Standard Deviation of Continuous Series

A continuous series groups data into classes such as 0-10, 10-20 and 20-30, and records how many items fall in each class. Once data is grouped like this, we no longer know the exact value of each item. A salary of Rs. 13,000 and a salary of Rs. 19,000 both sit inside the class 10-20 (in thousands), and the table does not tell them apart. We still want to know how spread out the data is, and the standard deviation still measures that. The only change is that every class must first be represented by one number.

\[ m = \frac{\text{lower limit} + \text{upper limit}}{2} \]

For the class 20-30, the mid-value is \( m = \frac{20 + 30}{2} = 25 \). Every student who scored somewhere from 20 up to 30 is treated as if they scored 25.

In every method below, \( N \) is the sum of the frequencies, \( N = \sum f \), and the mid-values \( m \) take the place of the unknown individual values. There are four methods. All four give the same standard deviation. They differ only in how heavy the arithmetic is, so the skill is choosing the one that keeps your numbers small.

• Actual mean method: deviations are measured from the true mean.
• Direct method: the mid-values are used as they are, with no mean subtracted.
• Short-cut method or assumed mean method: deviations are measured from a convenient assumed mean.
• Step deviation method: deviations from the assumed mean are divided by the class width to make them tiny.

In this method we first calculate the actual mean of the data. Then we measure how far each mid-value lies from that mean, square each distance, weight it by its frequency, find the average and take the square root. This is formula (i):

\[ \sigma = \sqrt{\frac{\sum f(m - \bar{X})^2}{N}} \]

Here \( m \) is the mid-value, \( N \) is the sum of the frequencies and \( \bar{X} \) is the mean, found as \( \bar{X} = \frac{\sum fm}{N} \). The method follows these steps:

• (a) Find the mid-value \( m \) of each class.
• (b) Multiply each frequency \( f \) by its mid-value to get \( fm \), and add these to get \( \sum fm \).
• (c) Find the mean \( \bar{X} = \frac{\sum fm}{N} \).
• (d) Subtract \( \bar{X} \) from each \( m \) and square the answer to get \( (m - \bar{X})^2 \).
• (e) Multiply each \( (m - \bar{X})^2 \) by its frequency to get \( f(m - \bar{X})^2 \).
• (f) Add these to get \( \sum f(m - \bar{X})^2 \).
• (g) Put the totals into formula (i).

Example 1. The monthly salaries of 100 employees of a company are given below, in thousands of rupees. Find the standard deviation by the actual mean method, and its coefficient.

Step 1: The mid-values are 5, 15, 25, 35, 45 and 55. Multiplying each by its frequency and adding gives \( \sum fm = 3100 \), with \( N = 100 \). So the mean is:

\[ \bar{X} = \frac{\sum fm}{N} = \frac{3100}{100} = 31 \]

Step 2: Since the mean is 31, each deviation is \( m - 31 \). Squaring each deviation and multiplying by its frequency gives this working table:

Step 3: Substituting \( \sum f(m - \bar{X})^2 = 16800 \) and \( N = 100 \) into formula (i):

\[ \sigma = \sqrt{\frac{16800}{100}} \]

Step 4: Dividing, then taking the square root:

\[ \sigma = \sqrt{168} = 12.96 \]

So the standard deviation of the salaries is 12.96 thousand rupees. A second number tells us how large this spread is compared with the mean.

Step 5: Dividing the standard deviation by the mean gives the coefficient:

\[ \frac{\sigma}{\bar{X}} = \frac{12.96}{31} = 0.418 \]

Therefore \( \sigma = 12.96 \) and its coefficient is 0.418. The actual mean method is comfortable here because the mean came out as the whole number 31. If the mean had been 31.47, every deviation would have been a messy decimal, and one of the next three methods would be wiser.

