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Mode of Grouped Data Questions in English
Class 10 Mathematics · Statistics · Mode of Grouped Data
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1
EasyMCQ
The wickets taken by a bowler in $10$ cricket matches are as follows:
$2, 6, 4, 5, 0, 2, 1, 3, 2, 3$
Find the mode of the data.
$2, 6, 4, 5, 0, 2, 1, 3, 2, 3$
Find the mode of the data.
A
$2$
B
$3$
C
$5$
D
$7$
Solution
(A) To find the mode,we first organize the data into a frequency distribution table:
The mode is the observation that occurs most frequently. From the table,the number of wickets $2$ appears $3$ times,which is the highest frequency.
Therefore,the mode of the given data is $2$.
| Number of wickets | $0$ | $1$ | $2$ | $3$ | $4$ | $5$ | $6$ |
| Number of matches | $1$ | $1$ | $3$ | $2$ | $1$ | $1$ | $1$ |
The mode is the observation that occurs most frequently. From the table,the number of wickets $2$ appears $3$ times,which is the highest frequency.
Therefore,the mode of the given data is $2$.
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View Solution2
EasyMCQ
$A$ survey conducted on $20$ households in a locality by a group of students resulted in the following frequency table for the number of family members in a household:
Find the mode of this data.
| Family size | $1-3$ | $3-5$ | $5-7$ | $7-9$ | $9-11$ |
| Number of families | $7$ | $8$ | $2$ | $2$ | $1$ |
Find the mode of this data.
A
$3.400$
B
$3.286$
C
$5.680$
D
$2.714$
Solution
(B) Here,the maximum class frequency is $8$,and the class corresponding to this frequency is $3-5$.
So,the modal class is $3-5$.
Now,
Modal class $= 3-5$,lower limit $(l)$ of modal class $= 3$,class size $(h) = 2$.
Frequency $(f_1)$ of the modal class $= 8$.
Frequency $(f_0)$ of the class preceding the modal class $= 7$.
Frequency $(f_2)$ of the class succeeding the modal class $= 2$.
Now,let us substitute these values in the formula:
Mode $= l + \left( \frac{f_1 - f_0}{2f_1 - f_0 - f_2} \right) \times h$
Mode $= 3 + \left( \frac{8 - 7}{2 \times 8 - 7 - 2} \right) \times 2 = 3 + \frac{2}{7} = 3 + 0.286 = 3.286$.
Therefore,the mode of the data above is $3.286$.
So,the modal class is $3-5$.
Now,
Modal class $= 3-5$,lower limit $(l)$ of modal class $= 3$,class size $(h) = 2$.
Frequency $(f_1)$ of the modal class $= 8$.
Frequency $(f_0)$ of the class preceding the modal class $= 7$.
Frequency $(f_2)$ of the class succeeding the modal class $= 2$.
Now,let us substitute these values in the formula:
Mode $= l + \left( \frac{f_1 - f_0}{2f_1 - f_0 - f_2} \right) \times h$
Mode $= 3 + \left( \frac{8 - 7}{2 \times 8 - 7 - 2} \right) \times 2 = 3 + \frac{2}{7} = 3 + 0.286 = 3.286$.
Therefore,the mode of the data above is $3.286$.
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View Solution3
DifficultMCQ
The following data provides information on the observed lifetimes (in $hours$) of $225$ electrical components:
Determine the modal lifetimes of the components.
| Lifetimes (in hours) | $0-20$ | $20-40$ | $40-60$ | $60-80$ | $80-100$ | $100-120$ |
| Frequency | $10$ | $35$ | $52$ | $61$ | $38$ | $29$ |
Determine the modal lifetimes of the components.
A
$65.625$
B
$60.100$
C
$64.475$
D
$70.500$
Solution
(A) From the data given above,it can be observed that the maximum class frequency is $61$,which belongs to the class interval $60-80$.
Therefore,the modal class is $60-80$.
Lower class limit $(l)$ of the modal class $= 60$.
Frequency $(f_1)$ of the modal class $= 61$.
Frequency $(f_0)$ of the class preceding the modal class $= 52$.
Frequency $(f_2)$ of the class succeeding the modal class $= 38$.
Class size $(h) = 20$.
Using the formula for mode:
$\text{Mode} = l + \left( \frac{f_1 - f_0}{2f_1 - f_0 - f_2} \right) \times h$
$= 60 + \left( \frac{61 - 52}{2(61) - 52 - 38} \right) \times 20$
$= 60 + \left( \frac{9}{122 - 90} \right) \times 20$
$= 60 + \left( \frac{9}{32} \right) \times 20$
$= 60 + \frac{180}{32} = 60 + 5.625 = 65.625$.
Therefore,the modal lifetime of the electrical components is $65.625 \text{ hours}$.
Therefore,the modal class is $60-80$.
Lower class limit $(l)$ of the modal class $= 60$.
