Calculate the mean deviation about the mean for the following frequency distribution: | Class interval | 0 - 4 | 4 - 8 | 8 - 12 | 12 - 16 | 16 - 20 | |---|---|---|---|---|---| | Frequency | 4 | 6 | 8 | 5 | 2 |

> [!TLDR] > Using class midpoints, the mean is $\bar x = 9.2$ and the mean deviation about the mean is $3.84$. ## Step-by-step solution **1. Midpoints and frequencies** | Class interval | Midpoint $x_i$ | Frequency $f_i$ | |---|---|---| | 0 - 4 | 2 | 4 | | 4 - 8 | 6 | 6 | | 8 - 12 | 10 | 8 | | 12 - 16 | 14 | 5 | | 16 - 20 | 18 | 2 | Total frequency $N = \sum f_i = 25$. **2. Mean** $$\bar x = \frac{\sum f_i x_i}{N} = \frac{2(4)+6(6)+10(8)+14(5)+18(2)}{25} = \frac{8+36+80+70+36}{25} = \frac{230}{25} = 9.2.$$ **3. Absolute deviations and weighted sum** | $x_i$ | $f_i$ | $\lvert x_i-\bar x\rvert$ | $f_i\lvert x_i-\bar x\rvert$ | |---|---|---|---| | 2 | 4 | 7.2 | 28.8 | | 6 | 6 | 3.2 | 19.2 | | 10 | 8 | 0.8 | 6.4 | | 14 | 5 | 4.8 | 24.0 | …

Calculate the mean deviation about the mean for th – Class 12 Mathematics

NCERT Exemplar · Q16

Calculate the mean deviation about the mean for the following frequency distribution:

Class interval	0 - 4	4 - 8	8 - 12	12 - 16	16 - 20
Frequency	4	6	8	5	2

Telangana TsbieLong· 5mImportance★★★★★est

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Concept understanding — Mean Deviation About Mean

Mean Deviation About Mean – The Intuition First

Imagine you have a small set of numbers: the marks of five students in a test: 4, 6, 8, 10, 12. The average (mean) is 8. Now, each student is some distance away from this average. The student who scored 4 is 4 marks below the mean; the one who scored 12 is 4 marks above. The student who scored 8 is exactly at the mean.

If you simply add these distances, the positives and negatives cancel out — you get zero. That's not useful. So instead, we ask: on average, how far is each data point from the mean? That's the mean deviation about mean.

Note

Mean deviation is a measure of spread or dispersion. It tells you how scattered the data is around the central value. A small mean deviation means most data points are close to the mean; a large one means they are spread out.

The Precise Definition

For a set of $n$ observations $x_1, x_2, \dots, x_n$ with mean $\bar{x}$ , the mean deviation about mean (often written as MD or M.D.) is:

\text{MD}(\bar{x}) = \frac{1}{n} \sum_{i=1}^{n} |x_i - \bar{x}|

That vertical bars mean absolute value — we take the distance without caring about direction. So every deviation is positive.

\text{Mean Deviation about Mean} = \frac{\sum |x_i - \bar{x}|}{n}

Step-by-Step Calculation

Let's use the marks example: 4, 6, 8, 10, 12.

Step 1: Find the mean.

\bar{x} = \frac{4+6+8+10+12}{5} = \frac{40}{5} = 8

Step 2: Find each absolute deviation $|x_i - \bar{x}|$ .

| $x_i$ | $x_i - \bar{x}$ | $|x_i - \bar{x}|$ |

|------|----------------|-------------------|

| 4 | -4 | 4 |

| 6 | -2 | 2 |

| 8 | 0 | 0 |

| 10 | 2 | 2 |

| 12 | 4 | 4 |

Step 3: Sum the absolute deviations.

4 + 2 + 0 + 2 + 4 = 12

Step 4: Divide by $n = 5$ .

\text{MD} = \frac{12}{5} = 2.4

So, on average, each student's mark is 2.4 marks away from the mean of 8.

Tip

Notice that the mean deviation is always less than or equal to the standard deviation (another measure of spread). For this data, standard deviation is about 2.83, which is larger than 2.4. This is because standard deviation squares deviations, giving more weight to extreme values.

Why Use Absolute Values?

You might wonder: why not just average the plain deviations (without absolute value)? Because the sum of $(x_i - \bar{x})$ is always zero — that's a property of the mean. The absolute value is the simplest way to make all deviations positive so they don't cancel.

Watch out

A common mistake: forgetting to take absolute values and getting zero. Always check: if your sum of deviations is zero, you forgot the absolute value.

When Is This Used?

Mean deviation is intuitive and easy to explain. It's used in:

Quality control (checking how consistent a manufacturing process is) …

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Concept: Mean Deviation About Mean measures the average absolute deviation of observations from their arithmetic mean.

