A class has average height of 68 inches and a standard division of 2 inches what height do you need to be taller than 90% of the
class please round your answer to the nearest thousandth.
1 answer:
Answer:
65.426
Step-by-step explanation:
Given that:
Mean height (m) = 68 inches
Standard deviation (s) = 2 inches
Height required to be taller than 90% of the class :
P(Z > x) = 0.9
The corresponding Zscore is - 1.282 (Z probability calculator)
Using the Zscore relation :
Zscore = (x - m) / s
-1.282 = (x - 68) / 2
-1.282 * 2 = x - 68
-2.564 = x - 68
-2.564 + 68 = x
x = 65.436
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Answer+Step-by-step explanation:
m = man's age
s = son's age
m=s+36
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Resolves to: s=10, m=46
So the man is 46 (=10+36).
In 8 years, son is 18 (=10+8), man is 54 (=3*18 and 46+8), thank you.
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