Frequency $(f_1)$ of the modal class $= 61$.
Frequency $(f_0)$ of the class preceding the modal class $= 52$.
Frequency $(f_2)$ of the class succeeding the modal class $= 38$.
Class size $(h) = 20$.
Using the formula for mode:
$\text{Mode} = l + \left( \frac{f_1 - f_0}{2f_1 - f_0 - f_2} \right) \times h$
$= 60 + \left( \frac{61 - 52}{2(61) - 52 - 38} \right) \times 20$
$= 60 + \left( \frac{9}{122 - 90} \right) \times 20$
$= 60 + \left( \frac{9}{32} \right) \times 20$
$= 60 + \frac{180}{32} = 60 + 5.625 = 65.625$.
Therefore,the modal lifetime of the electrical components is $65.625 \text{ hours}$.
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View Solution4
MediumMCQ
The given distribution shows the number of runs scored by some top batsmen of the world in one-day international cricket matches.
Find the mode of the data.
| Runs scored | Number of batsmen |
| $3000-4000$ | $4$ |
| $4000-5000$ | $18$ |
| $5000-6000$ | $9$ |
| $6000-7000$ | $7$ |
| $7000-8000$ | $6$ |
| $8000-9000$ | $3$ |
| $9000-10000$ | $1$ |
| $10000-11000$ | $1$ |
Find the mode of the data.
A
$3608.7$
B
$4985.50$
C
$4500$
D
$4608.7$
Solution
(D) From the given data,the maximum class frequency is $18$,which corresponds to the class interval $4000-5000$.
Therefore,the modal class is $4000-5000$.
Lower limit $(l)$ of the modal class $= 4000$.
Frequency $(f_1)$ of the modal class $= 18$.
Frequency $(f_0)$ of the class preceding the modal class $= 4$.
Frequency $(f_2)$ of the class succeeding the modal class $= 9$.
Class size $(h) = 1000$.
The formula for mode is:
$\text{Mode} = l + \left( \frac{f_1 - f_0}{2f_1 - f_0 - f_2} \right) \times h$
Substituting the values:
$\text{Mode} = 4000 + \left( \frac{18 - 4}{2(18) - 4 - 9} \right) \times 1000$
$\text{Mode} = 4000 + \left( \frac{14}{36 - 13} \right) \times 1000$
$\text{Mode} = 4000 + \left( \frac{14}{23} \right) \times 1000$
$\text{Mode} = 4000 + \frac{14000}{23}$
$\text{Mode} = 4000 + 608.695...$
$\text{Mode} \approx 4608.7$
Therefore,the mode of the given data is $4608.7$ runs.
Therefore,the modal class is $4000-5000$.
Lower limit $(l)$ of the modal class $= 4000$.
Frequency $(f_1)$ of the modal class $= 18$.
Frequency $(f_0)$ of the class preceding the modal class $= 4$.
Frequency $(f_2)$ of the class succeeding the modal class $= 9$.
Class size $(h) = 1000$.
The formula for mode is:
$\text{Mode} = l + \left( \frac{f_1 - f_0}{2f_1 - f_0 - f_2} \right) \times h$
Substituting the values:
$\text{Mode} = 4000 + \left( \frac{18 - 4}{2(18) - 4 - 9} \right) \times 1000$
$\text{Mode} = 4000 + \left( \frac{14}{36 - 13} \right) \times 1000$
$\text{Mode} = 4000 + \left( \frac{14}{23} \right) \times 1000$
$\text{Mode} = 4000 + \frac{14000}{23}$
$\text{Mode} = 4000 + 608.695...$
$\text{Mode} \approx 4608.7$
Therefore,the mode of the given data is $4608.7$ runs.
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View Solution5
MediumMCQ
$A$ student noted the number of cars passing through a spot on a road for $100$ periods each of $3\, \text{minutes}$ 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 | $7$ | $14$ | $13$ | $12$ | $20$ | $11$ | $15$ | $8$ |
A
$44.7$
B
$40$
C
$35.3$
D
$43.50$
Solution
(A) From the given data, it can be observed that the maximum class frequency is $20,$ which belongs to the $40-50$ class interval.
Therefore, the modal class $= 40-50$.
Lower limit $(l)$ of the modal class $= 40$.
Frequency $(f_1)$ of the modal class $= 20$.
Frequency $(f_0)$ of the class preceding the modal class $= 12$.
Frequency $(f_2)$ of the class succeeding the modal class $= 11$.
Class size $(h) = 10$.
Using the mode formula: $\text{Mode} = l + \left( \frac{f_1 - f_0}{2f_1 - f_0 - f_2} \right) \times h$.
$\text{Mode} = 40 + \left( \frac{20 - 12}{2(20) - 12 - 11} \right) \times 10$.
$\text{Mode} = 40 + \left( \frac{8}{40 - 23} \right) \times 10$.
