Statistics - Class 10 Maths

1. Measures of Central Tendency

A measure of central tendency is a single representative value that summarizes an entire set of numerical data by pinpointing the center or middle of its distribution.

Mean ($\bar{x}$)

The arithmetic average computed by summing all observations and dividing by the total frequency.

Median

The middle-most value separating the higher half from the lower half of an arranged distribution.

Mode

The observation that occurs with the highest frequency in a given distribution.

2. Mean of Grouped Data

For continuous grouped frequency distributions, individual observations are not known directly. Each class interval is represented by its midpoint, known as the class mark ($x_i$):

Class Mark Formula $$x_i = \frac{\text{Upper class limit} + \text{Lower class limit}}{2}$$

3. Direct Method

When numerical values of class marks $x_i$ and frequencies $f_i$ are small, the mean $\bar{x}$ is calculated directly:

Direct Method Formula $$\bar{x} = \frac{\sum f_i x_i}{\sum f_i} = \frac{\sum_{i=1}^n f_i x_i}{N} \qquad \left(N = \sum f_i\right)$$

Worked Example: Direct Method

Problem: The following table shows the distribution of marks scored by 30 students in a test. Find the mean marks using the Direct Method.

Class Interval	Class Mark ($x_i$)	Frequency ($f_i$)	$f_i x_i$
10 – 25	$17.5$	$2$	$35.0$
25 – 40	$32.5$	$3$	$97.5$
40 – 55	$47.5$	$7$	$332.5$
55 – 70	$62.5$	$6$	$375.0$
70 – 85	$77.5$	$6$	$465.0$
85 – 100	$92.5$	$6$	$555.0$
Total	—	$\sum f_i = 30$	$\sum f_i x_i = 1860.0$

Step 1: Sum the frequencies: $\sum f_i = 30$.

Step 2: Sum the product of frequencies and class marks: $\sum f_i x_i = 1860$.

Step 3: Compute the mean: $$\bar{x} = \frac{\sum f_i x_i}{\sum f_i} = \frac{1860}{30} = 62$$

Answer: The mean marks obtained by the students is $62$.

4. Assumed Mean Method (with Derivation)

When class marks $x_i$ are large, choose a convenient central value $a$ from the $x_i$ column as the assumed mean, and calculate deviations $d_i = x_i - a$.

Derivation of Assumed Mean Formula

1. Define the deviation of each class mark from the assumed mean $a$: $$d_i = x_i - a \implies x_i = a + d_i$$

2. Multiply both sides by the corresponding frequency $f_i$: $$f_i x_i = f_i(a + d_i) = f_i a + f_i d_i$$

3. Summing across all $n$ classes: $$\sum f_i x_i = \sum (f_i a) + \sum f_i d_i = a \sum f_i + \sum f_i d_i$$

4. Divide both sides by $\sum f_i$: $$\frac{\sum f_i x_i}{\sum f_i} = \frac{a \sum f_i}{\sum f_i} + \frac{\sum f_i d_i}{\sum f_i}$$

5. Since $\bar{x} = \dfrac{\sum f_i x_i}{\sum f_i}$ and $\bar{d} = \dfrac{\sum f_i d_i}{\sum f_i}$: $$\bar{x} = a + \frac{\sum f_i d_i}{\sum f_i}$$

Assumed Mean Formula $$\bar{x} = a + \frac{\sum f_i d_i}{\sum f_i} \qquad (d_i = x_i - a)$$

Worked Example: Assumed Mean Method

Problem: The following data gives the daily wages (in ₹) of 50 workers of a factory. Find the mean daily wage using the Assumed Mean Method.

Let assumed mean $a = 550$ (chosen from central class mark).

Daily Wages (₹)	$x_i$	$d_i = x_i - 550$	No. of Workers ($f_i$)	$f_i d_i$
500 – 520	$510$	$-40$	$12$	$-480$
520 – 540	$530$	$-20$	$14$	$-280$
540 – 560	$550$	$0$	$8$	$0$
560 – 580	$570$	$+20$	$6$	$+120$
580 – 600	$590$	$+40$	$10$	$+400$
Total	—	—	$\sum f_i = 50$	$\sum f_i d_i = -240$

Step 1: Select assumed mean $a = 550$.

Step 2: Compute $\sum f_i = 50$ and $\sum f_i d_i = (-480 - 280 + 0 + 120 + 400) = -240$.

Step 3: Substitute into the Assumed Mean formula: $$\bar{x} = a + \frac{\sum f_i d_i}{\sum f_i} = 550 + \left(\frac{-240}{50}\right) = 550 - 4.80 = 545.20$$

Answer: The mean daily wage of the workers is ₹545.20.

5. Step-Deviation Method (with Derivation)

When all class intervals have a common class size $h$, deviations $d_i$ share a common factor $h$. Define the reduced variable $u_i = \dfrac{x_i - a}{h}$.

