Question

In: Statistics and Probability

Use the following data X                      f(X) 0 0.10 1 0.15 2 0.30 3 0.20 4 0.15...

  1. Use the following data
  2. X                      f(X)

    0

    0.10

    1

    0.15

    2

    0.30

    3

    0.20

    4

    0.15

    5

    0.10

  3. Graph the probability distribution of X
  4. Calculate ?" and ?$.
  5. Calculate the interval (?" ± 2?"). Superimpose this interval on the graph of the probability distribution of X. Also, calculate the interval (?" ± ?"). What proportion of the measurements will fall within these intervals? Does this result agree with Chebyshev’s Theorem? The Empirical Rule?

Solutions

Expert Solution

The given PDF is graphed as a bar chart as:

The following table would be used for the computations now:

x f(x) xf(x) x^2*f(x)
0 0.1 0 0
1 0.15 0.15 0.15
2 0.3 0.6 1.2
3 0.2 0.6 1.8
4 0.15 0.6 2.4
5 0.1 0.5 2.5
1 2.45 8.05

The mean and standard deviation are computed using the above table last row of sum of columns as:

Therefore 2.45 is the required mean value here.

The standard deviation now is computed here as:

Therefore 1.43 is the required standard deviation here.

These are plotted on the graph as well as the ranges are plotted here as:

Mean - Std Dev = 2.45 - 1.43 = 1.02
Mean + Std Dev = 2.45 + 1.43 = 3.88
Mean - 2*Std Dev = 2.45 - 2*1.43 = -0.41
Mean + 2*Std Dev = 2.45 + 2*1.43 = 5.31

Note that the middle line represents the mean while the other two lines represent distances 1 standard deviation from mean from either side.

Note that P(X = 2) + P(X = 3) lies within 1 standard deviation of the mean that is 0.3 + 0.2 = 0.5 that is 50%

Also, we can see that within 2 standard deviations of the mean which is from (-0.41, 5.31), 100% of the observations lies within this range.

According to the empirical rule, 95% of the observations lies within 2 standard deviations of the mean and 68% lies within 1 standard deviation of the mean. Therefore empirical rule is not valid here ( anyways it is valid for only normal distribution )

According to chebyshev's inequality theorem, at least 1 - 1/k2 of the observations lies within k standard deviations of the mean.

For k = 1, 1 - 1/k2 = 0, therefore at least 0% lies within 1 standard deviation. As 50% lies this is applicable here.
Also for k = 2, 1 - 1/22 = 75%, as 100% lies therefore at least 75% is still valid. Therefore chebyshev's theorem is applicable for the given probability distribution here.


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