Solution:

First, find the class marks $x_i$ and compute the mean $\bar{x}$ :

Class	$x_i$	$f_i$	$f_i x_i$
0–4	2	4	8
4–8	6	6	36
8–12	10	8	80
12–16	14	5	70
16–20	18	2	36
Total		25	230

Mean: $\bar{x} = \frac{\sum f_i x_i}{\sum f_i} = \frac{230}{25} = 9.2$

Next, calculate absolute deviations $|x_i - \bar{x}|$ and their weighted sum:

| $x_i$ | $f_i$ | $|x_i - 9.2|$ | $f_i |x_i - 9.2|$ |

|-------|-------|---------------|---------------------|

| 2 | 4 | 7.2 | 28.8 |

| 6 | 6 | 3.2 | 19.2 | …

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Using class midpoints, the mean is $\bar x = 9.2$ and the mean deviation about the mean is $3.84$ .

Step-by-step solution

1. Midpoints and frequencies

Class interval	Midpoint $x_i$	Frequency $f_i$
0 - 4	2	4
4 - 8	6	6
8 - 12	10	8
12 - 16	14	5
16 - 20	18	2

Total frequency $N = \sum f_i = 25$ .

2. Mean

\bar x = \frac{\sum f_i x_i}{N} = \frac{2(4)+6(6)+10(8)+14(5)+18(2)}{25} = \frac{8+36+80+70+36}{25} = \frac{230}{25} = 9.2.

3. Absolute deviations and weighted sum

$x_i$	$f_i$	$\lvert x_i-\bar x\rvert$	$f_i\lvert x_i-\bar x\rvert$
2	4	7.2	28.8
6	6	3.2	19.2
10	8	0.8	6.4
14	5	4.8	24.0

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Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

TG EAPCET 2022Set eng-2022-07-18-AN1 markMCQ
$\dfrac{2}{3}$
›Reveal solutionSolution
The number of heads in 5 coin throws follows a Binomial distribution. The mean of a Binomial distribution is $np$ , which for this problem is $5 \times \frac{1}{2} = \frac{5}{2}$ .

When we perform a random experiment multiple times, and each trial has only two possible outcomes (often called "success" and "failure"), with the probability of success remaining constant for each trial, we are dealing with a sequence of Bernoulli trials. If we are interested in the total number of successes in a fixed number of such trials, the random variable follows a Binomial distribution.

In this problem, throwing a coin is a Bernoulli trial:

"Success" can be defined as getting a head.

"Failure" is getting a tail.

The probability of getting a head on a single throw is $P(\text{Head}) = \frac{1}{2}$ (assuming a fair coin). This probability remains constant for each throw.

The coin throws are independent events.

We are throwing 5 coins, so there are $n=5$ independent trials. The random variable is the number of heads, which is the number of successes in these $n$ trials. Therefore, this random variable follows a Binomial distribution.

Identify the type of distribution and its parameters:

The random variable $X$ , representing the number of heads in 5 coin throws, follows a Binomial distribution. A Binomial distribution is characterized by two parameters:

$n$ : The number of trials. Here, $n=5$ (since 5 coins are thrown).

$p$ : The probability of success in a single trial. Here, success is getting a head, so $p = \frac{1}{2}$ .

We can denote this as $X \sim B(n, p) = B(5, \frac{1}{2})$ .

Recall the formula for the mean of a Binomial distribution: …
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TG EAPCET 2021Set eng-2021-08-04-AN1 markMCQ
$Q. If 65 is the range of the ungrouped data 50, 70, 60, B, 20, 40, then the absolute difference of the possible values of B is (A) 90 (B) 20 (C) 80 (D) 65$
›Reveal solutionSolution
The range is the difference between the maximum and minimum values. Given the range 65, the unknown value $B$ must either be the new maximum or the new minimum, leading to two possible values whose absolute difference is 90.

The range of a data set is defined as the difference between the largest and the smallest observation. When a data set has one unknown value, that unknown could be the maximum, the minimum, or somewhere in between — and the given range tells us which case actually occurs.

Here the known values are 50, 70, 60, 20, and 40. Sorting them gives 20, 40, 50, 60, 70. So without $B$ , the current minimum is 20 and the current maximum is 70, giving a range of $70 - 20 = 50$ . But the problem says the range of the full data (including $B$ ) is 65, which is larger than 50. That means $B$ must lie outside the existing interval $[20, 70]$ — it must be either smaller than 20 or larger than 70.

Case 1: $B$ is the new minimum.

If $B < 20$ , then the minimum becomes $B$ and the maximum remains 70. The range is $70 - B = 65$ , so $B = 70 - 65 = 5$ .

Case 2: $B$ is the new maximum.

If $B > 70$ , then the maximum becomes $B$ and the minimum remains 20. The range is $B - 20 = 65$ , so $B = 20 + 65 = 85$ . …
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Step-by-step solution

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