Statistics - CBSE Class 10 Mathematics Notes

CHAPTER 13 : STATISTICS

13.1 Introduction

In your previous academic year (Class IX), you acquired knowledge regarding the categorization of raw data into both ungrouped and grouped frequency distributions. Furthermore, you were introduced to the graphical depiction of data using various chart types, including bar graphs, histograms (encompassing those with differing class widths), and frequency polygons. Beyond this, you delved into the analysis of ungrouped data through specific numerical indicators known as measures of central tendency, specifically the mean, median, and mode. This chapter aims to advance your understanding of these three statistical measures—mean, median, and mode—by transitioning their application from ungrouped to grouped data sets. Additionally, we will explore the notion of cumulative frequency, its distribution, and the methodology for constructing cumulative frequency curves, commonly referred to as ogives.

13.2 Mean of Grouped Data

As is generally understood, the mean (or arithmetic average) of a set of observations is computed by summing the values of all individual observations and then dividing this total by the aggregate number of observations. Recalling principles from Class IX, consider a scenario where $x_{1}, x_{2}, \ldots, x_{n}$ represent distinct observations, each paired with its corresponding frequency $f_{1}, f_{2}, \ldots, f_{n}$. This notation signifies that the observation $x_{1}$ appears $f_{1}$ times, $x_{2}$ appears $f_{2}$ times, and similarly for subsequent observations.

Consequently, the cumulative sum of the values across all observations is expressed as $f_{1}x_{1} + f_{2}x_{2} + \ldots +f_{n}x_{n}$, while the total count of observations is given by $f_{1} + f_{2} + \ldots +f_{n}$.

Therefore, the mean, denoted $\overline{x}$, for this dataset can be determined using the following formula:

$ \bar{x} = \frac{f_{1}x_{1} + f_{2}x_{2} + \cdots + f_{n}x_{n}}{f_{1} + f_{2} + \cdots + f_{n}} $

It is important to recall that this expression can be concisely represented through the use of the Greek letter $\Sigma$ (capital sigma), which signifies summation. Specifically,

$

\bar {x} = \frac {\sum_ {i = 1} ^ {n} f _ {i} x _ {i}}{\sum_ {i = 1} ^ {n} f _ {i}} $

This formula is often abbreviated to $\overline{x} = \frac{\sum f_i x_i}{\sum f_i}$, with the implicit understanding that the index $i$ ranges from 1 to $n$.

We will now apply this formula to compute the mean in the subsequent example.

Example 1: The scores achieved by 30 students enrolled in Class X at a particular educational institution on a Mathematics examination, graded out of a maximum of 100 points, are tabulated below. Determine the mean score attained by these students.

Solution: As a prerequisite for calculating the mean marks, it is necessary to compute the product of each observed mark, $x_{i}$, and its respective frequency, $f_{i}$. Therefore, we will arrange these values in a columnar format, as illustrated in Table 13.1.

Table 13.1

STATISTICS

We now proceed with the calculation:

$ \bar{x} = \frac{\sum f_i x_i}{\sum f_i} = \frac{1779}{30} = 59.3 $

Hence, the average marks obtained are 59.3.

In many real-world scenarios, datasets are often extensive, necessitating condensation into grouped data for effective analysis. Consequently, it becomes essential to transform ungrouped data into a grouped format and establish a methodology for determining its mean.

Let us convert the ungrouped data presented in Example 1 into grouped data by establishing class-intervals with a width of, for instance, 15. It is crucial to remember that when assigning frequencies to each class-interval, observations falling precisely on an upper class-limit are to be included in the subsequent class. For example, 4 students who scored exactly 40 marks would be counted within the 40-55 class-interval, not the 25-40 interval. Adhering to this convention, we can construct a grouped frequency distribution table (refer to Table 13.2).

Table 13.2

For each class-interval, a single representative value is required to characterize the entire class. The assumption made is that the frequency distribution within each interval is concentrated around its mid-point. Therefore, the mid-point (also known as the class mark) of each class is selected to represent the observations contained within that class. To determine the mid-point (or class mark) of a class, one calculates the average of its upper and lower limits. Specifically,

$ \text{Class mark} = \frac{\text{Upper class limit} + \text{Lower class limit}}{2} $

Referring to Table 13.2, for the class 10-25, the class mark is calculated as $\frac{10 + 25}{2}$, which equals 17.5. The class marks for the remaining intervals can be determined similarly and are compiled in Table 13.3. These class marks will serve as our $x_i$ values. Generally, for the $i$th class interval, we have a frequency $f_i$ corresponding to the class mark $x_i$. We can now proceed to calculate the mean using the same procedure as outlined in Example 1.

