Measure Of Central Tendency

The assumed mean method is used to find the arithmetic mean of grouped data. This method is recommended for large datasets because it simplifies calculations.

Let A be an assumed mean and $d_i$ be the deviation of each data point $x_i$ or class mark from A, then $d_i = x_i - A$.

$x_i = d_i + A \quad (4.2)$

Substituting equation (4.2) into the common mean formula $\bar{x} = \frac{\sum_{i=1}^{n} f_ix_i}{N}$ (4.1) gives:

$\bar{x} = \frac{\sum_{i=1}^{n} f_i(d_i + A)}{N} = \frac{\sum_{i=1}^{n} f_id_i + \sum_{i=1}^{n} f_iA}{N} = \frac{\sum_{i=1}^{n} f_id_i + A\sum_{i=1}^{n} f_i}{N}$

Since $\sum_{i=1}^{n} f_i = N$, we have:

$\bar{x} = \frac{\sum_{i=1}^{n} f_id_i + AN}{N} = \frac{\sum_{i=1}^{n} f_id_i}{N} + A$

Therefore, the arithmetic mean by the assumed mean method is given by:

$\bar{x} = A + \frac{\sum_{i=1}^{n} f_id_i}{N} \quad (4.3)$

Note: The assumed mean can be any class mark, but it's preferable to choose the class mark of the modal class or the central class.

Example 5

The following are tonnes of cement transported from the warehouse to retailers for 24 consecutive days:

66, 75, 79, 56, 61, 77, 92, 75, 78, 51, 82, 85, 59, 66, 91, 98, 90, 73, 71, 83, 85, 91, 77, 70.

Find the mean tonnes of the distribution using classes of width 10, starting with the class intervals $50 \le x < 60$, $60 \le x < 70$, ...

Solution:

Grouping the data and taking the assumed mean A to be 75 tonnes:

From the formula $\bar{x} = A + \frac{\sum_{i=1}^{n} f_id_i}{N}$:

$\bar{x} = 75 + \frac{50}{24} = 75 + 2.08333 \approx 77.1$

Therefore, the mean tonnes of the cement transported from the warehouse to retailers is approximately 77.1 tonnes.

The coding method (also called step deviation) simplifies calculations when the values of x, or both x and f, are large. It involves taking deviations from an arbitrary value A and then dividing by a common class size c.

For grouped or continuous frequency data with class intervals of class size c and deviation $d_i$, the coding number $u_i$ is calculated as:

$u_i = \frac{d_i}{c} \quad \text{for } i = 1, 2, 3, ..., n$

Since $d_i = x_i - A$, we have:

$u_i = \frac{x_i - A}{c} \quad (4.4)$

Rearranging equation (4.4) gives:

$x_i = cu_i + A \quad (4.5)$

Substituting equation (4.5) into the common mean formula $\bar{x} = \frac{\sum_{i=1}^{n} f_ix_i}{N}$ (4.1) gives:

$\bar{x} = \frac{\sum_{i=1}^{n} f_i(cu_i + A)}{N} = \frac{\sum_{i=1}^{n} f_icu_i + \sum_{i=1}^{n} f_iA}{N} = \frac{c\sum_{i=1}^{n} f_iu_i + A\sum_{i=1}^{n} f_i}{N}$

Since $\sum_{i=1}^{n} f_i = N$, we have:

$\bar{x} = \frac{c\sum_{i=1}^{n} f_iu_i + AN}{N} = A + \frac{c\sum_{i=1}^{n} f_iu_i}{N}$

Therefore, the arithmetic mean by the coding method is given by:

$\bar{x} = A + \frac{c\sum_{i=1}^{n} f_iu_i}{N}$

Example 1

Use the coding method to calculate the mean for the following frequency distribution:

Solution:

From the given distribution, $c = 10$ and let the assumed mean be $A = 35$.

Using the coding method formula, $\bar{x} = A + c\frac{\sum f_iu_i}{N}$:

$\bar{x} = 35 + 10\left(\frac{-8}{50}\right) = 35 - 1.6 = 33.4$

Therefore, the mean of the distribution is 33.4.

Example 2

The following is a frequency distribution table for the seed yields from the plots. Use the coding method to calculate the mean of the seed yields per plot.

Solution:

From the frequency distribution table, class size $c = 20$. Let $A = 94.5$ be the assumed mean.

Using the coding method formula, $\bar{x} = A + c\frac{\sum f_iu_i}{N}$:

$\bar{x} = 94.5 + 20\left(\frac{44}{35}\right) = 94.5 + 25.142857 \approx 119.6$

Therefore, the mean seed yields per plot is approximately 119.6 grams.

