
A student noted the number of cars passing through a spot on a road for 100 periods each of 3 minutes and summarized it in the table given below. Find the mode of the data.
Number of cars 0 – 10 10 – 20 20 – 30 30 – 40 40 – 50 50 – 60 60 – 70 70 – 80 Frequency 7 14 13 12 20 11 15 8
Number of cars | 0 – 10 | 10 – 20 | 20 – 30 | 30 – 40 | 40 – 50 | 50 – 60 | 60 – 70 | 70 – 80 |
Frequency | 7 | 14 | 13 | 12 | 20 | 11 | 15 | 8 |
Answer
486.3k+ views
Hint: We use the fact that if the data is in groups or class intervals, the modal interval corresponds to the highest frequency. We then use the formula for the mode $l + \dfrac{{{f_1} - {f_0}}}{{2{f_1} - {f_0} - {f_2}}}$. Substitute the values in this formula and do simplification to find the mode.
Complete step-by-step answer:
Let us write the formula of mode for grouped data.
Mode for grouped data is given as,
Mode $ = l + \left( {\dfrac{{{f_1} - {f_0}}}{{2{f_1} - {f_0} - {f_2}}}} \right) \times h$
Where $l$ is the lower modal class,
$h$ is the size of the class interval,
${f_1}$ is the frequency of modal class,
${f_0}$ is the frequency of the class preceding the modal class,
${f_2}$ is the frequency of the class succeeding the modal class,
The modal class is the interval with the highest frequency.
$ \Rightarrow $Modal class $ = 40 - 50$
The lower limit of the modal class is,
$ \Rightarrow l = 40$
The class-interval is,
$ \Rightarrow h = 50 - 40 = 10$
The frequency of the modal class is,
$ \Rightarrow {f_1} = 20$
The frequency of the class preceding the modal class is,
$ \Rightarrow {f_0} = 12$
The frequency of the class succeeding modal class is,
$ \Rightarrow {f_2} = 11$
Substitute these values in the mode formula,
$ \Rightarrow $ Mode $ = 40 + \dfrac{{20 - 12}}{{2\left( {20} \right) - 12 - 11}} \times 10$
Simplify the terms,
$ \Rightarrow $ Mode $ = 40 + \dfrac{8}{{40 - 23}} \times 10$
Subtract the values in the denominator and multiply the terms in the numerator,
$ \Rightarrow $ Mode $ = 40 + \dfrac{{80}}{{17}}$
Divide the numerator by denominator,
$ \Rightarrow $ Mode $ = 40 + 4.71$
Add the terms,
$\therefore $ Mode $ = 44.71$
Hence, the mode is 44.71.
Note: You may mistake the mode formula with that of the median. The Median formula is given as $l + \left( {\dfrac{{\dfrac{n}{2} - cf}}{f}} \right) \times h$. In this question, we wrote mode for grouped data. To find the mode for ungrouped data, we will find the observation which occurs the maximum number of times.
Complete step-by-step answer:
Let us write the formula of mode for grouped data.
Mode for grouped data is given as,
Mode $ = l + \left( {\dfrac{{{f_1} - {f_0}}}{{2{f_1} - {f_0} - {f_2}}}} \right) \times h$
Where $l$ is the lower modal class,
$h$ is the size of the class interval,
${f_1}$ is the frequency of modal class,
${f_0}$ is the frequency of the class preceding the modal class,
${f_2}$ is the frequency of the class succeeding the modal class,
The modal class is the interval with the highest frequency.
$ \Rightarrow $Modal class $ = 40 - 50$
The lower limit of the modal class is,
$ \Rightarrow l = 40$
The class-interval is,
$ \Rightarrow h = 50 - 40 = 10$
The frequency of the modal class is,
$ \Rightarrow {f_1} = 20$
The frequency of the class preceding the modal class is,
$ \Rightarrow {f_0} = 12$
The frequency of the class succeeding modal class is,
$ \Rightarrow {f_2} = 11$
Substitute these values in the mode formula,
$ \Rightarrow $ Mode $ = 40 + \dfrac{{20 - 12}}{{2\left( {20} \right) - 12 - 11}} \times 10$
Simplify the terms,
$ \Rightarrow $ Mode $ = 40 + \dfrac{8}{{40 - 23}} \times 10$
Subtract the values in the denominator and multiply the terms in the numerator,
$ \Rightarrow $ Mode $ = 40 + \dfrac{{80}}{{17}}$
Divide the numerator by denominator,
$ \Rightarrow $ Mode $ = 40 + 4.71$
Add the terms,
$\therefore $ Mode $ = 44.71$
Hence, the mode is 44.71.
Note: You may mistake the mode formula with that of the median. The Median formula is given as $l + \left( {\dfrac{{\dfrac{n}{2} - cf}}{f}} \right) \times h$. In this question, we wrote mode for grouped data. To find the mode for ungrouped data, we will find the observation which occurs the maximum number of times.
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