MATHEMATICS

Table 13.3

The summation of the values in the final column yields $\Sigma f_{i}x_{i}$. Consequently, the mean $\overline{x}$ for the provided data is calculated as:

$ \bar {x} = \frac {\sum f _ {i} x _ {i}}{\sum f _ {i}} = \frac {1860.0}{30} = 62 $

This approach to computing the mean is recognized as the Direct Method.

It is noticeable that both Table 13.1 and Table 13.3 utilize the identical dataset and apply the same formula for mean computation, yet they yield differing results. The reason for this divergence, and which value is more precise, warrants consideration. The discrepancy between the two outcomes stems from the mid-point assumption inherent in Table 13.3, where 59.3 represents the exact mean, while 62 is an approximation.

When the numerical magnitudes of $x_{i}$ and $f_{i}$ are substantial, calculating their product, $x_{i}f_{i}$, can prove to be a cumbersome and protracted process. Consequently, in such scenarios, it becomes pertinent to devise a computational simplification strategy.

While the values of $f_{i}$ remain immutable, we possess the capacity to transform each $x_{i}$ into a more manageable numerical value, thereby streamlining subsequent computations. The question then arises: how might this be achieved? A plausible approach involves subtracting a constant value from each $x_{i}$. Let us explore the efficacy of this technique.

The initial procedural step requires designating one of the $x_{i}$ values as the 'assumed mean,' represented by the variable 'a'. To optimize the reduction of computational effort, it is often advantageous to select 'a' as the $x_{i}$ value positioned centrally within the dataset $x_{1}, x_{2}, \ldots, x_{n}$. For instance, valid choices for 'a' could be $47.5$ or $62.5$. For the purpose of this illustration, we shall opt for $a = 47.5$.

Subsequently, the procedure involves determining the difference, denoted as $d_{i}$, between each $x_{i}$ and the assumed mean 'a'. This value $d_{i}$ quantifies the deviation of each $x_{i}$ from 'a'. Specifically, $d_{i} = x_{i} - a = x_{i} - 47.5$

The third phase entails computing the product of each $d_{i}$ with its respective frequency $f_{i}$, followed by aggregating all such products to obtain the sum of $f_{i}

d_{i}$ values. The detailed computations are presented in Table 13.4.

Table 13.4

Consequently, as derived from Table 13.4, the mean of these deviations, denoted as $\overline{d}$, is calculated as $\overline{d} = \frac{\sum f_i d_i}{\sum f_i}$.

Next, we proceed to establish the relationship connecting $\overline{d}$ and $\overline{x}$. Given that $d_{i}$ was computed by subtracting 'a' from each $x_{i}$, it logically follows that to recover the actual mean $\overline{x}$, 'a' must be added back to $\overline{d}$. This relationship can be elucidated through the following mathematical derivation: The mean of the deviations, $\overline{d}$, is expressed as $\overline{d} = \frac{\sum f_i d_i}{\sum f_i}$. Substituting the definition of $d_i$, we obtain $\overline{d} = \frac{\sum f_i (x_i - a)}{\sum f_i}$

$ \begin{array}{l} = \frac{\sum f_i x_i}{\sum f_i} - \frac{\sum f_i a}{\sum f_i} \ = \overline{x} - a \frac{\sum f_i}{\sum f_i} \ = \overline{x} - a \ \end{array} $

Thus, the relationship simplifies to $\overline{x} = a + \overline{d}$. This implies that the mean $\overline{x}$ can be computed as $\overline{x} = a + \frac{\sum f_i d_i}{\sum f_i}$.

By substituting the respective values for $a$, $\Sigma f_i d_i$, and $\Sigma f_i$ from Table 13.4 into the derived formula, the following calculation is performed:

$ \bar{x} = 47.5 + \frac{435}{30} = 47.5 + 14.5 = 62. $ Consequently, the arithmetic mean of the marks acquired by the students is determined to be 62.

The computational approach elaborated herein is formally recognized as the Assumed Mean Method.

Activity 1: Utilizing Table 13.3, compute the mean by successively designating each $x_i$ value (e.g., 17.5, 32.5, and so forth) as 'a'. What empirical observation do you make? It will become evident that the calculated mean remains invariant across all choices of 'a', consistently yielding a value of 62. (Consider the underlying mathematical rationale.)

Therefore, it can be conclusively stated that the resultant mean value is independent of the specific selection of 'a'.

It is noteworthy that in Table 13.4, the numerical entries within Column 4 exhibit a common factor, specifically 15. Consequently, by dividing all values in Column 4 by 15, we would derive proportionally smaller magnitudes, thereby simplifying their multiplication with $f_i$. (In this context, 15 represents the uniform class size for each interval.)