$\text{Mode} = 40 + \left( \frac{8}{17} \right) \times 10$.
$\text{Mode} = 40 + \frac{80}{17}$.
$\text{Mode} = 40 + 4.705 \approx 44.7$.
Therefore, the mode of this data is $44.7$ cars.
Therefore, the modal class $= 40-50$.
Lower limit $(l)$ of the modal class $= 40$.
Frequency $(f_1)$ of the modal class $= 20$.
Frequency $(f_0)$ of the class preceding the modal class $= 12$.
Frequency $(f_2)$ of the class succeeding the modal class $= 11$.
Class size $(h) = 10$.
Using the mode formula: $\text{Mode} = l + \left( \frac{f_1 - f_0}{2f_1 - f_0 - f_2} \right) \times h$.
$\text{Mode} = 40 + \left( \frac{20 - 12}{2(20) - 12 - 11} \right) \times 10$.
$\text{Mode} = 40 + \left( \frac{8}{40 - 23} \right) \times 10$.
$\text{Mode} = 40 + \left( \frac{8}{17} \right) \times 10$.
$\text{Mode} = 40 + \frac{80}{17}$.
$\text{Mode} = 40 + 4.705 \approx 44.7$.
Therefore, the mode of this data is $44.7$ cars.
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View Solution6
EasyMCQ
For the following distribution,the modal class is:
| Marks | Number of students |
| Below $10$ | $3$ |
| Below $20$ | $12$ |
| Below $30$ | $27$ |
| Below $40$ | $57$ |
| Below $50$ | $75$ |
| Below $60$ | $80$ |
A
$10-20$
B
$30-40$
C
$20-30$
D
$40-50$
Solution
(B) To find the modal class,we first convert the cumulative frequency distribution into a standard frequency distribution table:
The modal class is the class interval with the highest frequency.
From the table,the highest frequency is $30$,which corresponds to the class interval $30-40$.
Therefore,the modal class is $30-40$.
| Class Interval | Frequency |
| $0-10$ | $3$ |
| $10-20$ | $12 - 3 = 9$ |
| $20-30$ | $27 - 12 = 15$ |
| $30-40$ | $57 - 27 = 30$ |
| $40-50$ | $75 - 57 = 18$ |
| $50-60$ | $80 - 75 = 5$ |
The modal class is the class interval with the highest frequency.
From the table,the highest frequency is $30$,which corresponds to the class interval $30-40$.
Therefore,the modal class is $30-40$.
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View Solution7
MediumMCQ
The frequency distribution table of agricultural holdings in a village is given below:
Find the modal agricultural holdings of the village (in hectares).
| Area of land (in hectares) | $1-3$ | $3-5$ | $5-7$ | $7-9$ | $9-11$ | $11-13$ |
| Number of families | $20$ | $45$ | $80$ | $55$ | $40$ | $12$ |
Find the modal agricultural holdings of the village (in hectares).
A
$5.2$
B
$3.8$
C
$6.2$
D
$7.2$
Solution
(C) Here,the maximum class frequency is $80,$ and the class corresponding to this frequency is $5-7.$
So,the modal class is $5-7.$
$l$ (lower limit of modal class) $= 5$
$f_{1}$ (frequency of the modal class) $= 80$
$f_{0}$ (frequency of the class preceding the modal class) $= 45$
$f_{2}$ (frequency of the class succeeding the modal class) $= 55$
$h$ (class size) $= 2$
Mode $= l + \left( \frac{f_{1} - f_{0}}{2f_{1} - f_{0} - f_{2}} \right) \times h$
Mode $= 5 + \left( \frac{80 - 45}{2(80) - 45 - 55} \right) \times 2$
Mode $= 5 + \left( \frac{35}{160 - 100} \right) \times 2$
Mode $= 5 + \left( \frac{35}{60} \right) \times 2 = 5 + \frac{35}{30}$
Mode $= 5 + 1.166... \approx 6.17$ (Rounding to $6.2$ as per options).
Hence,the modal agricultural holdings of the village is $6.2$ hectares.
So,the modal class is $5-7.$
$l$ (lower limit of modal class) $= 5$
$f_{1}$ (frequency of the modal class) $= 80$
$f_{0}$ (frequency of the class preceding the modal class) $= 45$
$f_{2}$ (frequency of the class succeeding the modal class) $= 55$
$h$ (class size) $= 2$
Mode $= l + \left( \frac{f_{1} - f_{0}}{2f_{1} - f_{0} - f_{2}} \right) \times h$
Mode $= 5 + \left( \frac{80 - 45}{2(80) - 45 - 55} \right) \times 2$
Mode $= 5 + \left( \frac{35}{160 - 100} \right) \times 2$
Mode $= 5 + \left( \frac{35}{60} \right) \times 2 = 5 + \frac{35}{30}$
Mode $= 5 + 1.166... \approx 6.17$ (Rounding to $6.2$ as per options).
