# Class 9 – Statistics

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Topic Sub Topic Online Practice Test
Statistics
• Mean of grouped data
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Statistics
• Median of grouped data
• Mode of grouped data
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## Study Material

Introduction: We already learn how to find mean, median and mode of ungrouped data. In this synopsis we are going to learn about mean of grouped data. In this synopsis we are going to learn about mean of grouped data. Before going to discuss about mean of ungrouped data, we just recall mean of a regrouped data.

Mean of grouped data: We find the mean of grouped data in three ways.

1. Direct method

2. Short-cut (or) assumed mean method

3. Step deviation method

Direct method: As we know the mean (or average) of observations is the sum of the values of all the observations divided by the total number of observations. Let

be observations with respective frequencies  This means that observation

Occurs  times,  occurs  times, and so on.

Now, the sum of the values of all the observations  and the number of observations .

So, the mean  of the data is given by

Recall that we can write this in short, using the Greek letter Σ which means summation i.e.,

Example 1. The following table shows the weights of 15 members of an athletic team in a school.

 Weight (in kg) Number of athletes 42 4 45 3 46 5 48 2 49 1

Find the mean weight.

Solution: Form the above data, we may prepare the table given below:

 Weight (in kg) Number of athletes (frequency ) 42 45 46 48 49 4 3 5 2 1 168 135 230 96 49 = 15

Mean weight =  =  = 45.2 kg.

Example 2. The marks obtained by a set of students in an examination are given below:

 Marks 5 10 15 20 25 30 Number of students 6 4 6 12 x 4

If the mean of the above data is 18, calculate the numerical value of x.

Solution: From the above data, we may prepare the table given below:

 Marks Number of athletes (frequency) 5 10 15 20 25 30 6 4 6 12 X 4 30 40 90 240 25x 120 = (32 + x)

Mean =  =

But Mean = 18.

= 18 Þ 520 + 25x = 18 (32 + x) Þ 7x = 56 Þ x = 8

Hence, x = 8.

Short-cut Method: If the values of x or f are large, then direct method to calculate mean is not used because it will be time-consuming and difficult. In such a case, we take deviations, from an arbitrary point a called assumed mean, then

, where di = xi  a.

Example: Find the mean of given data in assumed mean method.

 Class 10-25 25-40 40-55 55-70 70-85 85-100 F 2 3 7 6 6 6

Solution:

 Class interval Number student (fi) Class Marks (xi) di = xi – 47.5 xi = a fidi 10-25 25-40 40-55 55-70 70-85 85-100 2 3 7 6 6 6 17.5 32.5 47.5(a) 62.5 77.5 92.5 -30 -15 0 15 30 45 -60 -45 0 90 180 270 Total

So, form the above table, the mean of the deviations,  =

Now, let us find the relation between d and

Since, in obtaining di we subtracted a from each xi so, in order to get the mean  We need to add a to . This can be explained mathematically as:

Mean of deviations, =

So,

-

d =   - a

Therefore

Substituting the values of a,  and  from the table, we get

x = 47.5 +

Therefore, the mean of the marks obtained by the students is 62.

Any number can be taken as assumed mean a but for the sale of fi.

Any number can be taken as assumed mean a but fo the sake of simplicity, 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.

Step deviation Method: When we use short-cut method to find mean, then the deviations, (di) are divisible by a common number (h). In such a case, the mean is reduced by taking

Where

Example: Using step deviation method, calculate the mean for the following data:

 Height (in cm) 135-140 140-145 145-150 150-155 155-160 160-165 165-170 170-175 Number of boys 4 9 18 28 24 10 5 2

Solution: Here, class size (c) = 5. Take assumed mean (A) = 152.5 Thus, from the given data, we may prepare the table given below:

 Class-Interval Class-mark xi Frequency fi fiui 135-140 140-145 145-150 150-155 155-165 165-170 170-175 137.5 142.5 147.5 152.7 = A 157.5 1652.5 167.5 172.5 4 9 18 28 24 10 5 2 -3 -2 -1 0 1 2 3 4 -12 -18 -18 0 24 20 15 8

Mean =  = (152.5 + 0.95) = 153.45 cm.

Example: The following table gives the frequency distribution of the percentage of marks obtained by 2300 students in a competitive examination.

 Marks obtained (in per cent) 11-20 21-30 31-40 41-50 51-60 61-70 71-80 Number of students 141 221 439 529 495 322 153

Find the mean percentage of marks of the group.

Solution: First we convert the given frequency distribution in continuous form (Exclusive class interval).

So, we may prepare the table given below:

 Class-Interval Class-Mark xi Frequency fi fiui 10.5-20.5 20.5-30.5 30.5-40.5 40.5-50.5 50.5-60.5 60.5-70.5 70.5-80.5 15.5 25.5 35.5 45.5 = A 55.5 65.5 75.5 141 221 439 529 495 322 153 -3 -2 -1 0 1 2 3 -423 -442 -439 0 495 644 459

Mean =  =  = 45.5 + 1.278 =46.778

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

The mean of 10 observations is 16.3. By an error, one observation is registered as 32 instead of 23. Find the correct mean.

Solution: Here n = 10,

But this value of  is incorrect as 32 is registered instead of 23.