Thus, we define $u_i = \frac{x_i - a}{h}$, where $a$ represents the assumed mean and $h$ denotes the class interval width.

Subsequently, we compute $u_i$ using this definition and proceed with the established methodology (specifically, determining $f_i u_i$ and subsequently $\Sigma f_i u_i$). With $h$ set to 15, we construct Table 13.5.

Table 13.5

Let us define the mean of $u_i$ as:

$ \bar{u} = \frac{\Sigma f_i u_i}{\Sigma f_i} $

At this juncture, we will once more establish the relationship between $\bar{u}$ and $\bar{x}$.

Given the expression:

$ u _ {i} = \frac {x _ {i} - a}{h} $

Consequently, we derive:

$ \begin{array}{l} \bar {u} = \frac {\sum f _ {i} \frac {(x _ {i} - a)}{h}}{\sum f _ {i}} = \frac {1}{h} \left[ \frac {\sum f _ {i} x _ {i} - a \sum f _ {i}}{\sum f _ {i}} \right] \ = \frac {1}{h} \left[ \frac {\sum f _ {i} x _ {i}}{\sum f _ {i}} - a \frac {\sum f _ {i}}{\sum f _ {i}} \right] \ = \frac {1}{h} [ \bar {x} - a ] \end{array} $

Thus,

$ h \bar {u} = \bar {x} - a $

Which simplifies to:

$ \bar {x} = a + h \bar {u} $

Therefore, the formula for the mean can be expressed as:

$ \bar {x} = a + h \left(\frac {\sum f _ {i} u _ {i}}{\sum f _ {i}}\right) $

By substituting the values for $a$, $h$, $\sum f_i u_i$, and $\sum f_i$ from Table 13.5, we obtain:

$ \begin{array}{l} \bar {x} = 47.5 + 15 \times \left(\frac {29}{30}\right) \ = 47.5 + 14.5 = 62 \end{array} $

Hence, the average marks secured by a student amount to 62.

The methodology outlined previously is designated as the Step-deviation method.

It is pertinent to observe that:

• The step-deviation method offers practical utility primarily when all $d_{i}$ values share a common factor.
• The calculated mean remains consistent across all three computational approaches.
• Both the assumed mean method and the step-deviation method represent streamlined variations of the direct method.
• The formula $\bar{x} = a + h\bar{u}$ retains its validity even if $a$ and $h$ deviate from the previously specified values, provided they are any non-zero quantities such that $u_{i} = \frac{x_{i} - a}{h}$.

We shall now demonstrate the application of these methods through an additional example.

Example 2: The subsequent table illustrates the percentage distribution of female educators within primary schools located in rural regions across diverse states and union territories (U.T.) of India. Determine the mean percentage of female teachers utilizing all three methods presented in this section.

Source : Seventh All India School Education Survey conducted by NCERT

Solution : We proceed by determining the class marks, $x_{i}$, for each class interval and tabulating them in a dedicated column (refer to Table 13.6):

Table 13.6

For this calculation, we designate $a = 50$ and $h = 10$. Consequently, the deviations $d_{i}$ are defined as $x_{i} - 50$, and the scaled deviations $u_{i}$ are given by $\frac{x_{i} - 50}{10}$.

These computed values for $d_{i}$ and $u_{i}$ are subsequently entered into Table 13.7.

Table 13.7

Based on the data presented in the preceding table, we derive the following summations: $\Sigma f_{i} = 35$, $\Sigma f_{i}x_{i} = 1390$, $\Sigma f _ {i} d _ {i} = - 3 6 0, \quad \Sigma f _ {i} u _ {i} = - 3 6.$

Applying the direct method, the mean $\overline{x}$ is computed as $\frac{\Sigma f_i x_i}{\Sigma f_i} = \frac{1390}{35} = 39.71$.

Employing the assumed mean method, the calculation yields:

$ \bar {x} = a + \frac {\Sigma f _ {i} d _ {i}}{\Sigma f _ {i}} = 5 0 + \frac {(- 3 6 0)}{3 5} = 3 9. 7 1 $

Utilizing the step-deviation method, the mean is determined by:

$ \bar {x} = a + \left(\frac {\Sigma f _ {i} u _ {i}}{\Sigma f _ {i}}\right) \times h = 5 0 + \left(\frac {- 3 6}{3 5}\right) \times 1 0 = 3 9. 7 1 $

Consequently, the average percentage of female educators in rural primary schools stands at 39.71.

