20% of 3 yards is how many fifths of 9 feet?
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Answer:
Step-by-step explanation:
The 90th percentile of a normally distributed curve occurs at 1.282 standard deviations. Similarly, the 10th percentile of a normally distributed curve occurs at -1.282 standard deviations.
To find the percentile for the television weights, use the formula:
, where
is the average of the set,
is some constant relevant to the percentile you're finding, and
is one standard deviation.
As I mentioned previously, 90th percentile occurs at 1.282 standard deviations. The average of the set and one standard deviation is already given. Substitute ,
, and
:
Therefore, the 90th percentile weight is 5.1282 pounds.
Repeat the process for calculating the 10th percentile weight:
The difference between these two weights is .