CBSE Class 10 Mathematics Statistics Notes Set 05

Introduction and Arithmetic Mean

Statistics is the study of the collection, organisation, analysis, interpretation and presentation of data.

(a) Measures of central tendency

A certain value representative of the whole data and signifying its characteristics is called a measure of central tendency or average. Mean, median and mode are the measures of central tendency.

(b) Arithmetic mean

The arithmetic mean (or simple mean) of a set of observations is obtained by dividing the sum of the values of observations by the number of observations.

(i) Mean of raw data. The mean of n observations x₁, x₂, x₃, …, xₙ is \[ \bar{x} = \dfrac{x_1 + x_2 + \dots + x_n}{n} = \dfrac{\sum_{i=1}^{n} x_i}{n}, \qquad n\bar{x} = \sum_{i=1}^{n} x_i . \] That is, the product of the mean and the number of items gives the sum of the observations.

(ii) Mean for an ungrouped frequency distribution. Let x₁, x₂, …, xₙ be n observations with respective frequencies f₁, f₂, …, fₙ. This means observation x₁ occurs f₁ times, x₂ occurs f₂ times and so on. Then the sum of the values of all the observations = f₁x₁ + f₂x₂ + … + fₙxₙ = \( \sum f_i x_i \) and the number of observations = f₁ + f₂ + … + fₙ = \( \sum f_i \). Therefore \[ \text{Mean } (\bar{x}) = \dfrac{\sum f_i x_i}{\sum f_i}. \] It is understood that i varies from 1 to n.

(iii) Mean of grouped data. To calculate the mean of grouped data, we have three methods: (I) Direct method, (II) Assumed mean method, (III) Step deviation method.

(I) Direct method. First, find the class marks of each class interval. These class marks are represented by xᵢ. For the iᵗʰ class interval, fᵢ is the frequency corresponding to the class mark xᵢ. The steps are:
1. Find the class mark (mid value) xᵢ of each class: \( x_i = \dfrac{\text{Lower limit} + \text{Upper limit}}{2} \).
2. Calculate fᵢxᵢ for each xᵢ.
3. Find Σfᵢxᵢ and Σfᵢ.
4. By using the formula \( \bar{x} = \dfrac{\sum f_i x_i}{\sum f_i} \), calculate the mean.

(II) Assumed mean method (shortcut method). When the numerical values of xᵢ and fᵢ are large, finding the product of xᵢ and fᵢ becomes tedious and time consuming. In such cases the assumed mean method is used. First, one among the xᵢ’s is chosen as the assumed mean and denoted by a (we may take a to be that xᵢ which lies in the centre, to reduce the calculation work). We subtract the assumed mean a from each value of xᵢ. The reduced value is dᵢ = xᵢ − a, called the deviation of xᵢ from the assumed mean a. Hence xᵢ = a + dᵢ and fᵢxᵢ = a fᵢ + dᵢ fᵢ, so \[ \sum f_i x_i = a \sum f_i + \sum d_i f_i = aN + \sum d_i f_i \quad [\because \textstyle\sum f_i = N]. \] Dividing both sides by N, \[ \bar{x} = a + \dfrac{\sum f_i d_i}{\sum f_i} \] where a = assumed mean and dᵢ = xᵢ − a. The steps are:
1. For each class interval, calculate the class mark xᵢ by xᵢ = ½ (lower class limit + upper class limit).
2. Choose a suitable value of xᵢ in the middle as the assumed mean a.
3. Calculate the deviation dᵢ = xᵢ − a for each i.
4. Calculate the product fᵢdᵢ for each i.
5. Find N = Σfᵢ.
6. Calculate the mean by the formula \( \bar{x} = a + \dfrac{\sum f_i d_i}{\sum f_i} \).

