STATISTICS-Notes

NCERT prescribes three standard methods for calculating the mean of grouped data. Each method is mathematically equivalent, but the choice depends on the nature of the data and ease of computation.

1. Direct Method:
   

   This is the most straightforward approach and is suitable when class marks and frequencies are small and manageable.
   If
   

   • \(x_i\) = class mark of the of \(i^{th}\) class
   • \(f_i\) = frequency of the \(i^{th}\) class
     then the mean \(\overline{x}\) is calculated as \[ \boxed{\;\boldsymbol{\overline{x} = \dfrac{\sum{f_i x_i}}{\sum{f_i}}}\;} \] This method clearly shows how each class contributes to the overall average, but it may involve lengthy calculations when values are large.

2. Assumed Mean Method:
   

   To simplify calculations, an assumed mean \(a\) is chosen, usually one of the class marks near the centre of the distribution. Deviations of class marks from this assumed mean are then calculated.
   Let \[d_i=x_i-a\] then mean is obtained using \[ \boxed{\;\boldsymbol{\overline{x} = a +\dfrac{\sum{f_i x_i}}{\sum{f_i}}}\;} \]

   This method reduces the size of numbers involved and is particularly useful when class marks are large but evenly spaced.
3. Step Deviation Method:
   When class intervals are equal and deviations are still large, the step deviation method provides further simplification by scaling down deviations.
   If
   • \(h\) = common width of the class intervals
   • \(u_i=\frac{x_i-a}{h}\)
     then the mean is calculated as \[ \boxed{\;\boldsymbol{\overline{x} = a + h \left(\dfrac{\sum{f_i x_i}}{\sum{f_i}}\right)}\;} \] This method is computationally efficient and widely used in examinations, especially when dealing with large data sets.

It is assumed that the frequency of each class interval is centred around its mid-point. So the mid-point (or class mark) of each class can be chosen to represent the observations falling in the class.

\[\small\boxed{\boldsymbol{x=\dfrac{\text{Upper class limit + Lower class limit}}{2}}}\]

In statistics, the mode represents the value that occurs most frequently in a data set. When data is presented in a grouped form, individual values are not explicitly available. Hence, the mode cannot be identified by simple inspection. Instead, a systematic method is used to estimate the mode of grouped data, which indicates the class interval containing the highest concentration of observations.

Modal Class

In a grouped frequency distribution, the modal class is the class interval with the maximum frequency. This class gives a rough idea of where the mode lies, but a formula is required to obtain a more precise value

Formula for Mode of Grouped Data

Let

• \(l\) = lower limit of the modal class
• \(h\) = size (width) of the class interval
• \(f_1\) = frequency of the modal class
• \(f_0\) = frequency of the class preceding the modal class
• \(f_2\) = frequency of the class succeeding the modal class

Then, the mode is given by the formula \[\boxed{\;\boldsymbol{\text{Mode}=l+\left(\dfrac{f_1-f_0}{2f_1-f_0-f_2}\right)h}\;}\] This formula estimates the value around which data is most densely clustered within the modal class.

Special Cases

• If two or more classes have the same highest frequency, the data is said to be bi-modal or multi-modal, and the mode may not be uniquely defined.
• If frequencies rise steadily and then fall, the mode is well-defined and meaningful.
• If the distribution is irregular, the mode may only give an approximate indication of concentration.

Uses of Mode

The mode is particularly useful when:

• Data contains extreme values that may distort the mean.
• The most common or typical value is required rather than an average.
• Data is qualitative or grouped into categories such as shoe sizes, grades, or income groups.

The median is a measure of central tendency that represents the middle value of a data set when the observations are arranged in order. In many real-life situations, data is organised into class intervals rather than listed individually. Such data is called grouped data, and in this case, the median cannot be identified by direct observation. Instead, it is determined using a structured approach based on cumulative frequencies.

