When tested, the lives (in hours) of 5 bulbs were noted as follows: 1357, 1090, 1666, 1494, 1623 The mean devi

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Q. When tested, the lives (in hours) of 5 bulbs were noted as follows: 1357, 1090, 1666, 1494, 1623 The mean deviations (in hours) from their mean is (A) 178 (B) 179 (C) 220 (D) 356

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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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Mean deviation about the mean is the average of the absolute differences between each data point and the mean. For the given bulb lifetimes, the mean is 1446 hours, and the mean deviation is 178 hours — option (A).

Why mean deviation about the mean?

Mean deviation measures how spread out the data is, but in a more intuitive way than variance or standard deviation. Instead of squaring differences (which amplifies large deviations), we take the absolute value of each difference. This gives us the average distance of each observation from the central value — here, the arithmetic mean.

The formula is simple:

\text{Mean Deviation} = \frac{1}{n} \sum_{i=1}^{n} |x_i - \bar{x}|

where $\bar{x}$ is the mean of the $n$ observations.

Let’s apply it step by step.

1. Find the mean $\bar{x}$

The bulb lifetimes are:

1357,\; 1090,\; 1666,\; 1494,\; 1623

Sum them:

1357 + 1090 = 2447

2447 + 1666 = 4113

4113 + 1494 = 5607

5607 + 1623 = 7230

Number of bulbs $n = 5$ , so

\bar{x} = \frac{7230}{5} = 1446

Tip

Always check your sum by adding in a different order or using a quick mental estimate: $1357 \approx 1400$ , $1090 \approx 1100$ , $1666 \approx 1700$ , $1494 \approx 1500$ , $1623 \approx 1600$ — sum ≈ $7300$ , so $1446$ is plausible.

2. Compute absolute deviations from the mean

For each bulb, subtract the mean and take the absolute value:

$|1357 - 1446| = |{-89}| = 89$ …

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Q.When tested, the lives (in hours) of 5 bulbs were noted as follows: 1357, 1090, 1666, 1494, 1623 The mean deviations (in hours) from their mean is (A) 178 (B) 179 (C) 220 (D) 356

Why mean deviation about the mean?

1. Find the mean xˉ\bar{x}xˉ

2. Compute absolute deviations from the mean

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Q.When tested, the lives (in hours) of 5 bulbs were noted as follows: 1357, 1090, 1666, 1494, 1623 The mean deviations (in hours) from their mean is
(A) 178
(B) 179
(C) 220
(D) 356

1. Find the mean $\bar{x}$

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