Chebyshev's Theorem and income

[PARAJON]

I would like your assistance with the following:



The mean income of a group of sample observations is $500; the standard deviation is $40. According to Chebyshev’s theorem, at least what percent of the incomes will lie between $400 and $600?

Possible answer:

At least 8/9 or 89% of all incomes will fall within 3 standard deviations of the mean

Is this correct...?

 

[Janitor]

400 and 600 are each 2.5 s.d.'s from the mean of 500. So if I use 1-1/x^2, I get 84%. Does this sound right?

 

[PARAJON]

i DON'T KNOW. I thought that 89% was the only answer

 

[selfAdjoint]

Where did you get your three standard deviations? Janitor is right; if the mean is 500, then the lower limit is 400 = 500 - 100, and the upper limit is 600 = 500 + 100. And if the sd is 40 then the spread in sd is 100/40 = 2.5 sd. So you work from that and get Janitor's result.

Was there an example in the book with sd = 3? You have to work it out each time for each problem.

 

[HallsofIvy]

As everyone else has pointed out "At least 8/9 or 89% of all incomes will fall within 3 standard deviations of the mean" is a TRUE statement but irrelevant because this problem is asking about 2.5 standard deviations, not 3.

 

[piro829]

What work did you use to get this answer?

 

[HallsofIvy]

Well, the main work was looking up Chebyshev's theorem! Once you know what it says, you do what Janitor already said:

400 and 600 are each 2.5 s.d.'s from the mean of 500. So if I use 1-1/x^2, I get 84%.

 

[TreaNesIadi]

Thanks All!

You guys just saved me with this! I was so confused with this theorum and my spanglish speaking teacher took but 3 minutes to go over it! Thank you so much to the one that posted the question and all the replies-- helped me heaps! Thanks!

 

[slimrs13]

Can anyone help me with this, using Chebychev's Theorem?


Old Faithful is a famous geyser at Yellowstone National
Park. From a sample with n = 32, the mean duration of Old Faithful's eruptions is
3.32 minutes and the standard deviation is 1.09 minutes. Using the Chebychev's
Theorem, determine at least how many of the eruptions lasted between 1.14
minutes and 5.5 minutes.

The answer in the book says 24...but I don't know how to come up with that.

 

[ethame]

you have to do as above but then go one step further...to get that 24...

First create a line as below to help you visual the standard deviations...

x-3s-------x-2s-------x-s-------x----------x+s--------x+2s------x+3s
.05--------1.14------2.23-------3.32------4.41--------5.5--------6.59

Chebystev's Theorem states that 3/4 of data lies between x-2s and x+2s
and 8/9 lies between x-3s and x+3s...

Add the data given in the original problem to see how many standard deviations you are within so then you can use the properties of the theorem just mentioned or use the equation: 1- (1/k^2)

According to your question: between 1.14 and 5.5 are 2 standard deviations (x-2s, x+2s)
therefore. We just said above that the theroem states that for this deviation 3/4 of data lies here. But if you want to double check you can use the equaiton:

1-(1/2^2) = 3/4 of the data is within 1.14min and 5.5min

NOW: 3/4 * 32 (number of values) = 24 !

 

[MRAguy]

I would like your assistance with the following:



The mean income of a group of sample observations is $500; the standard deviation is $40. According to Chebyshev’s theorem, at least what percent of the incomes will lie between $400 and $600?

Possible answer:

At least 8/9 or 89% of all incomes will fall within 3 standard deviations of the mean

Is this correct...?

Proof
2
I recall Chebyshev required 1 - 1 / k
2
1 - 1 / K = .89
2 2 2
.89 = 1 / K = K = 1 / .89 = K = 1.123595506 = k = 1.0599

Therefore: 40 +/- 1.0599 (500)


40 +/- 529.95 = 489.95 to 569.95

 

[MRAguy]

Proof
2
I recall Chebyshev required 1 - 1 / k
2
1 - 1 / K = .89
2 2 2
.89 = 1 / K = K = 1 / .89 = K = 1.123595506 = k = 1.0599

Therefore: 40 +/- 1.0599 (500) 40 +/- 529.95 = 489.95 to 569.95