribution.
12.10% score below 60 and 6.3% score above 95.
I have to find � = E[X] and σ^2 = Variance (X).
I have no idea what to do with this at all please help.
Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
Of randomly selected students with random variable x with normal distribution.
Draw a normal curve.
12.10% score below 60 and 6.3% score above 95.
sketch a left-tail of 12.10% to the left of 60
Find the z-value with that left tail:
invNorm(0.121) = -1.17
-----
sketch a right-tail of 6.3% to the right of 95
Find the z-value with a left tail of 93.7%
invNorm(0.937) = 1.53
--------

I have to find � = E[X] and σ^2 = Variance (X).
----
Form 2 equations using x = z*s + u to solve for s and u:
60 = -1.17*s + u
95 = 1.53*s + u
------
Subtract to solve for "s":
35 = 2.7s
s = 12.96
Variance = s^2 = 168
------
Solve for "u" using 60 = -1.17s + u
60 = -1.17*12.96 + u
u = 75.163
=======================
Cheers,
Stan H.

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SOLUTION: Of randomly selected students with random variable x with normal distribution. 12.10% score below 60 and 6.3% score above 95. I have to find � = E[X] and &#963;^2 = Variance (X)

SOLUTION: Of randomly selected students with random variable x with normal distribution. 12.10% score below 60 and 6.3% score above 95. I have to find � = E[X] and σ^2 = Variance (X)