Theoretical Mean Formula
\(\begin{array}{l}\mu\end{array} \) = \(\begin{array}{l}\frac{x+y}{2}\end{array} \) |
Standard Deviation Formula
\(\begin{array}{l}\sigma = \sqrt{ \frac{(y-x)^{2}}{12}}\end{array} \) |
Example of Uniform Distribution
Example 1: The data in the table below are 55 times a baby yawns, in seconds, of a 9-week-old baby girl.
| 10.4 | 19.6 | 18.8 | 13.9 | 17.8 | 16.8 | 21.6 | 17.9 | 12.5 | 11.1 | 4.9 |
| 12.8 | 14.0 | 22.8 | 20.8 | 15.9 | 16.3 | 13.4 | 17.1 | 14.5 | 19.0 | 22.8 |
| 1.3 | 0.7 | 8.9 | 11.9 | 10.9 | 7.3 | 5.9 | 3.7 | 17.9 | 19.2 | 9.8 |
| 5.8 | 6.9 | 2.6 | 5.8 | 21.7 | 11.8 | 3.4 | 2.1 | 4.5 | 6.3 | 10.7 |
| 8.9 | 9.7 | 9.1 | 7.7 | 10.1 | 3.5 | 6.9 | 7.8 | 11.6 | 13.8 | 18.6 |
- Solution:
- The sample mean = 11.49
- The sample standard deviation = 6.23.
As assumed, the yawn times in secs, it follows a uniform distribution between 0 to 23 seconds(Inclusive).
So, it is equally likely that any yawning time is from 0 to 23.
In this example,
The theoretical mean =
standard deviation =
standard deviation =
Example 2: The data given below is about the number of passengers on 35 different cabs. The sample mean and the sample standard deviation of the data are 7.9 and 4.33, respectively. The data follow a uniform distribution where all values between and including zero and 14 are equally likely. Identify the values of x and y. Calculate the theoretical mean and standard deviation.
| 1 | 7 | 3 | 5 | 6 |
| 12 | 11 | 10 | 13 | 4 |
| 10 | 13 | 12 | 10 | 0 |
| 4 | 4 | 0 | 11 | 4 |
| 11 | 14 | 6 | 2 | 4 |
| 2 | 11 | 10 | 6 | 11 |
| 14 | 13 | 12 | 9 | 4 |
Solution:
Given,
Sample mean = 7.9
Sample standard deviation = 4.33
From the given data, it is clear that the distribution lies in the interval between 0 and 14.
Thus, x = 0, y = 14
Theoretical mean = μ = (x + y)/2 = (0 + 14)/2 = 7
Theoretical standard deviation = σ = √[(x – y)2/12]
= √[(0 – 14)2/12
= √(196/2)
= √98
= 9.899
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