Meaning of Median in Grouped Data

For grouped data, the median divides the entire frequency distribution into two equal parts. This means that half of the observations lie below the median value and the remaining half lie above it. Since individual observations are unknown, the median obtained is an approximate value that lies within a specific class interval known as the median class.

Cumulative Frequency

To find the median, the concept of cumulative frequency is essential. The cumulative frequency of a class is the total number of observations up to and including that class. Cumulative frequencies help in locating the position of the median within the distribution.

Steps to Find the Median of Grouped Data

• Arrange the data into a frequency table with class intervals and corresponding frequencies.
• Compute the cumulative frequencies for all classes.
• Find the value of \(\frac{N}{2}\), where \(N\) is the sum of all frequencies.
• Identify the class whose cumulative frequency is just greater than \(\frac{N}{2}\). This class is called the median class.
• Apply the median formula to calculate the required value.

Formula for Median of Grouped Data

Let

• \(l\) = lower limit of the median class
• \(h\)= class width
• \(f\)= frequency of the median class
• \(cf\)= cumulative frequency of the class preceding the median class
• \(N\)= total frequency

Then, the median is given by \[\boxed{\;\boldsymbol{\text{Median}=l+\left(\dfrac{\frac{N}{2}-cf}{f}\right)h}\;}\]

Interpretation of the Formula

The term \(\frac{N}{2}−cf\) represents how many observations lie within the median class before reaching the middle position. Dividing by \(f\) gives the fractional position within the median class, and multiplying by \(h\) scales this position according to the class width. Adding this to \(l\) locates the median precisely within the class interval.

Key Characteristics of the Median

• The median is not affected by extreme values or outliers.
• It is particularly useful for skewed distributions.
• For grouped data, the median is an estimated value based on class intervals.

In a moderately symmetrical distribution, the median is related to the mean and mode by the empirical relation: \[\boxed{\;\boldsymbol{}\text{Mode}=3\text{(Median)}-2\text{(Mean)}\;}\] This relation helps in cross-checking calculations and understanding the overall nature of the distribution.

The marks obtained by 30 students of Class X of a certain school in a Mathematics paper consisting of 100 marks are presented in table below. Find the mean of the marks obtained by the students. \[\scriptsize \begin{array}{|c|c|} \hline \text{Marks Obtained } x_i&10&20&36&40&50&56&60&70&72&80&88&92&95\\\hline \text{Number of Students }f_i&1&1&3&4&3&2&4&4&1&1&2&3&1\\\hline \end{array} \]

Solution

Mean can be found by \[ \overline{x}=\dfrac{\sum(f_ix_i)}{\sum(f_i)} \]

Let us put \(x_i\) and \(f_ix_i\) is a table

Marks Obtained \(x_i\)	Number of Students \(f_i\)	\(f_ix_i\)
10	1	10
20	1	20
36	3	108
40	4	160
50	3	150
56	2	112
60	4	240
70	4	280
72	1	72
80	1	80
88	2	176
92	3	276
95	1	95
Total	\(\sum(f_i)=30\)	\(\sum(f_ix_i)=1779\)

Substitutng Values in Mean Formula \[ \begin{aligned} \overline{x}&=\dfrac{\sum{f_ix_i}}{\sum{f_i}}\\\\ &=\dfrac{1779}{30}\\\\ &=59.3 \end{aligned} \]

Let us convert this this ungrouped data into grouped data by forming class-interval of width, say 15

While allocating frequencies to each class-interval, students falling in any upper class-limit would be considered in the next class, e.g., 4 students who have obtained 40 marks would be considered in the class interval 40-55 and not in 25-40.

\[ \begin{array}{|c|c|} \hline \text{Class Interval }&10-25&25-40&40-55&55-70&70-85&85-100\\\hline \text{Number of Students }&2&3&7&6&6&6\\\hline \end{array} \] \[ \begin{aligned} \text{Class Mark }&=\dfrac{\text{Upper Class Limit + Lower Class Limit}}{2}\\\\ &=\dfrac{10+25}{2}\\\\ &=17.5 \end{aligned} \] Similarly, we can find class-interval of other class interval.
Let us put all data in a table