Observation: It is evident that all three computational methodologies yield an identical result. Therefore, the selection of the most suitable method is contingent upon the magnitude of the numerical values for $x_{i}$ and $f_{i}$. Should $x_{i}$ and $f_{i}$ be relatively small, the direct method represents an appropriate and efficient option. Conversely, when $x_{i}$ and $f_{i}$ are numerically substantial, recourse to either the assumed mean method or the step-deviation method is advisable. Furthermore, in scenarios where class intervals are disparate and $x_{i}$ values are numerically large, the step-deviation method remains applicable by judiciously selecting $h$ as a common divisor for all $d_{i}$ values.

Example 3: The following frequency distribution delineates the count of wickets secured by bowlers in one-day international cricket fixtures. Determine the mean number of wickets using an appropriate statistical technique. Additionally, elucidate the interpretative significance of this calculated mean.

Solution: In this particular dataset, it is observed that the class intervals are not uniform, and the values of $x_{i}$ are substantial. Despite these characteristics, we shall proceed with the application of the step-deviation method, setting the assumed mean $a = 200$ and the class height $h = 20$. The resultant calculations are presented in Table 13.8.

Table 13.8

Consequently, $\overline{u} = \frac{-106}{45}$. Thus, $\overline{x} = 200 + 20\left(\frac{-106}{45}\right) = 200 - 47.11 = 152.89$.

This result indicates that, for this cohort of 45 bowlers in one-day cricket, the mean number of wickets secured is 152.89.

We shall now assess your ability to apply the principles covered in this segment.

Activity 2 :

Form three distinct groups from your class and assign each group one of the subsequent tasks.

1. Gather the scores achieved by all students in your class in the most recent Mathematics examination administered by your institution. Subsequently, construct a grouped frequency distribution from the acquired data.
2. Compile the maximum daily temperatures observed in your city over a 30-day interval. Organize this information into a grouped frequency table.
3. Ascertain the heights (in cm) of all students within your class and then generate a grouped frequency distribution table based on this data.

Once each group has gathered its respective data and prepared the grouped frequency distribution tables, they are to calculate the mean for each dataset using a method deemed suitable by the group.

EXERCISE 13.1

1. As part of an environmental awareness initiative, a student group carried out a survey, gathering the subsequent data concerning the plant count across 20 residences within a specific neighborhood. Determine the average quantity of plants per household.

Identify the methodology employed for calculating the mean and provide a rationale for its selection.

2. Examine the subsequent distribution representing the daily earnings of 50 factory employees.

Calculate the average daily remuneration for the factory workers, utilizing a suitable statistical technique.

3. The subsequent distribution illustrates the daily pocket money received by children in a certain area. Given that the average pocket allowance is Rs 18, ascertain the value of the unknown frequency, denoted by $f$.

4. A medical practitioner at a hospital conducted examinations on thirty women, documenting and subsequently summarizing their heartbeats per minute as indicated below. Determine the average heart rate per minute for this cohort of women, selecting an appropriate computational method.

5. Within a retail marketplace, fruit sellers offered mangoes contained within packing boxes, where the quantity of mangoes varied across these containers. The ensuing data presents the distribution of mangoes based on the number of boxes.

Compute the average quantity of mangoes housed in a single packing box. Which specific statistical approach did you elect for this mean calculation?

6. The tabular data provided below delineates the daily food expenditures for 25 households situated within a particular community.

7. Data was gathered across 30 different locations within a specified urban area to determine the atmospheric concentration of $\mathrm{SO}_2$ (expressed in parts per million, or ppm), and this information is presented hereunder:

Calculate the average atmospheric concentration of $\mathrm{SO}_2$.

8. A teacher maintains the subsequent record of absenteeism for 40 students enrolled in a particular class throughout the entire academic term. Determine the average number of days a student was absent.

9. The following table gives the literacy rate (in percentage) of 35 cities. Find the mean literacy rate.

13.3 Mode of Grouped Data

As established in Class IX, the mode represents the observation value that appears with the highest frequency within a dataset. While the determination of the mode for ungrouped data has been previously covered, this section will focus on methodologies for ascertaining the mode in grouped data. It is conceivable for multiple values to exhibit the same highest frequency, a scenario termed multimodal data. Although grouped datasets can also be multimodal, our present scope will be confined to instances possessing a singular mode.

To begin, let us review the process of identifying the mode for ungrouped data using the subsequent example.

Example 4: Consider the number of wickets captured by a bowler across 10 cricket matches, presented as follows:

2 6 4 5 0 2 1 3 2 3

Find the mode of the data.