Hence,the modal agricultural holdings of the village is $6.2$ hectares.
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View Solution8
MediumMCQ
The monthly income of $100$ families is given as follows:
Calculate the modal income (in $Rs.$).
| Income (in $Rs.$) | Number of families |
| $0-5000$ | $8$ |
| $5000-10000$ | $26$ |
| $10000-15000$ | $41$ |
| $15000-20000$ | $16$ |
| $20000-25000$ | $3$ |
| $25000-30000$ | $3$ |
| $30000-35000$ | $2$ |
| $35000-40000$ | $1$ |
Calculate the modal income (in $Rs.$).
A
$11500$
B
$11675$
C
$11775$
D
$11875$
Solution
(D) In the given data,the highest frequency is $41,$ which corresponds to the modal class interval $10000-15000$.
Here,the lower limit of the modal class $l = 10000$,the frequency of the modal class $f_m = 41$,the frequency of the class preceding the modal class $f_1 = 26$,the frequency of the class succeeding the modal class $f_2 = 16$,and the class size $h = 5000$.
The formula for the mode is:
$\text{Mode} = l + \left( \frac{f_m - f_1}{2f_m - f_1 - f_2} \right) \times h$
Substituting the values:
$\text{Mode} = 10000 + \left( \frac{41 - 26}{2 \times 41 - 26 - 16} \right) \times 5000$
$= 10000 + \left( \frac{15}{82 - 42} \right) \times 5000$
$= 10000 + \left( \frac{15}{40} \right) \times 5000$
$= 10000 + 15 \times 125$
$= 10000 + 1875 = 11875$
Thus,the modal income is $Rs. 11875$.
Here,the lower limit of the modal class $l = 10000$,the frequency of the modal class $f_m = 41$,the frequency of the class preceding the modal class $f_1 = 26$,the frequency of the class succeeding the modal class $f_2 = 16$,and the class size $h = 5000$.
The formula for the mode is:
$\text{Mode} = l + \left( \frac{f_m - f_1}{2f_m - f_1 - f_2} \right) \times h$
Substituting the values:
$\text{Mode} = 10000 + \left( \frac{41 - 26}{2 \times 41 - 26 - 16} \right) \times 5000$
$= 10000 + \left( \frac{15}{82 - 42} \right) \times 5000$
$= 10000 + \left( \frac{15}{40} \right) \times 5000$
$= 10000 + 15 \times 125$
$= 10000 + 1875 = 11875$
Thus,the modal income is $Rs. 11875$.
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View Solution9
EasyMCQ
The weight of coffee in $70$ packets is shown in the following table:
Determine the modal weight (in $g$).
| Weight (in $g$) | Number of packets |
| $200-201$ | $12$ |
| $201-202$ | $26$ |
| $202-203$ | $20$ |
| $203-204$ | $9$ |
| $204-205$ | $2$ |
| $205-206$ | $1$ |
Determine the modal weight (in $g$).
A
$201.7$
B
$205.6$
C
$209.8$
D
$197.2$
Solution
(A) In the given data,the highest frequency is $26,$ which corresponds to the modal class interval $201-202$.
Here,the lower limit of the modal class $l = 201$,the frequency of the modal class $f_m = 26$,the frequency of the class preceding the modal class $f_1 = 12$,the frequency of the class succeeding the modal class $f_2 = 20$,and the class width $h = 1$.
The formula for the mode is:
$\text{Mode} = l + \left( \frac{f_m - f_1}{2f_m - f_1 - f_2} \right) \times h$
Substituting the values:
$\text{Mode} = 201 + \left( \frac{26 - 12}{2 \times 26 - 12 - 20} \right) \times 1$
$= 201 + \left( \frac{14}{52 - 32} \right)$
$= 201 + \frac{14}{20} = 201 + 0.7 = 201.7 \, g$.
Hence,the modal weight is $201.7 \, g$.
Here,the lower limit of the modal class $l = 201$,the frequency of the modal class $f_m = 26$,the frequency of the class preceding the modal class $f_1 = 12$,the frequency of the class succeeding the modal class $f_2 = 20$,and the class width $h = 1$.
The formula for the mode is:
$\text{Mode} = l + \left( \frac{f_m - f_1}{2f_m - f_1 - f_2} \right) \times h$
Substituting the values:
$\text{Mode} = 201 + \left( \frac{26 - 12}{2 \times 26 - 12 - 20} \right) \times 1$
$= 201 + \left( \frac{14}{52 - 32} \right)$
$= 201 + \frac{14}{20} = 201 + 0.7 = 201.7 \, g$.
Hence,the modal weight is $201.7 \, g$.