Class Interval	Number of Students \(f_i\)	Class Mark \(x_i\) \(\left(x_i=\frac{UCL+LCL}{2}\right)\)	\(f_ix_i\)
10-25	2	17.5	35.0
25-40	3	32.5	97.5
40-55	7	47.5	332.5
55-70	6	62.5	375
70-85	6	77.5	465
85-100	93.5	6	555.0
Total	\(\sum{f_i}=30\)		\(\sum{f_ix_i}=1860.0\)

Substituting values in Mean Formula \[ \begin{aligned} \overline{x}&=\dfrac{\sum{f_ix_i}}{\sum{f_i}}\\\\ &=\dfrac{1860}{30}\\\\ &=62 \end{aligned} \]

Let us solve the same question by assumed mean method

Let's assume mean (mid value) of \(x_i\), in this example we can take assumed mean value as 47.5 or 62.5
Let assumed mean \(a\) = 47.5

Class Interval	Number of Students \(f_i\)	Class Mark \(x_i\) \(\left(x_i=\frac{UCL+LCL}{2}\right)\)	Deviation \(d_i = x_i-47.5\)	\(f_id_i\)
10-25	2	17.5	-30	-60
25-40	3	32.5	-15	-45
40-55	7	47.5	0	0
55-70	6	62.5	15	90
70-85	6	77.5	30	180
85-100	93.5	6	45	270
Total	\(\sum{x_i}=30\)			\(\sum{f_id_i}=435\)

Substituting values in Assumed Mean Formula \[ \begin{aligned} \overline{x}&= a+\dfrac{\sum{f_id_i}}{\sum{f_i}}\\\\ &=47.5 + \dfrac{435}{30}\\\\ &=47.5 + 14.5\\\\ &=62.0 \end{aligned} \]

Applying Step Deviation Method

If we divide column-4 value with class size (h)
So, Let \[ u_i=\dfrac{x_i-a}{h} \] Let us form the table with this additional value \(u_i\)

Class Interval	Number of Students \(f_i\)	Class Mark \(x_i\) \(\left(x_i=\frac{UCL+LCL}{2}\right)\)	\(d_i = x_i-47.5\)	\(u_i=\dfrac{x_i-a}{h}\)	\(f_iu_i\)
10-25	2	17.5	-30	-2	-4
25-40	3	32.5	-15	-1	-3
40-55	7	47.5	0	0	0
55-70	6	62.5	15	1	6
70-85	6	77.5	30	2	12
85-100	93.5	6	45	3	18
Total	\(\sum{f_i}=30\)				\(\sum{f_iu_i}=29\)

Substituting Values in Step Deviation Method \[ \require{cancel} \begin{aligned} \overline{x} &= a + h \left(\dfrac{\sum{f_i u_i}}{\sum{f_i}}\right)\\\\ &=47.5+ \left(\cancel{15}\times\dfrac{29}{\cancelto{2}{30}}\right)\\\\ &=47.5+ \left(\dfrac{29}{2}\right)\\\\ &=47.5 + 14.5\\\\ &=26.0 \end{aligned} \]

A survey conducted on 20 households in a locality by a group of students resulted in the following frequency table for the number of family members in a household:

\[ \begin{array}{|c|c|} \hline \text{Family Size}&1-3&3-5&5-7&7-9&9-11\\\hline \text{Number of Families}&7&8&2&2&1\\\hline \end{array} \]

Solution

Here is Maximum Class Frequecy is 8, so modal class corresponding to this frequency is 3-5

• Modal Class=3-5
• Lower Limit of Modal Class \(l\)=3
• Class Size \(h\)=2
• Frequecy \(f_1\) of Modal Class = 8
• Frequecy \(f_0\) of class preceding the Modal Class = 7
• Frequecy \(f_2\) of class succeeding the Modal Class = 2

