Question

5. Find the mode of the following frequency distribution:
Class
0-100
100 - 200 200 - 300 300-400 400-500 500 - 600
Frequency
7
21
37
13
12
10
98 A mair of dice is tossed simultaneously
Find the probability that the​

Answer 1

Question:-

Find the mode of the following frequency distribution:

$\boxed{\begin{array}{c|c} \bf{Class} & \bf{Frequency} \\ \cline{1-2} \sf{\:\:0 - 100} & \sf{7} \\ \sf{100-200} & \sf{21} \\ \sf{200-300} & \sf{37} \\ \sf{300-400} & \sf{13} \\ \sf{400-500} & \sf{12} \\ \sf{500-600} & \sf{10}\end{array}}$

Solution:-

We know,

• $\dag{\underline{\boxed{\pink{\red{\bf{Mode = l + \bigg(\dfrac{f_1 - f_0}{2f_1 - f_0 - f_2}\bigg) \times h}}}}}}$

Where:-

• l = lower limit of the modal class
• h = height of the class
• f₀ = Frequency of the class preceding the modal class
• f₁ = Frequency of the modal class
• f₂ = Frequency of the class succeeding the modal class

Modal class:- The class in a given distribution table which has the greatest frequency is known as modal class.

Here,

The class 200-300 has the greatest frequency in the table. Hence it is the modal class.

From here we get:-

• l = 200
• h = 300 - 200 = 100
• f₀ = 21
• f₁ = 37
• f₂ = 13

Putting all the values in the formula:-

$= \sf{Mode = 200 + \dfrac{37-21}{2\times 37 - 21 - 13} \times 100}$

$= \sf{Mode = 200 + \dfrac{16}{74 - 34} \times 100}$

$= \sf{Mode = 200 + \dfrac{16}{40} \times 100}$

$= \sf{Mode = 200 + \dfrac{16}{4} \times 10}$

$= \sf{Mode = 200 + 4 \times 10}$

$= \sf{Mode = 200 + 40}$

$= \sf{Mode = 240}$

∴ Mode of the following data is 240

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Answer 2

Answer:

Given :-

$\begin{gathered}\begin{tabular}{|c|c|c|c|c|c|c|}\cline{1 - 7} \tt Class & \tt 0-100 & \tt 100-200 & \tt 200-300 & \tt 300-400 & \tt 400-500 & \tt 500-600\\ \cline{1-7}\tt Frequency & \tt 7 & \tt 21 & \tt 37 & \tt 13 & \tt 12 & \tt 10 \\\cline{1-7}\end{tabular}\end{gathered}$

To Find :-

• What is the mode.

Formula Used :-

$\clubsuit$ Mode Formula :

$\longmapsto \sf\boxed{\bold{\pink{Mode =\: l + \dfrac{f_1 - f_0}{2f_1 - f_0 - f_2} \times h}}}$

where,

• l = Lower limit of modal class
• f₁ = Frequency of the model class
• f₀ = Frequency of the class preceding the model class
• f₂ = Frequency of the class succeeding the model class
• h = Size of the class interval

Solution :-

$\boxed{\begin{array}{cccc}\sf Class\: Interval&\sf Frequency\\\frac{\qquad \qquad \qquad \qquad}{} &\frac{\qquad \qquad \qquad \qquad \qquad}{}\\\sf 0-100&\sf 7\\\\\sf 100-200&\sf 21\\\\\sf 200-300&\sf 37\\\\\sf 300-400&\sf 13\\\\\sf 400-500&\sf 12\\\\\sf 500-600&\sf 10\end{array}}$

Given :

• Lower limit of modal class (l) = 200
• Frequency of the modal class (f₁) = 37
• Frequency of the class preceding the model class (f₀) = 21
• Frequency of the class succeeding the model class (f₂) = 13
• Size of the class interval (h) = 300 - 200 = 100

According to the question by using the formula we get,

$\dashrightarrow \sf Mode =\: 200 + \dfrac{37 - 21}{2(37) - 21 - 13} \times 100$

$\dashrightarrow \sf Mode =\: 200 + \dfrac{16}{2 \times 37 - 34} \times 100$

$\dashrightarrow \sf Mode =\: 200 + \dfrac{16}{74 - 34} \times 100$

$\dashrightarrow \sf Mode =\: 200 + \dfrac{16}{40} \times 100$

$\dashrightarrow \sf Mode =\: 200 + \dfrac{160\cancel{0}}{4\cancel{0}}$

$\dashrightarrow \sf Mode =\: 200 + \dfrac{\cancel{160}}{\cancel{4}}$

$\dashrightarrow \sf Mode =\: 200 + 40$

$\dashrightarrow \sf\bold{\red{Mode =\: 240}}$

$\therefore$ The mode is 240.

$\rule{150}{2}$

IMPORTANT FORMULA :

$\clubsuit$ Mean Formula :

$\longmapsto \sf\boxed{\bold{\pink{Mean =\: \dfrac{\Sigma f_ix_i}{\Sigma f_i}}}}$

where,

• $\sf \Sigma f_ix_i$ = Sum of all the observations
• $\sf \Sigma f_i$ = Sum of frequencies or observations

$\clubsuit$ Median Formula :

$\longmapsto \sf\boxed{\bold{\pink{Median =\: l + \bigg\lgroup \dfrac{\frac{n}{2} - cf}{f}\bigg \rgroup \times h}}}$

where,

• l = Lower limit of median class
• n = Number of Observations
• cf = Cumulative frequency of the class preceding the median class
• f = Frequency of median class
• h = Size of the class interval (assuming class are of equal size)