Chebyshev’s Theorem shows you how to use the mean and standard deviation to find the percentage of total observations that fall within a given interval around the mean.

Chebyshev's theorem

For every number k greater than 1 at least (1 –

1
/
k2

) of the data values ​​are k standard deviations of the mean.

Important things to keep in mind about Chebyshev’s theorem

The distance between µ and µ + kσ is kσ, since µ + kσ – µ = kσ

The distance between µ and µ – kσ is kσ, since µ – (µ – kσ) = µ – µ – – kσ = 0 – – kσ = kσ

The answer to (1 –

1
/
k2

) is usually expressed as a percentage after you finish calculating.

Notice that we are saying that k is greater than 1 or at least 2. k must be 2 because we get 0 when k = 1.

If k = 1 then (1 –

1
/
k2

)

= (1 –

1
/
1

) = 1 – 1 = 0

How to find the standard deviation using Chebyshev’s theorem

Now let’s see how we can apply Chebyshev’s theorem.

For example, suppose you want to find the percentage of values ​​in a data set that are within 2 standard deviations of the mean.

Just replace k = 2 in the formula.

= (1 –

1
/
4th

) = 1 – 0.25 = 0.75

0.75 as percent is 75%

Therefore, 75% of the values ​​in a data set are within 2 standard deviations of the mean.

How do we use the mean and standard deviation to find the percentage of total observations that fall within a given interval around the mean?

Assuming µ = 39 and σ = 5, find the percentage of the values ​​that are between 29 and 49 of the mean.

29 ——– 39 ——————– 49

We just need to find k and there are 2 ways to do it.

First notice, as mentioned earlier, that the distance between 39 and 49 is equal to kσ

Since the distance between 39 and 49 is 10, kσ = 10

kσ = 10

σ = 5, so 5k = 10

Since 5 times 2 = 10, k = 2.

We calculated the percentage earlier when k = 2. We found it to be 75%.

Therefore, 75% of the values ​​are between 29 and 49 of the mean.

The second To find the answer, you need to find that µ + kσ = 49

29 ————————– 39 ———————- —- 49

µ – kσ µ µ + kσ

µ + kσ = 49

39 + kσ = 49

39-39 + kσ = 49-39

0 + kσ = 10

kσ = 10

5k = 10 and k = 2


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