Substitute these value in Formula: \[ \begin{aligned} \text{Mode}&=l+\left(\dfrac{f_1-f_0}{2f_1-f_0-f_2}\right)h\\\\ &=3+\dfrac{8-7}{2\times 8-7-2}\times 2\\\\ &=3+\dfrac{1}{16-9}\times 2\\\\ &=3+\dfrac{1}{7}\times 2\\\\ &=3+\dfrac{2}{7}\\\\ &=3+0.286\\\\ &=3.286 \end{aligned} \]

A survey regarding the heights (in cm) of 51 girls of Class X of a school was conducted and the following data was obtained:

Height (in cm)	Number of Girls
Less than 140	4
Less than 145	11
Less than 150	29
Less than 155	40
Less than 160	46
Less than 165	51

Find the median of the height.

Solution

To calculate Median of Height , we need to find class interval and their corresponding frequncies

Class Interval	Frequecy	Cumulative Frequecy
Less than 140	4	4
140-145	7	11
145-150	18	29
150-155	11	40
155-160	6	46
160-165	5	51

\(n=51\), therfore, \(\frac{n}{2}=25.5\), this observation lies in class 145-150

• \(l\) (Lower limt)=145
• \(cf\) (the cumulative frequency of the class preceding 145 - 150) = 11
• \(f\) (the frequency of the median class 145 - 150) = 18
• \(h\) = Class Size = 5

Substituting Median Formula \[ \begin{aligned} \text{Median}&=l+\left(\dfrac{\frac{n}{2}-cf}{f}\right)\times h\\\\ &=145+\left(\dfrac{\frac{51}{2}-11}{18}\right)\times 5\\\\ &=145+\left(\dfrac{25.5-11}{18}\right)\times 5\\\\ &=145+\left(\dfrac{14.5\times 5}{18}\right)\\\\ &=145+\left(\dfrac{72.5}{18}\right)\\\\ &=145+4.03\\\\ &=149.03 \end{aligned} \]

The median of the following data is 525. Find the values of x and y, if the total frequency is 100.

Class Interval	Frequecy
0-100	2
100-200	5
200-300	\(x\)
300-400	12
400-500	17
500-600	20
600-700	\(y\)
700-800	9
800-900	7
900-1000	4

Solution

Class Interval	Frequecy	Cumulative Frequecy
0-100	2	\(2\)
100-200	5	\(7\)
200-300	\(x\)	\(7+x\)
300-400	12	\(19+x\)
400-500	17	\(36+x\)
500-600	20	\(56+x\)
600-700	\(y\)	\(56+x+y\)
700-800	9	\(65+x+y\)
800-900	7	\(72+x+y\)
900-1000	4	\(76+x+y\)

It is given that \(n=100\)

\[ \begin{align} 76+x+y&=100\\ x+y&=100-76\\ x+y&=24\tag{1} \end{align} \] Median Class=500-600,

• \(l=500\)
• \(n=100\)
• \(cf=36+x\)
• \(f=20\)
• \(h=100\)

Substituting Values in Median Formula \[ \require{cancel} \begin{aligned} \text{Median}&=l+\left(\dfrac{\frac{n}{2}-cf}{f}\right)\times h\\\\ 525&=500+\left(\dfrac{\frac{100}{2}-(36+x)}{20}\right)\times 100\\\\ 525-500&=\left(\dfrac{50-36-x}{20}\right)\times 100\\\\ 525-500&=\left(\dfrac{14-x}{\cancel{20}}\right)\times \cancelto{5}{100}\\\\ \dfrac{\cancelto{5}{25}}{\cancel{5}}&=14-x\\\\\ 5&=14-x\\\\ \Rightarrow x&=9 \end{aligned} \]

Substituting \(x=9\) in equation-(1)

\[ \begin{aligned} x+y&=24\\ 9+y&=24\\ y&=24-9\\ &=15 \end{aligned} \]