Solution: We can construct a frequency distribution table for the provided data as follows:

Evidently, the bowler secured 2 wickets in the highest count of matches (specifically, 3). Consequently, the mode for this dataset is 2.

Within a grouped frequency distribution, direct identification of the mode through mere observation of frequencies is unfeasible. Instead, one can pinpoint the class interval possessing the highest frequency, which is designated as the modal class. The mode itself resides within this modal class and is calculated using the following formula:

$ \text{Mode} = l + \left(\frac {f _ {1} - f _ {0}}{2 f _ {1} - f _ {0} - f _ {2}}\right) \times h $

where: $l =$ the lower boundary of the modal class,

$h =$ the width of the class interval (presuming uniform class sizes),

$f_{1} =$ the frequency associated with the modal class,

$f_{0} =$ the frequency of the class interval immediately preceding the modal class,

$f_{2} =$ the frequency of the class interval immediately succeeding the modal class.

To demonstrate the application of this formula, let us examine the subsequent examples.

Example 5: A study involving 20 households within a specific locality, carried out by a student group, yielded the ensuing frequency table detailing the number of family members per household:

Find the mode of this data.

Solution: In this instance, the highest class frequency observed is 8, which corresponds to the class interval $3 - 5$. Therefore, the modal class is $3 - 5$.

Now, the modal class is $3 - 5$, the lower limit $(l)$ of the modal class is $3$, the class size $(h)$ is $2$, and the frequency $(f_{1})$ of the modal class is $8$.

The frequency $(f_0)$ of the class immediately preceding the modal class is $7$.

The frequency $(f_2)$ of the class immediately succeeding the modal class is $2$.

Now, by substituting these values into the formula:

$ \begin{array}{l} \text{Mode} = l + \left(\frac{8 - 7}{2 \times 8 - 7 - 2}\right) \times 2 = 3 + \frac{2}{7} = 3.286 \end{array} $

Consequently, the mode for the given data is $3.286$.

Example 6: Table 13.3 from Example 1 presents the mark distribution for 30 students in a mathematics examination. Determine the mode for this dataset. Additionally, compare and interpret both the mode and the mean.

Solution: Referring to Table 13.3 of Example 1, it is observed that the highest number of students (specifically, 7) scored marks within the 40 - 55 interval, thereby establishing 40 - 55 as the modal class. Consequently, the relevant parameters are:

• the lower limit $(l)$ of the modal class $= 40$,
• the class size $(h) = 15$,
• the frequency $(f_1)$ of the modal class $= 7$,
• the frequency $(f_0)$ of the class preceding the modal class $= 3$,
• the frequency $(f_2)$ of the class succeeding the modal class $= 6$.

Applying the mode formula:

$ \text{Mode} = l + \left(\frac{f_1 - f_0}{2f_1 - f_0 - f_2}\right) \times h, $

we calculate:

$ \text{Mode} = 40 + \left(\frac{7 - 3}{14 - 6 - 3}\right) \times 15 = 52 $

Thus, the modal mark is 52.

Furthermore, from Example 1, it is established that the mean mark is 62.

This indicates that the score of 52 was attained by the largest number of students, whereas the average score for a student was 62.

Remarks:

1. In Example 6, the mode is numerically less than the mean. However, in other contexts, the mode may be equivalent to or exceed the mean.
2. The selection between the mean and the mode as a measure of central tendency is contingent upon the specific requirements of the situation. One might seek the average performance of students (requiring the mean) or identify the most

frequently occurring score among students (requiring the mode).

Activity 3: Using the same groups and assigned scenarios from Activity 2, instruct each group to compute the mode of their respective datasets. They should then compare this mode with the mean and provide an interpretation for both statistical measures.

Remark: It is possible to compute the mode for grouped data even when class sizes are not uniform; however, this topic will not be addressed here.

EXERCISE 13.2

1. A hospital recorded the ages of patients admitted over a year, as presented in the table below:

Calculate both the mode and the mean for the provided dataset. Subsequently, compare and offer an interpretation of these two central tendency metrics.

2. The subsequent data illustrates the observed operational durations (in hours) for a collection of 225 electrical components:

Ascertain the modal lifetimes exhibited by these components.

3. The data presented outlines the distribution of total monthly household expenditures for 200 families residing in a particular village. Determine the modal monthly expenditure for these families, and additionally, calculate the mean monthly expenditure:

4. The ensuing distribution details the state-level teacher-student ratios observed in higher secondary educational institutions across India. Compute the mode and mean for this dataset, and then provide an interpretation of these two statistical measures.

5. The provided distribution displays the number of runs accumulated by several leading batsmen globally in one-day international cricket fixtures.