The median is a measure of central tendency representing the middle value of an ordered dataset. It divides the data into two equal halves. It's a positional value, depending on the data's position, not magnitude. The median is used for skewed distributions and is robust against outliers.

Median for grouped data

For grouped data, the median is found by:

1. Arranging class intervals in ascending or descending order.
2. Identifying the median class (the class interval containing half of the total observations).

The median ($M_d$) formula is derived from the cumulative frequency curve (ogive). Consider:

$L$: Lower real limit of the median class $f_m$: Frequency of the median class $\sum f_b$: Sum of frequencies of classes before the median class $N$: Total number of frequencies $c$: Class size

Figure 4.1 (Not included here, but assumed to be a standard ogive diagram)

From the (assumed) Figure 4.1, the tangent of angle $\theta$ in the right-angled triangle PQR is given by:

$\tan \theta = \frac{f_m}{c} \quad (4.7)$

Here, $f_m$ is the difference between half the total frequencies ($\frac{N}{2}$) and the cumulative frequency before the median class ($\sum f_b$):

$f_m = \frac{N}{2} - \sum f_b \quad (4.8)$

The difference between the median value ($M_d$) and the lower real limit ($L$) is related to the class size:

$d = M_d - L \quad (4.9)$

Substituting equations (4.8) and (4.9) into (4.7) would relate the change in cumulative frequency to the change in x on the ogive, leading to:

$\tan \theta = \frac{\frac{N}{2} - \sum f_b}{M_d - L} \quad (4.10)$

By comparing the ratios of changes (as in similar triangles on the ogive), we have the proportion:

$\frac{\frac{N}{2} - \sum f_b}{M_d - L} = \frac{f_m}{c}$

Cross-multiplying and rearranging to solve for $M_d$ gives the median formula for grouped data:

$M_d = L + c\left(\frac{\frac{N}{2} - \sum f_b}{f_m}\right)$

This is the final formula for calculating the median of grouped data.

Example 1

Calculate the median for the following data.

Solution:

Since $\frac{N}{2} = \frac{100}{2} = 50$, the median class is 7 – 8.

Therefore, $L = 7$, $\sum f_b = 38$, $c = 1$, and $f_m = 22$.

Using the median formula: $M_d = L + c\left(\frac{\frac{N}{2} - \sum f_b}{f_m}\right)$

$M_d = 7 + 1\left(\frac{50 - 38}{22}\right) = 7 + \frac{12}{22} = 7 + 0.545 \approx 7.55$

Therefore, the median is approximately 7.55 inches.

Example 2

Suppose fifty timber trees are grown using special soil. After some days their heights (to the nearest millimetres) are measured and grouped as shown in the following table:

Find the median measure of height of the trees.

Solution:

Since $\frac{N}{2} = \frac{50}{2} = 25$, the median class is 170 – 174.

Therefore, $L = 169.5$, $\sum f_b = 21$, $f_m = 9$, and $c = 5$.

$M_d = 169.5 + 5\left(\frac{25 - 21}{9}\right) = 169.5 + 5\left(\frac{4}{9}\right) = 169.5 + 2.22 \approx 171.7\text{ mm}$

Therefore, the median height of the trees is approximately 171.7 mm.

Quartiles divide a dataset into four equal parts. The lower quartile (or first quartile), denoted by $Q_1$, separates the lowest 25% of the data from the rest. The upper quartile (or third quartile), denoted by $Q_3$, separates the highest 25% of the data from the rest. The median (second quartile, $Q_2$) separates the data into two halves.

The formulas for computing quartiles are:

$Q_1 = L + c\left(\frac{\frac{N}{4} - \sum f_b}{f_m}\right) \quad (4.13)$

$Q_3 = L + c\left(\frac{\frac{3N}{4} - \sum f_b}{f_m}\right) \quad (4.14)$

Where:

$L$: Lower limit of the quartile class $\sum f_b$: Sum of frequencies preceding the quartile class $f_m$: Frequency of the quartile class $c$: Class size $N$: Total number of observations

The general formula for the $i^{th}$ quartile is:

$Q_i = L + c\left(\frac{\frac{iN}{4} - \sum f_b}{f_m}\right), \quad \text{for } i = 1, 2, 3.$

Interquartile range (IQR)

The interquartile range (IQR) measures the spread of the middle 50% of the data. It's the difference between the upper and lower quartiles:

$IQR = Q_3 - Q_1$

Semi-interquartile range (SIR)

The semi-interquartile range (SIR) is half of the interquartile range. It provides a measure of dispersion around the median:

$SIR = \frac{Q_3 - Q_1}{2}$

The procedure for finding $Q_1$ and $Q_3$ is the same as for the general $Q_i$ formula, just with different values of $i$.

