Tomi's Challenges - #21

Probability Level 3

Let

X = ( ∣ t − 15327846 ∣ ∣ t − 16327742 ∣ ∣ t − 17345235 ∣ ∣ t − 18478329 ∣ ∣ t − 19823877 ∣ ∣ t − 20034235 ∣ ∣ t − 21483984 ∣ ∣ t − 22534536 ∣ ∣ t − 23543789 ∣ ∣ t − 24234513 ∣ ) 1 10 \small X=(|t-15327846||t-16327742||t-17345235||t-18478329||t-19823877||t-20034235||t-21483984||t-22534536||t-23543789||t-24234513|)^{\frac{1}{10}}

where

t = 15327846 + 16327742 + 17345235 + 18478329 + 19823877 + 20034235 + 21483984 + 22534536 + 23543789 + 24234513 10 \small t=\frac{15327846+16327742+17345235+18478329+19823877+20034235+21483984+22534536+23543789+24234513}{10}

Out of the given options, choose the most precise upper bound of X X , without using a calculator.

X ≤ 9243450 X\leq 9243450 X ≤ 9243790 X\leq 9243790 X ≤ 8906950 X\leq 8906950 X ≤ 9243500 X\leq 9243500 X ≤ 8906700 X\leq 8906700 X ≤ 8906900 X\leq 8906900 X ≤ 9243300 X\leq 9243300 X ≤ 9243600 X\leq 9243600

Tomislav Franov by Tomislav Franov , 17, Croatia

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1 solution

Tomislav Franov
May 29, 2021

Using the fact that the geometric mean is always less or equal to the quadratic mean, we get:

X = ( ∣ t − 15327846 ∣ ∣ t − 16327742 ∣ ∣ t − 17345235 ∣ ∣ t − 18478329 ∣ ∣ t − 19823877 ∣ ∣ t − 20034235 ∣ ∣ t − 21483984 ∣ ∣ t − 22534536 ∣ ∣ t − 23543789 ∣ ∣ t − 24234513 ∣ ) 1 10 ≤ 1 10 ( ( t − 15327846 ) 2 + ( t − 16327742 ) 2 + ( t − 17345235 ) 2 + ( t − 18478329 ) 2 + ( t − 19823877 ) 2 + ( t − 20034235 ) 2 + ( t − 21483984 ) 2 + ( t − 22534536 ) 2 + ( t − 23543789 ) 2 + ( t − 24234513 ) 2 ) X=(|t-15327846||t-16327742||t-17345235||t-18478329||t-19823877||t-20034235||t-21483984||t-22534536||t-23543789||t-24234513|)^{\frac{1}{10}}\leq \sqrt{\frac{1}{10}((t-15327846)^2+(t-16327742)^2+(t-17345235)^2+(t-18478329)^2+(t-19823877)^2+(t-20034235)^2+(t-21483984)^2+(t-22534536)^2+(t-23543789)^2+(t-24234513)^2)}

But the quadratic mean is just the formula for the standard deviation of this data set with really large numbers and no outliers, which means that the standard deviation can be well approximated using the range:

R x = x m a x − x m i n 4 = 24234513 − 15327846 4 = 8906667 4 R_x=\frac{x_{max}-x_{min}}{4}=\frac{24234513-15327846}{4}=\frac{8906667}{4}

Therefore the best upper bound from the options above is the smallest, 8906700 8906700 .

1 Helpful 2 Interesting 1 Brilliant 0 Confused

The upper bound on the standard deviation is 1/2 the range, so 8906667/2. The closest number to this of those given is the smallest, 8906700.

G Silb - 1 week, 6 days ago

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https://www.thoughtco.com/range-rule-for-standard-deviation-3126231

Actually, I just came across this article. Shouldn't it be 8906667/4 instead?

Tomislav Franov - 1 week, 5 days ago

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No, range/4 is a very rough approximation for the standard deviation of a data set; it is not an upper bound. For example, the actual standard deviation of your data set is 3,050,512 which is roughly range/3 > range/4.

It seems like the thrust of your problem as you have stated it is for folks to recognize that you are talking about the standard deviation, and then to get a bound based on a simple hand calculation. The simplest bound is range/2. In the case where all the data points are at the endpoints of the interval, this is the actual standard deviation. For more clustered data, it is merely an upper bound (and range/4 is a guess for what the actual value would be).

Thanks for posting the problem!

G Silb - 1 week, 2 days ago

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@G Silb – Yeah but the problem stated to find the upper bound of X, not the upper bound of the standard deviation, so approximating the standard deviation could also be a way to do it, right?

Tomislav Franov - 1 week, 2 days ago

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