Empirical Rule for ungrouped data

Empirical Rule for ungrouped data

Empirical rule is the general rule of thumb that applies to bell shaped (symmetrical) distribution. The empirical rule can be stated as :

• $68$% of the data will fall within one standard deviation of the mean,
• $95$% of the data will fall within two standard deviations of the mean,
• $99.7$% of the data will fall within three standard deviations of the mean.

Formula

Let $x_1,x_2,\cdots, x_n$ be $n$ sample observations. If the distribution of $x$ is approximately symmetrical, then

$68$% of the data falls in $\overline{x}\pm 1 s_x$

$95$% of the data falls in $\overline{x}\pm 2 s_x$

$99.7$% of the data falls in $\overline{x}\pm 3 s_x$

where,

• $\overline{x}=\dfrac{1}{n}\sum_{i=1}^{n}x_i$ is the sample mean,
• $s_x =\sqrt{\dfrac{1}{n-1}\bigg(\sum_{i=1}^{n}x_i^2-\dfrac{\big(\sum_{i=1}^n x_i\big)^2}{n}\bigg)}$ is the sample standard deviation.

Example

The following data gives the hourly wage rates (in dollars) of 10 employees of a company.

20,21,24,25,18,22,24,22,20,22.

Check empirical rule for the given data.

Solution

Sample mean

The sample mean of $X$ is

$$ \begin{aligned} \overline{x} &=\frac{1}{n}\sum_{i=1}^n x_i\\ &=\frac{218}{10}\\ &=21.8\text{ dollars} \end{aligned} $$

The average of hourly wage rate is $21.8$ dollars.

Sample variance

Sample variance of $X$ is

$$ \begin{aligned} s_x^2 &=\dfrac{1}{n-1}\bigg(\sum_{i=1}^{n}x_i^2-\frac{\big(\sum_{i=1}^n x_i\big)^2}{n}\bigg)\\ &=\dfrac{1}{9}\bigg(4794-\frac{(218)^2}{10}\bigg)\\ &=\dfrac{1}{9}\big(4794-\frac{47524}{10}\big)\\ &=\dfrac{1}{9}\big(4794-4752.4\big)\\ &= \frac{41.6}{9}\\ &=4.6222 \end{aligned} $$

Sample standard deviation

The standard deviation is the positive square root of the variance.

The sample standard deviation is

$$ \begin{aligned} s_x &=\sqrt{s_x^2}\\ &=\sqrt{4.6222}\\ &=2.1499 \text{ dollars} \end{aligned} $$

Thus the standard deviation of hourly wage rate is $2.1499$ dollars.

Empirical Rule

$68$% of the data falls in $\overline{x}\pm 1 s_x$.

i.e., ($21.8\pm 1*2.1499$) contains $68$% of the data values.

$\Rightarrow$ ($19.6501, 23.9499$) contains $68$% of the data values.

$95$% of the data falls in $\overline{x}\pm 2 s_x$.

i.e., ($21.8\pm 2*2.1499$) contains $95$% of the data values.

$\Rightarrow$ ($17.5002, 26.0998$) contains $95$% of the data values.

$99.7$% of the data falls in $\overline{x}\pm 3 s_x$.

i.e., ($21.8\pm 3*2.1499$) contains $99.7$% of the data values.

$\Rightarrow$ ($15.3503, 28.2497$) contains $99.7$% of the data values.

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