Find percentage of a sample given the mean and standard deviation

In summary, the average person drinks 78L of beer per year with a standard deviation of 25L. Using Chebyshev's inequality, we can estimate that at least 81.40% of people would drink between 20 and 136L of beer per year.
  • #1
maiad
102
0

Homework Statement


People drink on average, 78L of beer per year with a standard deviation of 25L. at least what percentage of people would drink between 20 and 136L per year?

The Attempt at a Solution



I found the z-scores of both and they came up to be +/- 2.32. Compared these values to the z-table and got P(Z<2.32)=0.0102 and P(Z<2.32)=0.9898. Then i found the percentage as followed... 0.9898-0.0102 which is around 98%. Which is wrong. Is my assumption that the graph is symmetrical correct?
 
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  • #2
maiad said:

Homework Statement


People drink on average, 78L of beer per year with a standard deviation of 25L. at least what percentage of people would drink between 20 and 136L per year?

The Attempt at a Solution



I found the z-scores of both and they came up to be +/- 2.32. Compared these values to the z-table and got P(-2.32<Z)=0.0102 and P(Z<2.32)=0.9898. Then i found the percentage as followed... 0.9898-0.0102 which is around 97%. Which is wrong. Is my assumption that the graph is symmetrical correct?

Sure, this is implicitly given to be a normal distribution, which is symmetrical.

First of all, your left-tail probability makes no sense as you wrote it. Shouldn't it be ##p(z<-2.32) = 0.0102##?

When you subtract those two, the answer isn't "around 97%", it's a lot closer to 98%. Did you try being a lot more precise with your answer?

Because of the symmetry, you could just as easily have taken ##p(-2.32<z<2.32) = 1 - 2\times 0.0102##.
 
  • #3
Curious3141 said:
Sure, this is implicitly given to be a normal distribution, which is symmetrical.

First of all, your left-tail probability makes no sense as you wrote it. Shouldn't it be ##p(z<-2.32) = 0.0102##?

When you subtract those two, the answer isn't "around 97%", it's a lot closer to 98%. Did you try being a lot more precise with your answer?

Because of the symmetry, you could just as easily have taken ##p(-2.32<z<2.32) = 1 - 2\times 0.0102##.

You're correct. The left tail is wrong as you stated. But even with more precision in my numbers, it's still a fair bit off from the actual answer which is 81.40%
 
  • #4
  • #5
I wouldn't implicitly consider it normal unless stated. Anyway, one can exploit some of available probabilistic inequalities. Since the Chebyshev's inequality is usually taught, try this one, maiad. Most likely, you will succeed to get your "correct" value :-)
 
  • #6
camillio said:
I wouldn't implicitly consider it normal unless stated. Anyway, one can exploit some of available probabilistic inequalities. Since the Chebyshev's inequality is usually taught, try this one, maiad. Most likely, you will succeed to get your "correct" value :-)

That actually does work out to the expected answer. And I think you're quite correct - considering the phrase "at least" in the question, they aren't expecting any assumption about the underlying distribution.

Maiad - please take Camillio's advice, Chebyshev's inequality seems to be what's expected.
 

1. How do you calculate the percentage of a sample given the mean and standard deviation?

To find the percentage of a sample given the mean and standard deviation, you need to first calculate the Z-score of the sample using the formula (sample mean - population mean) / standard deviation. Then, you can use a Z-score table or a statistical software to find the corresponding percentage.

2. Can the percentage of a sample be more than 100%?

No, the percentage of a sample cannot be more than 100%. The percentage represents a portion or fraction of the whole, so it cannot exceed the total value of 100%.

3. Is the percentage of a sample always the same as the mean?

No, the percentage of a sample and the mean are two different measures. The mean represents the average value of the sample, while the percentage represents the proportion of the sample in relation to the whole population.

4. What does the standard deviation tell us about the sample?

The standard deviation measures the spread of the sample data around the mean. It tells us how much the individual data points deviate from the average, giving us a sense of the variability or dispersion of the sample.

5. Can you calculate the percentage of a sample without the mean and standard deviation?

No, in order to calculate the percentage of a sample, you need to have both the mean and the standard deviation. These two measures are essential for determining the relative position of the sample within the population and cannot be calculated without them.

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