Derivation of Step-Deviation Formula

1. Define the scaled step-deviation: $$u_i = \frac{x_i - a}{h} \implies x_i - a = h u_i \implies x_i = a + h u_i$$

2. Multiply by frequency $f_i$: $$f_i x_i = f_i a + h(f_i u_i)$$

3. Take the algebraic sum over all classes: $$\sum f_i x_i = a \sum f_i + h \sum f_i u_i$$

4. Divide throughout by $\sum f_i$: $$\frac{\sum f_i x_i}{\sum f_i} = a + h \left(\frac{\sum f_i u_i}{\sum f_i}\right)$$

5. Recognizing $\bar{x}$ on the left-hand side: $$\bar{x} = a + \left(\frac{\sum f_i u_i}{\sum f_i}\right) \times h$$

Step-Deviation Formula $$\bar{x} = a + \left(\frac{\sum f_i u_i}{\sum f_i}\right) \times h \qquad \left(u_i = \frac{x_i - a}{h}\right)$$

Worked Example: Step-Deviation Method

Problem: A study was conducted on 45 localities to determine the concentration of $\text{SO}_2$ in parts per million (ppm). Calculate the mean concentration using the Step-Deviation Method.

Let assumed mean $a = 0.10$ and class width $h = 0.04$.

$\text{SO}_2$ Conc. (ppm)	$x_i$	$u_i = \frac{x_i - 0.10}{0.04}$	Frequency ($f_i$)	$f_i u_i$
0.00 – 0.04	$0.02$	$-2$	$4$	$-8$
0.04 – 0.08	$0.06$	$-1$	$9$	$-9$
0.08 – 0.12	$0.10$	$0$	$9$	$0$
0.12 – 0.16	$0.14$	$+1$	$14$	$+14$
0.16 – 0.20	$0.18$	$+2$	$5$	$+10$
0.20 – 0.24	$0.22$	$+3$	$4$	$+12$
Total	—	—	$\sum f_i = 45$	$\sum f_i u_i = 19$

Step 1: Identify $a = 0.10$ and common width $h = 0.04$.

Step 2: Compute $\sum f_i = 45$ and $\sum f_i u_i = -8 - 9 + 0 + 14 + 10 + 12 = 19$.

Step 3: Apply the Step-Deviation formula: $$\bar{x} = a + \left(\frac{\sum f_i u_i}{\sum f_i}\right) \times h = 0.10 + \left(\frac{19}{45}\right) \times 0.04$$ $$\bar{x} \approx 0.10 + 0.0169 = 0.1169 \approx 0.117\text{ ppm}$$

Answer: The mean concentration of $\text{SO}_2$ is approximately $0.117\text{ ppm}$.

6. Median

The median is the value of the middle-most observation when the data is arranged in ascending or descending order of magnitude.

Ungrouped Data Review:
• If $n$ is odd: $\text{Median} = \left(\dfrac{n + 1}{2}\right)^{\text{th}}\text{ observation}$.
• If $n$ is even: $\text{Median} = \dfrac{\left(\frac{n}{2}\right)^{\text{th}}\text{ obs} + \left(\frac{n}{2} + 1\right)^{\text{th}}\text{ obs}}{2}$.

7. Median for Grouped Data

To compute the median for continuous grouped frequency distributions:

Steps to Calculate Grouped Median

Step 1: Construct the cumulative frequency ($cf$) column.

Step 2: Determine the total frequency $n = \sum f_i$ and find $\dfrac{n}{2}$.

Step 3: Locate the median class — the class interval whose cumulative frequency is just greater than (or equal to) $\dfrac{n}{2}$.

Step 4: Apply the interpolation formula: $$\text{Median} = l + \left(\frac{\frac{n}{2} - cf}{f}\right) \times h$$

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

$l =$ lower limit of the median class.
$n =$ total number of observations ($\sum f_i$).
$cf =$ cumulative frequency of the class preceding the median class.
$f =$ actual frequency of the median class.
$h =$ class size (assuming equal class width).

8. Mode of Grouped Data & Modal Class

The modal class is the class interval corresponding to the maximum frequency in the given distribution. The mode lies inside this modal class and is computed by:

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

$l =$ lower limit of the modal class.
$h =$ size 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.

9. Empirical Relationship Between Mean, Median, and Mode

For moderately skewed frequency distributions, Karl Pearson established the following empirical connection between the three measures of central tendency:

Empirical Formula $$\text{Mode} = 3\,\text{Median} - 2\,\text{Mean}$$

Alternate Formulations: $$\text{Mean} - \text{Mode} = 3(\text{Mean} - \text{Median})$$ $$\text{Mode} - \text{Median} = 2(\text{Median} - \text{Mean})$$

10. Cumulative Frequency Curves (Ogives)

An ogive (pronounced oh-jive) is a smooth graph obtained by plotting cumulative frequencies against class boundary limits.

(i) Less Than Series Ogive

Formed by cumulating frequencies from the lowest class upward. Points plotted are: $(\text{Upper Class Limit}, \text{Cumulative Frequency})$. It is a rising S-curve from bottom-left to top-right.

(ii) More Than (Greater Than) Series Ogive

Formed by cumulating frequencies from the highest class downward. Points plotted are: $(\text{Lower Class Limit}, \text{Cumulative Frequency})$. It is a falling curve from top-left to bottom-right.

The intersection point of the 'less than' and 'more than' ogives projects directly down to the median on the horizontal axis.

Obtaining Median Graphically:
• Method 1 (Two Ogives): Draw both ogives on the same scale. The abscissa ($x$-coordinate) of their point of intersection gives the median.
• Method 2 (Single Ogive): Locate $\dfrac{N}{2}$ on the vertical $y$-axis, draw a horizontal line to intersect the ogive at point $P$, then drop a perpendicular from $P$ to the $x$-axis. The foot of this perpendicular is the median.