Explanation Definition: 1. If a data set contains n data values, the mean x ¯ of the data set is x ¯ = ∑ x n . 2. If x is a data value in a set whose mean is x ¯ , then x − x ¯ is called x’s deviation from the mean. 2. We denote the standard deviation of a sample of n data values by s = ∑ ( x − x ¯ ) 2 n − 1 . 4. We calculate the standard deviation, s, of a sample that is given as a frequency distribution as s = ∑ ( x − x ¯ ) 2 ⋅ f n − 1 . Here f is the frequency of data value x , and n = ∑ f is the number of data values in the distribution. Calculation: We will consider here two examples to prove that it is not true that the more the numbers there are in a distribution, the larger the standard deviation. Suppose we have the following data. Data value, x Frequency, f 17 1 18 3 19 2 20 2 21 1 22 1 Find the mean of the given data set. Begin by finding the sum of the given data. Sum of 19, 18, 20, 19, 21, 20, 22, 18, 18, and 17 is 192. There are 10 values in the given data. Therefore, the mean is x ¯ = 192 10 = 19.2. Now find the standard deviation. Observe that there are some repeated values in the given data. Therefore, we will use a frequency table to compute the standard deviation. The frequency of 17 is 1. That is, 17 comes only once in the data set. Similarly, the frequency of 18 is 3. The frequency of 19 is 2. The frequency of 20 is 2. The frequency of 21 is 1. The frequency of 22 is 1. Now find the product x ⋅ f for each data value. Notice that 17 ⋅ 1 = 17 , 18 ⋅ 3 = 54 , 19 ⋅ 2 = 38 , 20 ⋅ 2 = 40 , 21 ⋅ 1 = 21 , and 22 ⋅ 1 = 22. The deviation is the difference between each product value and the mean. Therefore, the deviation of 17 from the mean 19.2 is 17 − 19.2 = − 2.2. The deviation of 18 from the mean 19.2 is 18 − 19.2 = − 1.2. The deviation of 19 from the mean 19.2 is 19 − 19.2 = − 0.2. The deviation of 20 from the mean 19.2 is 20 − 19.2 = 0.8. The deviation of 21 from the mean 19.2 is 21 − 19.2 = 1.8. The deviation of 22 from the mean 19.2 is 22 − 19.2 = 2.8. Square each of the deviation value. Data value, x Frequency, f Product, x ⋅ f Deviation, ( x − 19.2 ) Deviation squared, ( x − 19.2 ) 2 17 1 17 ⋅ 1 = 17 17 − 19.2 = − 2.2 ( − 2.2 ) 2 = 4.84 18 3 18 ⋅ 3 = 54 18 − 19.2 = − 1.2 ( − 1.2 ) 2 = 1.44 19 2 19 ⋅ 2 = 38 19 − 19.2 = − 0.2 ( − 0.2 ) 2 = 0.04 20 2 20 ⋅ 2 = 40 20 − 19.2 = 0.8 ( 0.8 ) 2 = 0.64 21 1 21 ⋅ 1 = 21 21 − 19.2 = 1.8 ( 1.8 ) 2 = 3.24 22 1 22 ⋅ 1 = 22 22 − 19.2 = 2.8 ( 2.8 ) 2 = 7.84 ∑ f = 10 ∑ ( x ⋅ f ) = 192 Now multiply each deviation squared value by the frequency of the corresponding data value. For data value 17, ( 17 − 19.2 ) 2 ⋅ 1 = 4.84 ⋅ 1 = 4.84. For data value 18, ( 18 − 19.2 ) 2 ⋅ 3 = 1.44 ⋅ 3 = 4.32. For data value 19, ( 19 − 19.2 ) 2 ⋅ 2 = 0.04 ⋅ 2 = 0.08. For data value 20, ( 20 − 19.2 ) 2 ⋅ 2 = 0.64 ⋅ 2 = 1.28. For data value 21, ( 21 − 19.2 ) 2 ⋅ 1 = 3.24 ⋅ 1 = 3.24. For data value 22, ( 22 − 19.2 ) 2 ⋅ 1 = 7.84 ⋅ 1 = 7.84. The sum of these values is 4.84 + 4.32 + 0.08 + 1.28 + 3.24 + 7.84 = 21.6 Remember that ∑ f is the number of data values that we have been representing by n . The standard deviation is therefore s = ( x − 19.2 ) 2 ⋅ f n − 1 = 21.6 9 = 2.4 = 1.549193338483. So, we say that when the mean is 19.2, the standard deviation is 1.549193338483. Now, consider another data. Number, x Frequency, f 2 1 3 1 4 0 5 2 6 3 7 4 8 4 9 3 10 2 First find the mean of the given data set. To compute the mean, let us find all products x ⋅ f of the given values and their respective frequencies. For x = 2 and f = 1 , the product x ⋅ f = 2 ⋅ 1 = 2. For x = 3 and f = 1 , the product x ⋅ f = 3 ⋅ 1 = 3