Identify the mode within this dataset.

6. A student recorded the count of vehicles traversing a specific road location over 100 distinct 3-minute intervals, subsequently summarizing this information in the table below. Determine the mode of this dataset:

13.4 Median of Grouped Data

As previously covered in Class IX, the median serves as a central tendency metric that identifies the central value within a dataset. To determine the median for ungrouped data, one must initially sort the observations' values in ascending sequence. Subsequently, if the total number of observations, $n$, is odd, the median corresponds to the $\left(\frac{n + 1}{2}\right)$ th observation. Conversely, if $n$ is an even number, the median is computed as the arithmetic mean of the $\frac{n}{2}$ th and the $\left(\frac{n}{2} + 1\right)$ th observations.

Consider an instance where we need to determine the median for the subsequent dataset, which represents the scores, out of 50, achieved by 100 students in an examination:

Initially, the marks are organized in ascending sequence, and a frequency table is constructed as presented below:

Table 13.9

In this case, $n = 100$, which is an even number. Consequently, the median will be the average of the $\frac{n}{2}$th and the $\left(\frac{n}{2} + 1\right)$th observations; specifically, the 50th and 51st observations. To pinpoint these observations, we proceed as follows:

Table 13.10

We now augment the preceding frequency table with an additional column illustrating this information, designating it as the cumulative frequency column.

Table 13.11

Referring to the table provided, we observe that:

$ 50 \text{th observaton is } 28 \quad (\text{Why?}) $

$ 51 \text{st observation is } 29 $

Therefore,

$ \text{Median} = \frac{28 + 29}{2} = 28.5 $

Note: The combination of Column 1 and Column 3 from Table 13.11 constitutes what is termed a Cumulative Frequency Table. A median score of 28.5 indicates that approximately half of the students achieved scores below 28.5, while the other half scored above 28.5.

We shall now explore the methodology for calculating the median of grouped data, using the subsequent scenario as an illustration.

Let us consider a grouped frequency distribution detailing the marks, out of 100, secured by 53 students in a particular assessment, presented as follows:

Table 13.12

Based on the preceding table, attempt to address the subsequent inquiries:

What is the count of students who achieved a score below 10? The straightforward response is 5.

To determine the number of students achieving scores below 20, one must consider those who scored in the 0-10 range in addition to those who scored within the 10-20 interval. Consequently, the aggregate number of students with marks under 20 amounts to 5 + 3, which equals 8. This value, 8, represents the cumulative frequency for the class interval 10-20.

Through an analogous process, the cumulative frequencies for the remaining class intervals can be calculated, encompassing the number of students scoring below 30, below 40, and so forth, up to below 100. These are presented in Table 13.13:

Table 13.13

The aforementioned arrangement of data is termed a 'less than' cumulative frequency distribution. In this context, the values 10, 20, 30, ..., up to 100, denote the upper boundaries for their corresponding class intervals.

A comparable tabulation can be constructed for the number of students whose scores are greater than or equal to 0, greater than or equal to 10, greater than or equal to 20, and so forth. Referencing Table 13.12, it is evident that all 53 students achieved marks of 0 or higher. Given that 5 students obtained scores within the 0-10 interval, it logically follows that 53 - 5 = 48 students scored 10 marks or more. Proceeding with this methodology, the count of students scoring 20 or above is 48 - 3 = 45; for 30 or above, it is 45 - 4 = 41; and so on, as illustrated in Table 13.14.

Table 13.14

The preceding tabular representation is referred to as a 'more than' cumulative frequency distribution. In this configuration, the values 0, 10, 20, ..., up to 90, designate the lower boundaries of their corresponding class intervals.

To ascertain the median for grouped data, one may employ either of these cumulative frequency distributions.

By integrating the data from Tables 13.12 and 13.13, Table 13.15 is derived and presented hereafter:

Table 13.15

In the context of grouped data, it is often challenging to ascertain the precise middle observation merely by examining cumulative frequencies, as this central value typically resides within a specific

class interval. Consequently, it becomes imperative to identify a particular value within a class that effectively partitions the entire distribution into two equal halves. The critical question then arises: which class precisely contains this median value?

To identify this crucial class, one must first compute the cumulative frequencies for all classes and determine the value of $\frac{n}{2}$. Subsequently, the class whose cumulative frequency is greater than (and closest to) $\frac{n}{2}$ is located. This designated class is referred to as the median class. Considering the preceding distribution, where $n = 53$, the value of $\frac{n}{2}$ is $26.5$. In this instance, the class interval $60 - 70$ possesses a cumulative frequency of 29, which is demonstrably greater than (and nearest to) $\frac{n}{2}$, or 26.5.

