Class 11 — Mathematics (NCERT)

Measures of Dispersion: Variance and Standard Deviation

Class 11

✓By the end of this lesson students will be able to understand the need for measures of dispersion.
✓By the end of this lesson students will be able to define and calculate variance for ungrouped and grouped data.
✓By the end of this lesson students will be able to define and calculate standard deviation for ungrouped and grouped data.
✓By the end of this lesson students will be able to interpret the meaning and properties of variance and standard deviation.
✓By the end of this lesson students will be able to compare the variability of two or more distributions using the coefficient of variation.

Key concepts

Measures of Dispersion

Measures of central tendency (like mean, median, mode) give us an idea about the central value of the data. However, they do not tell us anything about how the data is spread out or dispersed. Measures of dispersion quantify the extent to which the values in a distribution are spread out from the average. Common measures include Range, Quartile Deviation, Mean Deviation, Variance, and Standard Deviation. Variance and Standard Deviation are the most important and widely used measures of dispersion.

Variance (σ²)

Variance is defined as the average of the squares of the deviations of the observations from their arithmetic mean. It is denoted by σ² (sigma squared). A larger variance indicates that the data points are spread out over a wider range, while a smaller variance indicates that the data points are clustered closer to the mean. The unit of variance is the square of the unit of the observations.

For ungrouped data: σ² = (1/N) Σ(xi - x̄)² OR σ² = (1/N) Σxi² - (x̄)²\nFor grouped data: σ² = (1/N) Σfi(xi - x̄)² OR σ² = (1/N) Σfi xi² - (x̄)² (where N = Σfi)

Standard Deviation (σ)

Standard deviation is the positive square root of the variance. It is denoted by σ (sigma). It is the most important and widely used measure of dispersion. Standard deviation is expressed in the same units as the data, which makes it easier to interpret compared to variance. A low standard deviation indicates that the data points tend to be close to the mean, while a high standard deviation indicates that the data points are spread out over a wider range.

σ = √Variance = √σ²

Coefficient of Variation (C.V.)

The Coefficient of Variation is a relative measure of dispersion, used to compare the variability of two or more distributions, especially when their means are different or when the units of measurement are different. It is a unitless measure.

C.V. = (Standard Deviation / Mean) × 100 = (σ / x̄) × 100

Key facts to remember

1Measures of dispersion indicate the spread or variability of data.
2Variance (σ²) is the average of the squares of the deviations from the mean.
3Standard Deviation (σ) is the positive square root of the variance.
4Standard deviation is expressed in the same units as the original data.
5Both variance and standard deviation are always non-negative.
6Standard deviation is independent of change of origin but dependent on change of scale.
7A larger standard deviation implies greater variability in the data.
8The Coefficient of Variation (C.V.) is a relative measure of dispersion, useful for comparing different datasets.

Worked examples

Example 1

Calculate the variance and standard deviation for the following data: 6, 8, 10, 12, 14, 16, 18, 20.

IStep 1: Calculate the arithmetic mean (x̄).

IIx̄ = (6 + 8 + 10 + 12 + 14 + 16 + 18 + 20) / 8 = 104 / 8 = 13

IIIStep 2: Calculate Σxi².

IVΣxi² = 6² + 8² + 10² + 12² + 14² + 16² + 18² + 20²

V= 36 + 64 + 100 + 144 + 196 + 256 + 324 + 400 = 1520

VIStep 3: Calculate Variance (σ²) using the formula σ² = (1/N) Σxi² - (x̄)².

VIIN = 8

VIIIσ² = (1/8) × 1520 - (13)²

9= 190 - 169 = 21

10Step 4: Calculate Standard Deviation (σ) by taking the positive square root of the variance.

11σ = √21 ≈ 4.58

Answer

Variance (σ²) = 21, Standard Deviation (σ) ≈ 4.58

Using the computational formula (Σxi² / N) - x̄² often simplifies calculations.

Example 2

Calculate the variance and standard deviation for the following discrete frequency distribution:

IStep 1: Prepare a table to calculate Σfi xi and Σfi xi².

II| xi | fi | fi xi | fi xi² |

III|----|----|-------|--------|

IV| 4 | 3 | 12 | 48 |

V| 6 | 5 | 30 | 180 |

VI| 8 | 9 | 72 | 576 |

VII| 10 | 5 | 50 | 500 |

VIII| 12 | 3 | 36 | 432 |

9|----|----|-------|--------|

10| | N=25 | Σfi xi = 200 | Σfi xi² = 1736 |

11Step 2: Calculate the arithmetic mean (x̄).

12x̄ = Σfi xi / N = 200 / 25 = 8

13Step 3: Calculate Variance (σ²) using the formula σ² = (1/N) Σfi xi² - (x̄)².

14σ² = (1/25) × 1736 - (8)²

15= 69.44 - 64 = 5.44

16Step 4: Calculate Standard Deviation (σ).

17σ = √5.44 ≈ 2.33

Answer

Variance (σ²) = 5.44, Standard Deviation (σ) ≈ 2.33

Ensure to correctly calculate fi xi and fi xi² for each class.

Example 3

Calculate the variance and standard deviation for the following continuous frequency distribution:

IStep 1: Determine the mid-points (xi) for each class interval.

III|----------------|----|----------------|-------|--------|

IV| 0-10 | 2 | 5 | 10 | 50 |

V| 10-20 | 3 | 15 | 45 | 675 |

VI| 20-30 | 5 | 25 | 125 | 3125 |

VII| 30-40 | 7 | 35 | 245 | 8575 |

VIII| 40-50 | 3 | 45 | 135 | 6075 |

9|----------------|----|----------------|-------|--------|

10| | N=20 | | Σfi xi = 560 | Σfi xi² = 18500 |

11Step 2: Calculate the arithmetic mean (x̄).

12x̄ = Σfi xi / N = 560 / 20 = 28

13Step 3: Calculate Variance (σ²) using the formula σ² = (1/N) Σfi xi² - (x̄)².

14σ² = (1/20) × 18500 - (28)²

15= 925 - 784 = 141

16Step 4: Calculate Standard Deviation (σ).

17σ = √141 ≈ 11.87

Answer

Variance (σ²) = 141, Standard Deviation (σ) ≈ 11.87

For continuous frequency distributions, mid-points of the class intervals are used as xi.

Common mistakes

✗Forgetting to take the square root at the end to find the standard deviation from the variance.
✗Using median or mode instead of the arithmetic mean for calculating deviations.
✗Making calculation errors, especially with squaring numbers or summing large values.
✗Incorrectly applying the formulas for grouped versus ungrouped data.
✗Not using the mid-points of class intervals for continuous frequency distributions.
✗Confusing the formulas for variance and standard deviation, or using the wrong formula for the given data type.

Exam tips

★Memorise all the formulas for variance and standard deviation for both ungrouped and grouped data.
★Show all steps clearly in your calculations, as partial marks are often awarded for correct intermediate steps.
★Use a calculator carefully and double-check your entries to avoid numerical errors.
★Practice a variety of problems, including those with decimals and large numbers, to build confidence and speed.
★Understand the difference between variance and standard deviation and when to use each measure.

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