If a normal distribution has a mean of 132 and a standard devistion of 20, what is the value that has a z score of 3.6
1 answer:
z = (x - μ)/σ
Substituting given values, you have
3.6 = (x - 132)/20
(20)(3.6) = x - 132 . . . . multiply by 20
72 + 132 = x . . . . . . . . .add 132
204 = x
The value that has a z-score of 3.6 is 204.
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I think 60
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12 and 15
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