(III) Step-deviation method. The main objective of subtracting an arbitrary number from each observation xᵢ in the shortcut method is to reduce the numbers so that the calculation becomes simpler. We can make these differences still smaller if they are divisible by a common non-zero number, say h. Let \( u_i = \dfrac{x_i - a}{h} \), so xᵢ = a + h uᵢ and fᵢxᵢ = a fᵢ + h uᵢ fᵢ. Then Σfᵢxᵢ = aN + h Σuᵢfᵢ, and dividing by N, \[ \bar{x} = a + \left( \dfrac{\sum f_i u_i}{\sum f_i} \right) h \] where a = assumed mean, uᵢ = (xᵢ − a)/h and h = width of the class interval. The steps are:
1. For each class interval, calculate the class mark xᵢ = ½ (lower class limit + upper class limit).
2. Choose a suitable value of xᵢ in the middle as the assumed mean a.
3. Calculate the deviation dᵢ = xᵢ − a for each i.
4. Calculate uᵢ = (xᵢ − a)/h, where h is the class size or common factor of all dᵢ.
5. Find fᵢuᵢ for each uᵢ.
6. Find Σfᵢuᵢ and Σfᵢ.
7. Calculate the mean by the formula \( \bar{x} = a + \left( \dfrac{\sum f_i u_i}{\sum f_i} \right) h \).

(c) Properties of mean

1. The sum of deviations from the mean is zero, i.e. \( \sum_{i=1}^{n} (x_i - \bar{x}) = 0 \).
2. If a constant real number ‘a’ is added to each of the observations, then the new mean will be \( \bar{x} + a \).
3. If a constant real number ‘a’ is subtracted from each of the observations, then the new mean will be \( \bar{x} - a \).
4. If each observation is multiplied by a constant real number ‘a’, then the new mean will be \( a\bar{x} \).
5. If each observation is divided by a non-zero constant ‘a’, then the new mean will be \( \dfrac{\bar{x}}{a} \).

(d) Merits of arithmetic mean

1. It is rigidly defined, simple, easy to understand and easy to calculate.
2. It is based upon all the observations.
3. Its value being unique, we can use it to compare different sets of data.
4. It is least affected by sampling fluctuations.
5. Mathematical analysis of mean is possible. So, it is relatively reliable.

(e) Demerits of arithmetic mean

1. It cannot be determined by inspection nor can it be located graphically.
2. It cannot be used for qualitative characteristics such as intelligence, honesty, beauty etc.
3. It cannot be obtained if a single observation is missing.
4. It is affected very much by extreme values. In case of extreme items, the arithmetic mean gives a distorted picture of the distribution and no longer remains representative of the distribution.
5. It may lead to wrong conclusions if the details of the data from which it is computed are not given.
6. It cannot be calculated if the extreme class is open, e.g. below 10 or above 90.
7. It cannot be used in the study of ratios, rates etc.

(f) Uses of arithmetic mean

1. It is used for calculating average marks obtained by a student.
2. It is extensively used in practical statistics and to obtain estimates.
3. It is used by businessmen to find out profit per unit article, output per machine, average monthly income and expenditure etc.

Solved Examples

Example 1. Find the mean of the factors of 10.
Answer:
Factors of 10 are 1, 2, 5 and 10.
\( \bar{x} = \dfrac{1 + 2 + 5 + 10}{4} = \dfrac{18}{4} = \) 4.5

Example 2. The mean of marks scored by 100 students was found to be 40. Later on it was discovered that a score of 53 was misread as 83. Find the correct mean.
Answer:
n = 100, \( \bar{x} \) = 40, so \( 40 = \dfrac{1}{100} \sum x_i \) and the incorrect value of Σxᵢ = 4000.
Correct value of Σxᵢ = 4000 − 83 + 53 = 3970.
Correct mean = \( \dfrac{3970}{100} \) = 39.7.

Check Your Progress

1. Find the mean of the data 16, 20, 26, 40, 50, 60, 70, 30.
2. The weight in kg of 9 members in a school boxing team are 54, 59, x, 53, 73, 49, 50, 58, 45. If the average is 56 then x is
3. The mean of x₁, x₂, …, x₁₀ is M. When xᵢ, i = 1, 2, …, 10 is replaced by xᵢ + 10, find the new mean.
4. Find the arithmetic mean age of the data below.

5. The following set of raw data shows the lengths, in millimetres, measured to the nearest mm, of 40 leaves taken from plants of a certain species. This is the table of frequency distribution. Choosing a suitable assumed mean, compute the mean by the step-deviation method.

Answers
1. 39
2. 63
3. M + 10
4. 17.28
5. 43