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View Solution10
MediumMCQ
Find the mode of the following frequency distribution:
| Class | $0-100$ | $100-200$ | $200-300$ | $300-400$ | $400-500$ | $500-600$ |
|---|---|---|---|---|---|---|
| Frequency | $7$ | $21$ | $37$ | $13$ | $12$ | $10$ |
A
$120$
B
$240$
C
$560$
D
$340$
Solution
(B) The modal class is the class with the highest frequency. Here,the highest frequency is $37$,which corresponds to the class interval $200-300$.
Thus,the modal class is $200-300$.
We use the formula for mode: $Z = l + \left( \frac{f_1 - f_0}{2f_1 - f_0 - f_2} \right) \times h$
Where:
$l$ (lower limit of modal class) $= 200$
$h$ (class size) $= 100$
$f_1$ (frequency of modal class) $= 37$
$f_0$ (frequency of the class preceding the modal class) $= 21$
$f_2$ (frequency of the class succeeding the modal class) $= 13$
Substituting these values into the formula:
$Z = 200 + \left( \frac{37 - 21}{2(37) - 21 - 13} \right) \times 100$
$Z = 200 + \left( \frac{16}{74 - 34} \right) \times 100$
$Z = 200 + \left( \frac{16}{40} \right) \times 100$
$Z = 200 + 0.4 \times 100$
$Z = 200 + 40 = 240$
Therefore,the mode of the given frequency distribution is $240$.
Thus,the modal class is $200-300$.
We use the formula for mode: $Z = l + \left( \frac{f_1 - f_0}{2f_1 - f_0 - f_2} \right) \times h$
Where:
$l$ (lower limit of modal class) $= 200$
$h$ (class size) $= 100$
$f_1$ (frequency of modal class) $= 37$
$f_0$ (frequency of the class preceding the modal class) $= 21$
$f_2$ (frequency of the class succeeding the modal class) $= 13$
Substituting these values into the formula:
$Z = 200 + \left( \frac{37 - 21}{2(37) - 21 - 13} \right) \times 100$
$Z = 200 + \left( \frac{16}{74 - 34} \right) \times 100$
$Z = 200 + \left( \frac{16}{40} \right) \times 100$
$Z = 200 + 0.4 \times 100$
$Z = 200 + 40 = 240$
Therefore,the mode of the given frequency distribution is $240$.
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View Solution11
Medium
The mode of the following data is $33 \frac{1}{3}$ and the total frequency is $100$. Find the missing frequencies $x$ and $y$.
| Class | $0-10$ | $10-20$ | $20-30$ | $30-40$ | $40-50$ | $50-60$ |
| Frequency | $7$ | $12$ | $x$ | $28$ | $y$ | $9$ |
Solution
(A) Here,the mode $33 \frac{1}{3}$ lies in the class $30-40$.
Therefore,$30-40$ is the modal class.
Now,$c$ (class length) $= 10$,$l$ (lower limit of the modal class) $= 30$,$f_1$ (frequency of the modal class) $= 28$,$f_0$ (frequency of the class preceding the modal class) $= x$,and $f_2$ (frequency of the class succeeding the modal class) $= y$.
Substituting the values in the mode formula $Z = l + \left( \frac{f_1 - f_0}{2f_1 - f_0 - f_2} \right) \times c$:
$33 \frac{1}{3} = 30 + \left( \frac{28 - x}{2(28) - x - y} \right) \times 10$
$3 \frac{1}{3} = \frac{28 - x}{56 - x - y} \times 10$
$\frac{10}{3} = \frac{28 - x}{56 - x - y} \times 10$
$\frac{1}{3} = \frac{28 - x}{56 - x - y}$
$56 - x - y = 84 - 3x$
$2x - y = 28$ ... $(1)$
Moreover,the total frequency is $100$.
$7 + 12 + x + 28 + y + 9 = 100$
$56 + x + y = 100$
$x + y = 44$ ... $(2)$
Adding equations $(1)$ and $(2)$:
$3x = 72 \implies x = 24$
Substituting $x = 24$ in equation $(2)$:
$24 + y = 44 \implies y = 20$
Thus,the missing frequencies are $x = 24$ and $y = 20$.
Therefore,$30-40$ is the modal class.
Now,$c$ (class length) $= 10$,$l$ (lower limit of the modal class) $= 30$,$f_1$ (frequency of the modal class) $= 28$,$f_0$ (frequency of the class preceding the modal class) $= x$,and $f_2$ (frequency of the class succeeding the modal class) $= y$.
Substituting the values in the mode formula $Z = l + \left( \frac{f_1 - f_0}{2f_1 - f_0 - f_2} \right) \times c$:
$33 \frac{1}{3} = 30 + \left( \frac{28 - x}{2(28) - x - y} \right) \times 10$
$3 \frac{1}{3} = \frac{28 - x}{56 - x - y} \times 10$
$\frac{10}{3} = \frac{28 - x}{56 - x - y} \times 10$
$\frac{1}{3} = \frac{28 - x}{56 - x - y}$
$56 - x - y = 84 - 3x$
$2x - y = 28$ ... $(1)$
Moreover,the total frequency is $100$.
