Question 687161
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
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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
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I have to find µ = E[X] and σ^2 = Variance (X).
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Form 2 equations using x = z*s + u to solve for s and u:
60 = -1.17*s + u
95 = 1.53*s + u
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Subtract to solve for "s":
35 = 2.7s
s = 12.96
Variance = s^2 = 168
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Solve for "u" using 60 = -1.17s + u
60 = -1.17*12.96 + u
u = 75.163
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Cheers,
Stan H.