# Class 10 – Statistics

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## Online Tests

Topic Sub Topic Online Practice Test
Statistics
• Mean of grouped data
• Median and mode of grouped data
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Statistics
• G.C.F and L.C.F curves
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## Study Material

 Introduction:Statistics is the study of collection, analysis, interpretation, and organization of data. In applying statistics to, e.g., a scientific, industrial, or societal problem, it is conventional to begin with a statistical population or a statistical model process to be studied. Populations can be diverse topics such as “all persons living in a country” or “every atom composing a crystal”. Statistics deals with all aspects of data including the planning of data collection in terms of the design of surveys and experiments.    Mean of ungrouped data: The arithmetic mean of a statistical data is defined as the quotient of the sum of all the terms or entries divided by the number of items. If  are the items given, then This is usually denoted by .  i) As some values of x will be less and some more than , in some sense lies at the centre of all the values justifying that it is the measure of central tendency. ii) If to each item x is added a number k, then the new mean = . iii) If each item is multiplied by a number k, then the new mean = k x . Example:Find the arithmetic mean of numbers 7, 6, 10, 2, 5, 8, 9. Solution: . Example:The sum of 15 observations of a data is 420. Find the mean. Solution:Here n = 15, Properties of arithmetic mean: 1.              Algebraic sum of deviations of a set of values from their arithmetic mean is zero. 2.              The sum of the squares of the deviations of a set of values is minimum when taken about the mean.  Example: Using the formula, prove that . Solution:

i.e., the algebraic sum of deviations of a set of values from their arithmetic mean is zero.

Example: The mean of a data is 9. If each observation is multiplied by 3 and then 1 is added to each result, find the mean of the new observations.

Solution:Let the observations be .

Given

Each observation is multiplied by 3 and then 1 is added i.e., . Then the mean is given by

Now

Mean of new observations = 28.

Mean of an ungrouped frequency distribution:

i)               Direct method: If the entries  occurs  times respectively then the arithmetic mean is

The above formula can be stated in simpler form as

Where
is the total frequency.

Example:The frequency distribution of the number of heads obtained in tossing five coins 100 times is given below. Find the mean of the data.

 Number of heads 0 1 2 3 4 5 Frequency 1 7 20 64 5 3

Solution:To find the mean from the given frequency distribution we prepare the table as below.

 Number of heads Frequency 0 1 0 1 7 7 2 20 40 3 64 192 4 5 20 5 3 15 Total N = 100

ii)             Assumed mean method: The frequencies and the values of the variable are quite large numbers. The product will also be large. We can’t do anything with the frequencies but we can change each xito a smaller number, so that our calculations become easy.
In this method assume one of the observations which is convenient as assumed mean (a). Then find the deviation (d) of other observations from the assumed mean.

Any number can be taken as assumed mean (a) but for accurate values, it is generally taken as that value of the variable which has the greatest frequency in the frequency distribution or which is near about the middle of the frequency distribution.

Consider the following data to calculate the mean for the frequency distribution by assumed mean method.

 xi 73 72 71 70 69 68 67 66 65 fi 2 4 6 10 11 7 5 4 1

From the above table. It is clear that, the variable which has greatest frequency is 69 and also it is in the middle of the frequency distribution.

The assumed mean (a) = 69.

Solution:Let the assumed mean a = 70.

 xi fi di = xi – a fidi 73 2 73-69 = 4 8 72 4 72-69 = 3 12 71 6 71-69 = 2 12 70 10 70-69 = 1 10 69 11 69-69 = 0 0 68 7 68-69 = -1 -7 67 5 67-69 = -2 -10 66 4 66-69 = -3 -12 65 1 65-69 = -4 -4

From above table, the mean of the deviations,

Now let us find the relation between .

Since, in obtaining di we subtract ‘a’ from each xi

So, in order to get the mean  we need to add ‘a’ to d.

This can be explained mathematically as:

Mean of deviations,

So,

Therefore

Now substituting the value of a,  and from the table, we get

Mean of the grouped data:

i)               Direct method: Sometimes, data is so large that it is difficult to study the data and also difficult to find measures of central tendency (Mean, Mode, Median). In this situation we make group of the data with suitable class intervals called as grouped data.

Example:

 Marks 10 20 26 30 34 40 48 50 53 60 70 75 80 Number of students 2 4 3 2 5 6 7 2 5 1 6 7 3

Now, convert the data of above table into grouped data by forming class intervals of width say 10.

While allocating frequencies to each class interval, students whose score is equal to any upper class boundary would be considered in the next class. e.g., 4 students who have obtained 20 marks would be considered in the next class. i.e., 20-30 and not in 10-20.

 Class interval 10-20 20-30 30-40 40-50 50-60 60-70 70-80 80-90 Number of students 2 7 7 13 7 1 13 3

In a grouped data, it is assumed that the frequency of each class is concentrated at its mid-value.

Mid-value (or) class mark

 Class interval Number of students (fi) Class marks (xi) fixi 10-20 2 15 30 20-30 7 25 175 30-40 7 35 245 40-50 13 45 585 50-60 7 55 385 60-70 1 65 65 70-80 13 75 975 80-90 3 85 255

This new method of finding mean is known as direct method.

ii)             Assumed mean method:

Example:Calculate the mean daily wage of the workers given below:

 Daily wages (in Rs) 80-100 100-120 120-140 140-160 160-180 Number of workers 20 30 20 40 90

Solution:

 Daily wages Number of workers Class mark xi fixi di =xi – a fidi 80-100 20 90 1800 - 40 - 800 100-120 30 110 3300 - 20 - 600 120-140 20 130 (a) 2600 0 0 140-160 40 150 6000 20 800 160-180 90 170 15300 40 3600

Let the assumed mean a = 130.

Direct method:

Assumed mean method: We use the formula derived for finding the mean of ungrouped data using assumed mean here.

iii)            Step-deviation method:According to the formula , we have to find  to calculate the mean. Some times when the frequencies are large in number, this becomes cumbersome. This can be simplified if the class interval of each class of grouped data is same. As in above example, if we divide all the values of deviations by 20, we would get smaller numbers which we then multiply with fi. Here 20 is the class size of each class integral.

So, let  where a is the assumed mean and h is the class size.

 Daily wages Number of workers Class mark xi fixi di = xi – a fidi fiui 80-100 20 90 1800 - 40 - 800 -2 -40 100-120 30 110 3300 - 20 - 600 -1 -30 120-140 20 130 (a) 2600 0 0 0 0 140-160 40 150 6000 20 800 1 40 160-180 90 170 15300 40 3600 2 180

Now, we calculate ui

Let

Here, again the let us find the relation between

We have

So,

Therefore,

Step-deviation Method is a very short method and should always be used for grouped data where class interval sizes are equal.

Example:Find the mean of the following frequency distribution by step-deviation method.

 Class interval 0-19 20-39 40-59 60-79 80-99 100-119 Frequency 9 16 24 15 4 2

Solution:Here class length are equal.

Class length = h = 20. a = 69.5, h = 20.

 Class interval Frequency fi Class mark xi fiui 0-19 9 9.5 - 3 - 27 20-39 16 29.5 - 2 - 32 40-59 24 49.5 - 1 - 24 60-79 15 69.5 0 0 80-99 4 89.5 1 4 100-119 2 109.5 2 4

Mean of the combined distributions: When two sets of scores have been combined into a single distribution, then the mean of the combined distribution is the weighted mean of the means of the components, the weight being the total frequencies in those components.

In other words,
Where
is the mean of the combined distribution.

are the means of the component distributions.

are the total frequencies of the component distributions.