$7 + 12 + x + 28 + y + 9 = 100$
$56 + x + y = 100$
$x + y = 44$ ... $(2)$
Adding equations $(1)$ and $(2)$:
$3x = 72 \implies x = 24$
Substituting $x = 24$ in equation $(2)$:
$24 + y = 44 \implies y = 20$
Thus,the missing frequencies are $x = 24$ and $y = 20$.
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View Solution12
MediumMCQ
Find the mode of the following frequency distribution:
| Class | $20-30$ | $30-40$ | $40-50$ | $50-60$ | $60-70$ | $70-80$ | $80-90$ |
| Frequency | $8$ | $12$ | $27$ | $43$ | $55$ | $37$ | $18$ |
A
$640$
B
$62$
C
$63$
D
$64$
Solution
(D) To find the mode of a grouped frequency distribution,we use the formula:
$Mode = l + \left( \frac{f_1 - f_0}{2f_1 - f_0 - f_2} \right) \times h$
$1$. Identify the modal class (the class with the highest frequency). Here,the highest frequency is $55$,which corresponds to the class interval $60-70$.
$2$. From the modal class:
Lower limit $(l)$ = $60$
Frequency of the modal class $(f_1)$ = $55$
Frequency of the class preceding the modal class $(f_0)$ = $43$
Frequency of the class succeeding the modal class $(f_2)$ = $37$
Class size $(h)$ = $70 - 60 = 10$
$3$. Substitute the values into the formula:
$Mode = 60 + \left( \frac{55 - 43}{2(55) - 43 - 37} \right) \times 10$
$Mode = 60 + \left( \frac{12}{110 - 80} \right) \times 10$
$Mode = 60 + \left( \frac{12}{30} \right) \times 10$
$Mode = 60 + \left( \frac{12}{3} \right)$
$Mode = 60 + 4 = 64$
Therefore,the mode of the given frequency distribution is $64$.
$Mode = l + \left( \frac{f_1 - f_0}{2f_1 - f_0 - f_2} \right) \times h$
$1$. Identify the modal class (the class with the highest frequency). Here,the highest frequency is $55$,which corresponds to the class interval $60-70$.
$2$. From the modal class:
Lower limit $(l)$ = $60$
Frequency of the modal class $(f_1)$ = $55$
Frequency of the class preceding the modal class $(f_0)$ = $43$
Frequency of the class succeeding the modal class $(f_2)$ = $37$
Class size $(h)$ = $70 - 60 = 10$
$3$. Substitute the values into the formula:
$Mode = 60 + \left( \frac{55 - 43}{2(55) - 43 - 37} \right) \times 10$
$Mode = 60 + \left( \frac{12}{110 - 80} \right) \times 10$
$Mode = 60 + \left( \frac{12}{30} \right) \times 10$
$Mode = 60 + \left( \frac{12}{3} \right)$
$Mode = 60 + 4 = 64$
Therefore,the mode of the given frequency distribution is $64$.
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View Solution13
MediumMCQ
Find the mode of the following frequency distribution:
| Class | $0-20$ | $20-40$ | $40-60$ | $60-80$ | $80-100$ | $100-120$ |
| Frequency | $8$ | $9$ | $26$ | $23$ | $20$ | $14$ |
A
$57$
B
$55$
C
$56$
D
$54$
Solution
(A) To find the mode of a grouped frequency distribution,we use the formula:
$Mode = l + \left( \frac{f_1 - f_0}{2f_1 - f_0 - f_2} \right) \times h$
$1$. Identify the modal class: The class with the highest frequency is $40-60$ (frequency $26$).
$2$. Here,the lower limit of the modal class $l = 40$,the size of the class interval $h = 20$.
$3$. The frequency of the modal class $f_1 = 26$.
$4$. The frequency of the class preceding the modal class $f_0 = 9$.
$5$. The frequency of the class succeeding the modal class $f_2 = 23$.
$6$. Substituting these values into the formula:
$Mode = 40 + \left( \frac{26 - 9}{2(26) - 9 - 23} \right) \times 20$
$Mode = 40 + \left( \frac{17}{52 - 32} \right) \times 20$
$Mode = 40 + \left( \frac{17}{20} \right) \times 20$
$Mode = 40 + 17 = 57$
Thus,the mode is $57$.
$Mode = l + \left( \frac{f_1 - f_0}{2f_1 - f_0 - f_2} \right) \times h$
$1$. Identify the modal class: The class with the highest frequency is $40-60$ (frequency $26$).
$2$. Here,the lower limit of the modal class $l = 40$,the size of the class interval $h = 20$.
$3$. The frequency of the modal class $f_1 = 26$.
$4$. The frequency of the class preceding the modal class $f_0 = 9$.
$5$. The frequency of the class succeeding the modal class $f_2 = 23$.