Therefore, the class $60 - 70$ is established as the median class.

Upon successfully identifying the median class, the following formula is employed for the calculation of the median:

$ \mathrm {Median} = l + \left(\frac {\frac {n}{2} - \mathrm {cf}}{f}\right) \times h, $

where the variables are defined as:

$l =$ the lower boundary of the median class,

$n =$ the total count of observations,

cf = the cumulative frequency of the class immediately preceding the median class,

$f =$ the frequency of the median class itself,

$h =$ the uniform class size (assuming equal class widths).

By substituting the specified values, namely $\frac{n}{2} = 26.5$, $l = 60$, $\mathrm{cf} = 22$, $f = 7$, and $h = 10$, into the aforementioned formula, we derive:

$ \begin{array}{l} \mathrm {Median} = 60 + \left(\frac {26.5 - 22}{7}\right) \times 10 \ = 60 + \frac {45}{7} \ = 66.4 \ \end{array} $

This result indicates that approximately half of the students achieved scores below 66.4, while the remaining half attained scores exceeding 66.4.

Example 7: A study was conducted to ascertain the heights (in cm) of 51 female students from Class X at a particular school, yielding the subsequent dataset:

Determine the median height.

Solution: To compute the median height, it is first necessary to establish the class intervals and their corresponding frequencies from the provided cumulative frequency distribution.

Given that the distribution is of the "less than" type, the values 140, 145, 150, ..., 165 represent the upper limits of the respective class intervals. Consequently, the class intervals should be structured as: below 140, 140 - 145, 145 - 150, ..., 160 - 165. From the initial distribution, it is evident that 4 girls have a height less than 140 cm, implying that the frequency for the class interval "below 140" is 4. Furthermore, 11 girls have heights less than 145 cm, and 4 girls have heights less than 140 cm. Therefore, the count of girls whose height falls within the interval 140 - 145 cm is $11 - 4 = 7$. Following this pattern, the frequency for the 145 - 150 cm interval is $29 - 11 = 18$, and for the 150 - 155 cm interval, it is $40 - 29 = 11$, and so forth. Thus, our frequency distribution table, incorporating the given cumulative frequencies, is constructed as follows:

Table 13.16

Given $n = 51$, it follows that $\frac{n}{2} = \frac{51}{2} = 25.5$. This particular observation is situated within the class interval 145 - 150. Consequently, we identify the following parameters:

$l$ (the lower boundary of the median class) $= 145$

cf (the cumulative frequency of the class preceding the median class 145 - 150) = 11,

$f$ (the frequency associated with the median class 145 - 150) = 18,

$h$ (the width of the class interval) $= 5$

Applying the formula for the median, Median $= l + \left(\frac{\frac{n}{2} - \mathrm{cf}}{f}\right) \times h$, we compute:

$ \begin{array}{l} \text {Median} = 145 + \left(\frac {25.5 - 11}{18}\right) \times 5 \ = 145 + \frac {72.5}{18} = 149.03. \ \end{array} $

Therefore, the calculated median height for the girls is $149.03\mathrm{cm}$.

This outcome implies that approximately $50%$ of the girls possess a height below this value, while the remaining $50%$ are taller.

Example 8: Consider the following dataset, where the median is given as 525. Determine the values of $x$ and $y$, given that the aggregate frequency totals 100.

Solution: To determine the unknown frequencies, we first construct the cumulative frequency distribution:

We are informed that the total frequency, $n$, equals 100.

Thus, the sum of all frequencies, $76 + x + y$, must be equal to 100, leading to the equation: $x + y = 24$ (1)

The median value is 525, which falls within the class interval $500 - 600$. This identifies the median class.

From this median class, we extract the following values: $l = 500$ (lower limit), $f = 20$ (frequency of the median class), $\mathrm{cf} = 36 + x$ (cumulative frequency of the preceding class), and $h = 100$ (class size).

Employing the formula for the median:

$ \text{Median} = l + \left(\frac{\frac{n}{2} - \mathrm{cf}}{f}\right) h, \text{ we get} $

Substituting the known values into the formula:

$ 525 = 500 + \left(\frac{50 - 36 - x}{20}\right) \times 100 $

This simplifies to: $525 - 500 = (14 - x) \times 5$

Further simplification yields: $25 = 70 - 5x$

Rearranging the terms: $5x = 70 - 25 = 45$

Therefore, the value of $x$ is: $x = 9$

Substituting $x = 9$ into equation (1), we determine: $9 + y = 24$

Consequently, the value of $y$ is: $y = 15$

Having now examined all three primary measures of central tendency, it is pertinent to consider which measure is most appropriate for specific analytical contexts.