Example 1

The following distribution table gives the amount of time (in minutes) spent on the internet each morning by a group of social workers.

Find the interquartile range for the frequency distribution table.

Solution:

The interquartile range is given by the formula, $IQR = Q_3 - Q_1$.

First quartile ($Q_1$):

$\frac{N}{4} = \frac{60}{4} = 15$. The cumulative frequency just greater than or equal to 15 is 17. The corresponding class is 13-15.

$L = 12.5$, $\sum f_b = 4$, $f_m = 13$, $c = 3$.

$Q_1 = 12.5 + 3\left(\frac{15 - 4}{13}\right) = 12.5 + 3\left(\frac{11}{13}\right) \approx 12.5 + 2.54 \approx 15.04$

Third quartile ($Q_3$):

$\frac{3N}{4} = \frac{3 \times 60}{4} = 45$. The cumulative frequency just greater than or equal to 45 is 60. The corresponding class is 22-24.

$L = 21.5$, $\sum f_b = 36$, $f_m = 24$, $c = 3$.

$Q_3 = 21.5 + 3\left(\frac{45 - 36}{24}\right) = 21.5 + 3\left(\frac{9}{24}\right) \approx 21.5 + 1.13 \approx 22.63$

Interquartile range (IQR):

$IQR = Q_3 - Q_1 = 22.63 - 15.04 = 7.59$

Therefore, the interquartile range is approximately 7.59 minutes.

Example 2

The following table shows the distribution of maximum load, in tonnes, produced by a certain metropolitan company.

Use the distribution table to find each of the following:

(a) First quartile (b) Third quartile (c) Interquartile range (d) Semi-interquartile range

Solution:

$N = 60$, $c = 0.5$

(a) First quartile ($Q_1$):

$\frac{N}{4} = \frac{60}{4} = 15$. The cumulative frequency just greater than or equal to 15 is 19. The corresponding class is 10.25-10.75.

$L = 10.25$, $\sum f_b = 10$, $f_m = 9$.

$Q_1 = 10.25 + 0.5\left(\frac{15 - 10}{9}\right) \approx 10.25 + 0.28 \approx 10.53 \text{ tonnes}$

(b) Third quartile ($Q_3$):

$\frac{3N}{4} = \frac{3 \times 60}{4} = 45$. The cumulative frequency just greater than or equal to 45 is 49. The corresponding class is 11.75-12.25.

$L = 11.75$, $\sum f_b = 44$, $f_m = 5$, and $c = 0.5$.

$Q_3 = 11.75 + 0.5\left(\frac{45 - 44}{5}\right) = 11.75 + 0.5\left(\frac{1}{5}\right) = 11.75 + 0.1 = 11.85 \text{ tonnes}$

Therefore, 75% of the maximum load is less than or equal to 11.85 tonnes.

(c) Interquartile range (IQR):

$IQR = Q_3 - Q_1 = 11.85 - 10.53 = 1.32 \text{ tonnes}$

Therefore, the interquartile range is 1.32 tonnes.

(d) Semi-interquartile range (SIR):

$SIR = \frac{Q_3 - Q_1}{2} = \frac{1.32}{2} = 0.66 \text{ tonnes}$

Therefore, the semi-interquartile range is 0.66 tonnes.

Percentiles divide the total frequency into one hundred equal parts. The general formula for the $i^{th}$ percentile is:

$P_i = L + c\left(\frac{\frac{iN}{100} - \sum f_b}{f_m}\right) \quad (4.15)$

Where:

$L$: Lower limit of the percentile class $f_m$: Frequency of the percentile class $c$: Class size $N$: Total number of observations $\sum f_b$: Sum of the frequencies of all classes below the $i^{th}$ percentile class

Example 1

Given the following distribution table:

Compute each of the following:

(a) The 50th percentile (b) The 70th percentile

Solution:

(a) Computation of $P_{50}$:

Position of $P_{50}$: $\frac{50}{100} \times 49 = 24.5$

The 24.5th position is in the cumulative frequency 26. The percentile class is 15-20.

$L = 15$, $\sum f_b = 11$, $f_m = 15$, $c = 5$.

$P_{50} = 15 + 5\left(\frac{24.5 - 11}{15}\right) = 15 + 5\left(\frac{13.5}{15}\right) = 15 + 4.5 = 19.5$

(b) Computation of $P_{70}$:

Position of $P_{70}$: $\frac{70}{100} \times 49 = 34.3$

The 34.3rd position is in the cumulative frequency 36. The percentile class is 20-25.