$6$. Substituting these values into the formula:
$Mode = 40 + \left( \frac{26 - 9}{2(26) - 9 - 23} \right) \times 20$
$Mode = 40 + \left( \frac{17}{52 - 32} \right) \times 20$
$Mode = 40 + \left( \frac{17}{20} \right) \times 20$
$Mode = 40 + 17 = 57$
Thus,the mode is $57$.
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View Solution14
MediumMCQ
Find the mode of the following frequency distribution:
| Class | $0-10$ | $10-20$ | $20-30$ | $30-40$ | $40-50$ | $50-60$ | $60-70$ | $70-80$ |
|---|---|---|---|---|---|---|---|---|
| Frequency | $6$ | $8$ | $17$ | $24$ | $42$ | $30$ | $15$ | $8$ |
A
$45$
B
$46$
C
$47$
D
$48$
Solution
(B) The formula for the mode of a grouped frequency distribution is given by:
$Mode = l + \left( \frac{f_1 - f_0}{2f_1 - f_0 - f_2} \right) \times h$
$1$. Identify the modal class: The class with the highest frequency is $40-50$,so the modal class is $40-50$.
$2$. Identify the parameters:
- Lower limit of the modal class $(l)$ = $40$
- Frequency of the modal class $(f_1)$ = $42$
- Frequency of the class preceding the modal class $(f_0)$ = $24$
- Frequency of the class succeeding the modal class $(f_2)$ = $30$
- Class size $(h)$ = $10$
$3$. Substitute the values into the formula:
$Mode = 40 + \left( \frac{42 - 24}{2(42) - 24 - 30} \right) \times 10$
$Mode = 40 + \left( \frac{18}{84 - 54} \right) \times 10$
$Mode = 40 + \left( \frac{18}{30} \right) \times 10$
$Mode = 40 + \left( \frac{18}{3} \right) = 40 + 6 = 46$
Thus,the mode is $46$.
$Mode = l + \left( \frac{f_1 - f_0}{2f_1 - f_0 - f_2} \right) \times h$
$1$. Identify the modal class: The class with the highest frequency is $40-50$,so the modal class is $40-50$.
$2$. Identify the parameters:
- Lower limit of the modal class $(l)$ = $40$
- Frequency of the modal class $(f_1)$ = $42$
- Frequency of the class preceding the modal class $(f_0)$ = $24$
- Frequency of the class succeeding the modal class $(f_2)$ = $30$
- Class size $(h)$ = $10$
$3$. Substitute the values into the formula:
$Mode = 40 + \left( \frac{42 - 24}{2(42) - 24 - 30} \right) \times 10$
$Mode = 40 + \left( \frac{18}{84 - 54} \right) \times 10$
$Mode = 40 + \left( \frac{18}{30} \right) \times 10$
$Mode = 40 + \left( \frac{18}{3} \right) = 40 + 6 = 46$
Thus,the mode is $46$.
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View Solution15
MediumMCQ
Find the mode of the following frequency distribution:
| Class | $0-15$ | $15-30$ | $30-45$ | $45-60$ | $60-75$ | $75-90$ | $90-105$ |
| Frequency | $4$ | $5$ | $23$ | $45$ | $66$ | $42$ | $15$ |
A
$67.5$
B
$77$
C
$67$
D
$66$
Solution
(C) The modal class is the class with the highest frequency. Here,the highest frequency is $66$,which corresponds to the class interval $60-75$.
Formula for mode: $\text{Mode} = l + \left( \frac{f_1 - f_0}{2f_1 - f_0 - f_2} \right) \times h$
Here:
$l$ (lower limit of modal class) $= 60$
$h$ (class size) $= 15$
$f_1$ (frequency of modal class) $= 66$
$f_0$ (frequency of the class preceding the modal class) $= 45$
$f_2$ (frequency of the class succeeding the modal class) $= 42$
Substituting the values:
$\text{Mode} = 60 + \left( \frac{66 - 45}{2(66) - 45 - 42} \right) \times 15$
$\text{Mode} = 60 + \left( \frac{21}{132 - 87} \right) \times 15$
$\text{Mode} = 60 + \left( \frac{21}{45} \right) \times 15$
$\text{Mode} = 60 + \left( \frac{21}{3} \right) = 60 + 7 = 67$
Formula for mode: $\text{Mode} = l + \left( \frac{f_1 - f_0}{2f_1 - f_0 - f_2} \right) \times h$
Here:
$l$ (lower limit of modal class) $= 60$
$h$ (class size) $= 15$
$f_1$ (frequency of modal class) $= 66$
$f_0$ (frequency of the class preceding the modal class) $= 45$
$f_2$ (frequency of the class succeeding the modal class) $= 42$
Substituting the values:
$\text{Mode} = 60 + \left( \frac{66 - 45}{2(66) - 45 - 42} \right) \times 15$
$\text{Mode} = 60 + \left( \frac{21}{132 - 87} \right) \times 15$
$\text{Mode} = 60 + \left( \frac{21}{45} \right) \times 15$
$\text{Mode} = 60 + \left( \frac{21}{3} \right) = 60 + 7 = 67$
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View Solution16
MediumMCQ
The mode of the following frequency distribution is $37$. Find the missing frequency $x$.
