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

Mathematics

1 answer:

nika2105 [10]3 years ago

4 0

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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&\stackrel{mL}{amount}&\stackrel{\%}{concentration}&\stackrel{amount}{concentration}\\
&------&------&------\\
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\textit{second brand}&y&0.14&0.14y\\
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x+y=230\implies \boxed{y}=230-x\\
0.09x+0.14y=29.9\\
----------\\
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\end{cases}
\\\\\\
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\\\\\\
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