The direct method uses neither the actual mean nor an assumed mean. The mid-values are used straight away, together with their squares. It suits data whose mid-values are small numbers, because then the squares stay manageable. This is formula (ii):

\[ \sigma = \sqrt{\frac{\sum fm^2}{N} - \left(\frac{\sum fm}{N}\right)^2} \]

Here \( m \) is the mid-value and \( N \) is the sum of the frequencies. The term \( fm^2 \) means \( f \times m^2 \). The quickest way to get it is to multiply the \( fm \) column by \( m \) once more.

• (a) Find the mid-value \( m \) of each class.
• (b) Multiply each \( f \) by its \( m \) to get \( fm \).
• (c) Multiply each \( fm \) by its \( m \) again to get \( fm^2 \).
• (d) Add the columns to get \( \sum fm \) and \( \sum fm^2 \).
• (e) Put the totals into formula (ii).

Example 2. The marks obtained by 20 students are given below. Find the standard deviation by the direct method.

Step 1: The mid-values, the products \( fm \) and the products \( fm^2 \) are worked out in a table:

Step 2: Substituting \( N = 20 \), \( \sum fm = 460 \) and \( \sum fm^2 = 13300 \) into formula (ii):

\[ \sigma = \sqrt{\frac{13300}{20} - \left(\frac{460}{20}\right)^2} \]

Step 3: Dividing inside the root. Since \( \frac{13300}{20} = 665 \) and \( \frac{460}{20} = 23 \), this becomes:

\[ \sigma = \sqrt{665 - 23^2} = \sqrt{665 - 529} \]

Step 4: Subtracting and taking the square root:

\[ \sigma = \sqrt{136} = 11.66 \]

So the standard deviation of the marks is 11.66.

When the actual mean comes out as an awkward decimal, subtracting it from every mid-value is slow and easy to get wrong. Instead, we pick a convenient mid-value near the middle of the table and treat it as if it were the mean. Deviations measured from it are small whole numbers. The formula then corrects for the fact that this value is not the true mean.

This is formula (iii):

\[ \sigma = \sqrt{\frac{\sum fd^2}{N} - \left(\frac{\sum fd}{N}\right)^2} \]

Here \( d = m - A \), \( m \) is the mid-value, \( N \) is the sum of the frequencies and \( A \) is the assumed mean. The steps are:

• (a) Find the mid-value \( m \) of each class.
• (b) Take the middle mid-value as the assumed mean \( A \), and find each deviation \( d = m - A \).
• (c) Multiply each \( d \) by its frequency to get \( fd \).
• (d) Multiply each \( fd \) by its \( d \) again to get \( fd^2 \).
• (e) Add the columns to get \( \sum fd \) and \( \sum fd^2 \).
• (f) Put the totals into formula (iii).

Example 3. The marks of 60 students are given below. Find the standard deviation by the short-cut method.

Step 1: The mid-values are 25, 35, 45, 55 and 65. The middle one, 45, is taken as the assumed mean, so \( d = m - 45 \). The working table is:

Step 2: Substituting \( N = 60 \), \( \sum fd = -130 \) and \( \sum fd^2 = 9100 \) into formula (iii):

\[ \sigma = \sqrt{\frac{9100}{60} - \left(\frac{-130}{60}\right)^2} \]

Step 3: Dividing inside the root. Since \( \frac{9100}{60} = 151.67 \) and \( \frac{-130}{60} = -2.167 \), whose square is 4.694, this becomes:

\[ \sigma = \sqrt{151.67 - 4.694} \]

Step 4: Subtracting and taking the square root:

\[ \sigma = \sqrt{146.97} = 12.12 \]

So the standard deviation of the marks is 12.12. Notice that we never had to find the actual mean to get this answer.

Suppose the mid-values are large, such as 50, 150, 250, 350 and 450. Even after subtracting an assumed mean of 250, the deviations are -200, -100, 0, 100 and 200, and their squares run into tens of thousands. When all the classes have the same width, every one of these deviations is a multiple of that width. Dividing by the width shrinks them to -2, -1, 0, 1 and 2. We work with these tiny numbers and multiply back by the width at the very end.