$L = 20$, $\sum f_b = 26$, $f_m = 10$, $c = 5$.

$P_{70} = 20 + 5\left(\frac{34.3 - 26}{10}\right) = 20 + 5\left(\frac{8.3}{10}\right) = 20 + 4.15 = 24.15$

Example 2

Find the 20th percentile in the following information:

Solution:

Computation of $P_{20}$:

Position of $P_{20}$: $\frac{20}{100} \times 18 = 3.6$

The 3.6th value is in the cumulative frequency 8. The class interval is 4-6.

$L = 4$, $\sum f_b = 3$, $f_m = 5$, $c = 2$.

$P_{20} = 4 + 2\left(\frac{3.6 - 3}{5}\right) = 4 + 2\left(\frac{0.6}{5}\right) = 4 + 0.24 = 4.24$

Example 3

(a) The maximum mass of lower 35% of the children ($P_{35}$):

Position of $P_{35}$: $\frac{35}{100} \times 60 = 21$

The 21st value falls within the cumulative frequency of 24. The corresponding class is 20-29.

$L = 19.5$, $\sum f_b = 4$, $f_m = 20$, $c = 10$.

$P_{35} = 19.5 + 10\left(\frac{21 - 4}{20}\right) = 19.5 + 10\left(\frac{17}{20}\right) = 19.5 + 8.5 = 28 \text{ kg}$

(b) The minimum mass of upper 35% of the children ($P_{65}$):

This is equivalent to the 65th percentile.

Position of $P_{65}$: $\frac{65}{100} \times 60 = 39$

The 39th value falls within the cumulative frequency of 50. The corresponding class is 40-49.

$L = 39.5$, $\sum f_b = 38$, $f_m = 12$, $c = 10$.

$P_{65} = 39.5 + 10\left(\frac{39 - 38}{12}\right) = 39.5 + 10\left(\frac{1}{12}\right) \approx 39.5 + 0.83 \approx 40.33 \text{ kg}$

(c) The limits for the mass of middle 30% of the children:

The middle 30% lies between the 35th and 65th percentiles. Therefore, the limits are $P_{35}$ and $P_{65}$, which are 28 kg and approximately 40.33 kg.

Therefore, the limits for the mass of the middle 30% of the children are approximately 28 kg and 40.33 kg.

The mode of a distribution is the most frequently occurring value. It represents the point of highest frequency density.

Consider the mode $M_0$ with:

$L$: Lower real limit of the modal class $U$: Upper real limit of the modal class $t_1$: Difference between the modal class frequency and the frequency of the next lower class $t_2$: Difference between the modal class frequency and the frequency of the next higher class $c$: Class size

From the Figure, and using similar triangles, the mode $M_0$ can be calculated as follows:

$M_0 = L + c\left(\frac{t_1}{t_1 + t_2}\right) \quad (4.18)$

Example 1

Compute the mode for the following distribution.

Solution:

The highest frequency is 72, in the class interval 21 – 28. This is the modal class.

$L = 21$, $t_1 = 72 - 36 = 36$, $t_2 = 72 - 51 = 21$, $c = 7$.

$M_0 = 21 + 7\left(\frac{36}{36 + 21}\right) = 21 + 7\left(\frac{36}{57}\right) \approx 21 + 4.42 \approx 25.42$

Therefore, the mode is approximately 25.42.

Example 2

Find the mode for the following distribution.

Solution:

The highest frequency is 12. The modal class is 60 – 80.

$L = 60$, $t_1 = 12 - 10 = 2$, $t_2 = 12 - 6 = 6$, $c = 20$.

$M_0 = 60 + 20\left(\frac{2}{2 + 6}\right) = 60 + 20\left(\frac{2}{8}\right) = 60 + 5 = 65$

Therefore, the mode is 65.

Example 3

The number of words in each of the first eight sentences of a book were counted. The results were recorded as follows.

Estimate the modal length of the sentences correct to one decimal place.

Solution:

The highest frequency is 23. The modal class is 17 – 20. The class boundaries are 16.5-20.5

$L = 16.5$, $t_1 = 23 - 21 = 2$, $t_2 = 23 - 4 = 19$, $c = 4$.

$M_0 = 16.5 + 4\left(\frac{2}{2 + 19}\right) = 16.5 + 4\left(\frac{2}{21}\right) \approx 16.5 + 0.38 \approx 16.9$

Therefore, the modal length of the sentences is approximately 16.9.