Consider the following data about the heights of plants in Jonathan's garden:

3cm, 4cm, 5cm, 7cm, 11cm

Now, let's calculate the **mean - μ -** of these values.

μ = (3 + 4 + 5 + 7 + 11)/5 = 6cm

If we use this value to describe the mean height of plants, we immediately run into difficulties; because, it does not represent the true nature of heights of these plants - some are as short asÂ 3 cm and some are as tall as 11 cm.

Therefore, the mean in this case, to say the least, is a bit misleading. This leads to a need ofÂ another value that helps us to understand the distribution of data in a given situation.

Now let's see how much each value of data has deviated ( going away ) from the mean:

x | 3 | 4 | 5 | 7 | 11 |
---|---|---|---|---|---|

μ | 6 | 6 | 6 | 6 | 6 |

(x - μ) | -3 | -2 | -1 | 1 | 5 |

Let's find the average of these deviations from the mean value:

Σ(x - μ) / 5 = (-3 + -2 + -1 + 1 + 5 )/5 = 0

The *deviations* turned out to be zero, not because of lack of deviations; it is because, the deviations turned out to be
*negative* and *positive* which in the end led to be cancelled out.

Now, in order to deal with issue, let's square the deviations to remove the *negative* signs, which is as follows:

x | 3 | 4 | 5 | 7 | 11 |
---|---|---|---|---|---|

μ | 6 | 6 | 6 | 6 | 6 |

(x - μ) | -3 | -2 | -1 | 1 | 5 |

(x - μ)^{2} | 9 | 4 | 1 | 1 | 25 |

Since we squared the deviations, just to deal with negative values, it's time we
reversed the process: let's find the square root of the following result:

√(Σ(x - μ)^{2})/5 = √(40/5) = 2.8

This is called the *standard deviation* of the above set of
data representing the heights of plants in Jonathan's garden. It gives us a clearer picture of data
distribution along with the mean. With the value of the standard deviation, the data can be described
in the following way:

The mean height of the plants in Jonathan's garden is 6cm and the standard deviation is 2. 8. That means the heights of most plants falls into the range from (6-2.8) = 3.2cm to (6+2.8)=8.8cm.

The example shows how important the Standard deviation is to get a clear picture about a set of data. Without it, talking about data is like, recalling the fate of Titanic without the iceberg!!

So, the formula for standard deviation is as follows:

σ = √Σ(x - μ)^{2}/n

where n is the total frequency.

Calculator-friendly formula for Standard Deviation

σ = √(Σ(x - μ)^{2})/n

σ = √(Σ(x^{2} - 2xμ + μ^{2})/n

σ = √(Σ(x^{2} - Σ2xμ + Σμ^{2})/n

σ = √(Σ(x^{2} - 2μΣx + Σμ^{2})/n

σ = √(Σ(x^{2} - 2μnμ + nμ^{2})/n

σ = √(Σ(x^{2} - 2nμ^{2} + nμ^{2})/n

σ = √(Σ(x^{2} - nμ^{2})/n

σ = √(Σx^{2}/n) - μ^{2}

σ = √(Σx^{2}/n) - μ^{2}

To find the standard deviation in grouped data, we change the method

slightly - **σ = √(Σf(x - μ) ^{2})/n**, where f is the frequency of each
class and n is the total frequency.

**E.g.**

The frequency of shoe sizes of students in a certain class is as follows:

shoe-size(x) | frequency(f) |
---|---|

3 | 3 |

4 | 5 |

5 | 10 |

6 | 8 |

7 | 4 |

μ = Σfx/n = 5.2

σ = √(Σf(x - μ)^{2})/n = √(Σf(x - 5.2;)^{2})/30 = 2.3

**E.g.**

Marks(x) | frequency(f) |
---|---|

0 - 20 | 3 |

21 - 40 | 6 |

41 - 60 | 9 |

61 - 80 | 8 |

81 - 100 | 4 |

μ = Σfx/n = 52.7

σ = √(Σf(x - μ)^{2})/n = √(Σf(x - 52.7)^{2})/30 = 2.55

I am sure, you have got a good understanding of the concept of **standard deviation** by now.

**Now, in order to complement what you have just learnt, work out the following questions:**

- The time taken by 10 engineers to install a satellite dish, in minutes, is as follows:

51, 49, 56, 60, 52, 58, 49, 56, 52, 57

Find the mean and the standard deviation. - When a die is thrown, the numbers turn out as follows:

Number frequency (f) 1 3 2 7 3 10 4 14 5 8 6 2 Find the mean and the standard deviation.

- The weights of some chicks obtained by a farmer are as follows:

weight (x) frequency (f) 0 - 20 7 21 - 40 11 41 - 60 3 61 - 80 7 81 - 100 2 Find the mean and the standard deviation.

- The standard deviation of a certain set of data is 4.2. What would be the next standard deviation, if each data was increased by 5 ?
- The standard deviation of a certain set of data is 4.2. What would be the next standard deviation, if each data was multiplied by 5 ?

Maths is challenging; so is finding the right book. K A Stroud, in this book, cleverly managed to make all the major topics crystal clear with plenty of examples; popularity of the book speak for itself - 7^{th} edition in print.