This is formula (iv):

\[ \sigma = \sqrt{\frac{\sum fd'^2}{N} - \left(\frac{\sum fd'}{N}\right)^2} \times h \]

Here \( m \) is the mid-value, \( N \) is the sum of the frequencies, \( h \) is the class width (the size of each class interval) and \( A \) is the assumed mean. The steps are:

• (a) Find the mid-value \( m \) of each class.
• (b) Take the middle mid-value as the assumed mean \( A \), subtract it from each mid-value and divide by the class width \( h \) to get \( d' \).
• (c) Multiply each \( d' \) by its frequency to get \( fd' \).
• (d) Multiply each \( fd' \) by its \( d' \) again to get \( fd'^2 \).
• (e) Add the columns to get \( \sum fd' \) and \( \sum fd'^2 \).
• (f) Put the totals into formula (iv), remembering the final multiplication by \( h \).

Example 4. The hourly wages of 25 workers are given below. Find the standard deviation by the step deviation method.

Step 1: The mid-values are 50, 150, 250, 350 and 450. Taking \( A = 250 \) and \( h = 100 \) gives \( d' = \frac{m - 250}{100} \). The working table is:

Step 2: Substituting \( N = 25 \), \( \sum fd' = -11 \), \( \sum fd'^2 = 39 \) and \( h = 100 \) into formula (iv):

\[ \sigma = \sqrt{\frac{39}{25} - \left(\frac{-11}{25}\right)^2} \times 100 \]

Step 3: Dividing inside the root. Since \( \frac{39}{25} = 1.56 \) and \( \left(\frac{-11}{25}\right)^2 = (-0.44)^2 = 0.1936 \), this becomes:

\[ \sigma = \sqrt{1.56 - 0.1936} \times 100 \]

Step 4: Subtracting and taking the square root:

\[ \sigma = \sqrt{1.3664} \times 100 = 1.1689 \times 100 \]

Step 5: Multiplying by the class width:

\[ \sigma = 116.89 \]

So the standard deviation of the hourly wages is Rs. 116.89.

Four methods, one answer. As a check, take the salary data of Example 1 again and use the step deviation method with \( A = 35 \) and \( h = 10 \). The step deviations are -3, -2, -1, 0, 1 and 2, which give \( \sum fd' = -40 \) and \( \sum fd'^2 = 184 \) with \( N = 100 \). Substituting into formula (iv):

\[ \sigma = \sqrt{\frac{184}{100} - \left(\frac{-40}{100}\right)^2} \times 10 \]

Simplifying inside the root, \( 1.84 - 0.16 = 1.68 \), so:

\[ \sigma = \sqrt{1.68} \times 10 = 12.96 \]

Before you start a question, look at the mid-values and ask which method will give the lightest working. The table below puts the four side by side.

If a question names the method, use that method, even if another would be quicker. The examiner is checking that you can carry out the method asked for.

Standard deviation is measured in the same units as the data. That makes it unfair for comparing two groups whose means are very different. A spread of Rs. 110 is huge for pocket money that averages Rs. 200, but tiny for a salary that averages Rs. 40,000. To compare fairly, we express the standard deviation as a percentage of the mean.

\[ \text{C.V.} = \frac{\sigma}{\bar{X}} \times 100\% \]

Example 5. The salaries of the employees of two industries are analysed in the table below. Find which industry has more uniformity in its salary distribution.

Here the mean salary of industry A is \( \bar{X}_1 = 50000 \) and its standard deviation is \( \sigma_1 = 100 \). For industry B, \( \bar{X}_2 = 40000 \) and \( \sigma_2 = 110 \).

Step 1: Finding the coefficient of variation of industry A:

\[ \text{C.V.}_A = \frac{100}{50000} \times 100\% = 0.2\% \]

Step 2: Finding the coefficient of variation of industry B:

\[ \text{C.V.}_B = \frac{110}{40000} \times 100\% = 0.275\% \]

Step 3: Comparing the two. Since \( 0.2\% < 0.275\% \), industry A has the smaller coefficient of variation. So the salaries of industry A are more uniform than those of industry B.