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Advanced Math Topics

Advanced Math Topics. 3.6 Chebyshev’s Theorem: Applying the Standard Deviation. We learned how to calculate the standard deviation (s) and we learned that a high value of s means the data is spread far from the mean and a low

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Advanced Math Topics

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  1. Advanced Math Topics 3.6 Chebyshev’s Theorem: Applying the Standard Deviation

  2. We learned how to calculate the standard deviation (s) and we learned that a high value of s means the data is spread far from the mean and a low value of s means the data is close to the mean. This next theorem gives us more concrete information of this idea. Chebyshev’s Theorem In any population or sample… Let k be any number equal to or greater than 1, Then the proportion of any distribution that lies Within k standard deviations of the mean is at least… 1 - 1 k2

  3. What does this theorem mean? Computing some values of k… What does this look like? 1 Remember k = the standard deviation = s 1 - k (Std. deviations) k2 1 At least 75% of the data will lie in this interval. 1 - = 0 = 0% 1 12 1 x + 2s x – 2s x 1 - = 3/4 = .75 = 75% 2 22 At least 88.89% of the data will lie in this interval. 1 1 - = 8/9 = .8889 = 88.89% 3 32 x + 3s x – 3s x

  4. What if we had values for the mean and standard deviations? x = 50 s = 18 Let’s use k = 2. 75 Then at least ______% of the data will lie within 2 standard deviations of the mean . 75% of the data or more lie within the interval x – 2s and x + 2s 50 – 2(18) and 50 + 2(18) 14 and 86 14 50 86 Chebyshev’s Theorem states that at least 75% of the data will lie in this interval.

  5. 1) In one month, the average file claimed with an insurance company was $831. The standard deviation was $150. Find the interval that will contain all values that are within 2.5 standard deviations of the mean. 831 – 2.5(150) and 831 + 2.5(150) $456 and $1206 At least what % of the data will lie between $456 and $1206? 1 1 - = .84 = 84% (2.5)2 At least 84% of the claims will be in the interval $456-$1206.

  6. 2) All G.E. light bulbs have a mean life of 1600 hours with a standard deviation of 100 hours. At least what % of the bulbs will have a mean life between 1450 and 1750 hours? How far is the upper and lower limit of the interval from the mean? How many standard deviations from the mean is this? 1750 – 1600 = 150 and 1600 – 1450 = 150 We know that k standard deviations is 150 thus k(100) = 150. k = 1.5 1 1 - = .5556 = 55.56% (1.5)2 At least 55.56% of the bulbs will have a life in the interval of 1450-1750 hours.

  7. Pafnuty Lvovich Chebyshev

  8. HW • P. 139 #1-5 Let’s do #1 together.

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