The arithmetic mean stands out as the most commonly employed measure of central tendency due to its comprehensive nature, incorporating every observation in its calculation. Furthermore, it inherently falls within the range defined by the dataset's extreme values—that is, between the minimum and maximum observations. Its utility extends to facilitating comparisons between multiple distributions. For instance, by contrasting the average (mean) examination scores of students from various educational institutions, one can infer which institution demonstrates superior academic performance.

Nonetheless, the mean exhibits a susceptibility to being unduly influenced by outliers or extreme values within the data. For example, while the mean effectively represents data from classes with relatively uniform frequencies, its representativeness diminishes significantly when one class possesses a markedly different frequency (e.g., a frequency of 2) compared to others (e.g., frequencies of 20, 25, 20, 21, 18). In such scenarios, the mean fails to accurately depict the underlying distribution of the data, rendering it an unsuitable representative.

When the focus is on identifying a representative value within a dataset, rather than individual data points, and the presence of extreme values might skew the result, the median serves as a more suitable indicator of central tendency than the mean. Examples include determining the characteristic productivity level of a workforce or the typical income within a nation.

The mode is the optimal statistical measure for identifying the most common or prevalent category or value within a dataset. This applies, for instance, when seeking to ascertain the most viewed television program, the consumer product with the highest demand, or the most frequently chosen vehicle color among the populace.

Remarks :

1. An empirical relationship exists among the three primary measures of central tendency:

$ 3 \text{ Median} = \text{Mode} + 2 \text{ Mean} $

2. While it is possible to compute the median for grouped data characterized by unequal class intervals, this specific methodology will not be elaborated upon in the current context.

EXERCISE 13.3

1. The subsequent frequency distribution details the monthly electricity usage for 68 consumers within a particular area. Determine the median, mean, and mode for this dataset, and subsequently analyze their relationships.

2. Given that the median of the distribution presented below is 28.5, ascertain the numerical values of $x$ and $y$.

3. A life insurance agent compiled the subsequent data concerning the age distribution of 100 policyholders. Compute the median age, considering that policies are exclusively issued to individuals aged 18 years or older but under 60 years.

4. The lengths of 40 plant leaves were measured with precision to the nearest millimeter, and the resulting data is presented in the table below:

Find the median length of the leaves.

(Hint: To accurately compute the median, the provided data must be transformed into continuous classes, as the median formula presupposes such a classification. Consequently, the class intervals will be adjusted to 117.5 - 126.5, 126.5 - 135.5, and so forth, up to 171.5 - 180.5.)

5. The ensuing table illustrates the distribution of the operational lifespan for 400 neon lamps:

Find the median life time of a lamp.

6. A sample of 100 surnames was arbitrarily selected from a local telephone directory, yielding the subsequent frequency distribution detailing the count of English alphabet letters within these surnames:

Calculate the median number of letters present in these surnames. Additionally, ascertain the mean number of letters and the modal length of the surnames.

7. Presented below is the distribution representing the weights of 30 students within a specific class. Determine the median weight for this group of students.

13.5 Summary

This chapter has covered the subsequent key concepts:

1. The calculation of the mean for grouped datasets can be achieved through several methodologies:

(i) The direct method: $\overline{x} = \frac{\sum f_i x_i}{\sum f_i}$

(ii) The assumed mean method: $\overline{x} = a + \frac{\sum f_i d_i}{\sum f_i}$

(iii) The step deviation method: $\overline{x} = a + \left(\frac{\sum f_i u_i}{\sum f_i}\right) \times h$ ,

These calculations proceed under the premise that a class's frequency is centrally located at its midpoint, known as its class mark.

2. The mode for grouped data is derivable using the following expression:

$ \operatorname {Mode} = l + \left(\frac {f _ {1} - f _ {0}}{2 f _ {1} - f _ {0} - f _ {2}}\right) \times h $

wherein the constituent symbols retain their conventional interpretations.

3. The cumulative frequency for a given class interval is defined as the aggregate of the frequencies of all class intervals preceding that particular class.

4. The median for grouped data is determined by employing the formula:

$ \operatorname {Median} = l + \left(\frac {\frac {n}{2} - \operatorname {cf}}{f}\right) \times h, $

with all symbols carrying their standard significances.

A NOTE TO THE READER

When computing the mode and median for grouped data, it is imperative to verify that the class intervals are continuous prior to the application of the respective formulas. This prerequisite also extends to the construction of an ogive. Moreover, it is important to note that for ogives, the scaling adopted for the two axes may not necessarily be uniform.