Grouping Data: Definition, Advantages and Sample Questions

Ans:

X is the midpoint of the class. It is adding the class limits and divide by 2.

\(\bar{x} = \frac{\sum f x}{\sum f}\)

= 832/50

=16.64

Ans: To find the mean of the marks obtained by the students in the Science paper, find the product of each xi and their corresponding frequency fi.

Thus, by using the formula, \(\bar{x} = \frac{\displaystyle\sum_{i=1}^{n} f_{i}x_{i}}{\displaystyle\sum_{i=1}^{n} f_{i}}\),

\(\bar{x}\) = 1779/30

\(\bar{x}\) = 59.3

Hence, the mean of the marks obtained is 59.3.

Ans. 1st Step: Construct the cumulative frequency distribution

\(\frac{n}{2} = \frac{50}{2} = 25\), class median is the 3rd class.

So, F = 22, f_m = 12, L_m = 20.5 and i = 10

Therefore,

Median = L_m + (n/2+F)/ f_m . i

= 21.5 +(25-22)/12. 10

= 24

Thus, 25 persons take less than 24 minutes to travel to work and another 25 persons take more than 24 minutes to travel to work.

Ans: To find the median height, firstly, find the class intervals and their corresponding frequencies.

The class is below 140, 140-145, 145-150, 150-155, 155-160 and 160-165.

From the given distribution, it is observed that:

4 boys are below 140. Therefore, the frequency of class intervals below 140 is 4. 

11 boys are there with heights less than 145, and 4 boys with height less than 140

Hence, the frequency distribution for the class interval 140-145=11-4=7

Similarly, the frequency of 145 -150= 29–11=18

Frequency of 150-155=40-29=11

Frequency of 155–160=46-40=6

Frequency of 160-165=51-46=5

Therefore, the frequency distribution table along with the cumulative frequencies are given below:

Here, n= 51.

Therefore, n/2=51/2=25.5

Thus, the observations lie between the class interval 145-150, i.e, the median class.

Therefore,

Lower class limit=145

Class size, h=5

Frequency of the median class, f=18

Cumulative frequency of the class preceding the median class, cf=11.

The formula to find the median of the grouped data is:

Median = \(l + (\frac{\frac{n}{2}-cf}{f}) \times h\) Now, substituting the values in the formula, 

Median = \(145+(\frac{25.5-11}{18}) \times5\)

Median = \(145 + (\frac{72.5}{18})\)

Median = 145 + 4.03

Median = 149.03.

Therefore, the median height for the given data is 149. 03 cm.

Ans: From the given table, it is observed that the maximum class frequency is 8, and the corresponding class interval is 3-5.

Therefore, the modal class for the given data is 3-5.

The lower limit of modal class, l=3

Class size, h=2

Frequency of modal class, f₁ = 8

Frequency of class proceeding to modal class, f₀ = 7

Frequency of class succeeding to modal class, f₂ = 2

The formula to find the mode of the grouped data is:

Mode = \(l + (\frac{f_{1}-f_{0}}{2f_{1}-f_{0}-f_{2}})×h\)

Now, substituting the values in the mode formula,

Mode = \(3 + (\frac{8-7}{2(8)-7-2}) \times 2\)

Mode = \(3+(\frac{2}{7})\)

Mode = \(\frac{(21+2)}{7}\)

Mode = \(\frac{23}{7}\)

Mode = 3.286.

Therefore, the mode of the given grouped data is 3.286.

Ans:

Add up all the (x_if_i) values to obtain ∑x_if_i. Add up all the fi values to get ∑f_i

Now, use the mean formula.

\(\bar{X}\) = ∑x_if_i / ∑f_i = \(\frac{2750}{50}\) = 55

Mean = 55.

Ans. 

Let the frequency of the class 30-40 be f₁ and that of the class 50-60 be f₂. The total frequency is 229
12 + 30 + f₁ + 65 + f₂ + 25 + 18 = 229 ⇒ f₁ + f₂ = 79
Given that the median is 46, clearly 46 lies in the class 40-50. So 40-50 is the median class
∴ l = 40,h = 10, f = 65 and C = 42 + f₁ , N = 229

Median = \(l+\frac{(\frac{N}{2}–Cf)}{f} \times h\) 

⇒ \(46 = 40+\frac{(\frac{229}{2}–(42+f1))}{65} \times 10\) 

⇒ \(46 = 40 + \frac{(145–2f1)}{13 }\)

⇒ \(6 = \frac{(145–2f1)}{13}\)

⇒ 2f₁ = 67

⇒ f₁ = 33.5 or 34 

Since f₁+f₂=79

f₂=45

Hence f₁=34 and f₂=45

Ans: When the data is presented in the form of class intervals, the mid-point of each class (also called class mark) is considered for calculating the mean.

The formula for mean remains the same as discussed above.

Note:

\(\text{Class Mark} = \frac{\text{Upper limit + Lower limit}}{2}\)

\(\bar{X} = \frac{\sum x_{i}f_{i}}{\sum f_{i}} = \frac{1390}{35} = 39.71\)

 ∑f_i = 35 and ∑x_if_i = 35

Mean = 39.71

Ans:

For a frequency distribution,

So, total number of observations = (1 + 4 + 7 + 5 + 9 + 8) = 34

It is an even number.

So, median is average of values at \((\frac{n}{2})^{th}\) and \((\frac{n}{2} + 1)^{th}\).

∴ Here, median is the average of the values at \((\frac{34}{2})^{th}\) or 17th position and \((\frac{34}{2}+1)^{th}\) or 18th position.

So, value of x at 17^th position is 13 and at 18^th position is 14.

∴Median = \(\frac{(13+14)}{2}\) = \(\frac{22}{7}\) = 13.5.

Mode for a frequency distribution is the value that is the most common frequent value.

Here 14 has frequency 9 which is highest. So, mode for the frequency distribution is 14.

Ans: Empirical relationship between mean, median, and mode is:

Mode = 3 Median – 2 Mean

⇒ Mode – Mean = 3 Median – 2 Mean – Mean

⇒ Mode – Mean = 3 Median – 3 Mean

⇒ Mode – Mean = 3 [Median – Mean]