| Class | $0-10$ | $10-20$ | $20-30$ | $30-40$ | $40-50$ | $50-60$ | $60-70$ |
| Frequency | $4$ | $7$ | $x$ | $18$ | $15$ | $8$ | $7$ |
A
$13$
B
$12$
C
$11$
D
$10$
Solution
(C) The mode is given as $37$. Since $37$ lies in the class interval $30-40$,the modal class is $30-40$.
The formula for mode is: $\text{Mode} = l + \left( \frac{f_1 - f_0}{2f_1 - f_0 - f_2} \right) \times h$
Here,$l = 30$ (lower limit of modal class),$h = 10$ (class size),$f_1 = 18$ (frequency of modal class),$f_0 = x$ (frequency of preceding class),and $f_2 = 15$ (frequency of succeeding class).
Substituting the values: $37 = 30 + \left( \frac{18 - x}{2(18) - x - 15} \right) \times 10$
$37 - 30 = \left( \frac{18 - x}{36 - 15 - x} \right) \times 10$
$7 = \left( \frac{18 - x}{21 - x} \right) \times 10$
$7(21 - x) = 10(18 - x)$
$147 - 7x = 180 - 10x$
$10x - 7x = 180 - 147$
$3x = 33$
$x = 11$.
The formula for mode is: $\text{Mode} = l + \left( \frac{f_1 - f_0}{2f_1 - f_0 - f_2} \right) \times h$
Here,$l = 30$ (lower limit of modal class),$h = 10$ (class size),$f_1 = 18$ (frequency of modal class),$f_0 = x$ (frequency of preceding class),and $f_2 = 15$ (frequency of succeeding class).
Substituting the values: $37 = 30 + \left( \frac{18 - x}{2(18) - x - 15} \right) \times 10$
$37 - 30 = \left( \frac{18 - x}{36 - 15 - x} \right) \times 10$
$7 = \left( \frac{18 - x}{21 - x} \right) \times 10$
$7(21 - x) = 10(18 - x)$
$147 - 7x = 180 - 10x$
$10x - 7x = 180 - 147$
$3x = 33$
$x = 11$.
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View Solution17
Medium
The mode of the following frequency distribution is $46$ and the total frequency is $150$. Find the missing frequencies $x$ and $y$.
| Class | $0-10$ | $10-20$ | $20-30$ | $30-40$ | $40-50$ | $50-60$ | $60-70$ | $70-80$ |
| Frequency | $6$ | $8$ | $17$ | $x$ | $42$ | $30$ | $y$ | $8$ |
Solution
(A) Given total frequency $N = 150$.
Sum of frequencies: $6 + 8 + 17 + x + 42 + 30 + y + 8 = 150$
$111 + x + y = 150 \implies x + y = 39$ --- $(1)$
Since the mode is $46$,the modal class is $40-50$.
Here,$l = 40, f_1 = 42, f_0 = x, f_2 = 30, h = 10$.
Mode formula: $\text{Mode} = l + \left( \frac{f_1 - f_0}{2f_1 - f_0 - f_2} \right) \times h$
$46 = 40 + \left( \frac{42 - x}{2(42) - x - 30} \right) \times 10$
$6 = \left( \frac{42 - x}{84 - 30 - x} \right) \times 10$
$6 = \frac{10(42 - x)}{54 - x} \implies 3(54 - x) = 5(42 - x)$
$162 - 3x = 210 - 5x$
$2x = 48 \implies x = 24$
Substituting $x = 24$ in $(1)$: $24 + y = 39 \implies y = 15$.
Thus,$x = 24$ and $y = 15$.
Sum of frequencies: $6 + 8 + 17 + x + 42 + 30 + y + 8 = 150$
$111 + x + y = 150 \implies x + y = 39$ --- $(1)$
Since the mode is $46$,the modal class is $40-50$.
Here,$l = 40, f_1 = 42, f_0 = x, f_2 = 30, h = 10$.
Mode formula: $\text{Mode} = l + \left( \frac{f_1 - f_0}{2f_1 - f_0 - f_2} \right) \times h$
$46 = 40 + \left( \frac{42 - x}{2(42) - x - 30} \right) \times 10$
$6 = \left( \frac{42 - x}{84 - 30 - x} \right) \times 10$
$6 = \frac{10(42 - x)}{54 - x} \implies 3(54 - x) = 5(42 - x)$
$162 - 3x = 210 - 5x$
$2x = 48 \implies x = 24$
Substituting $x = 24$ in $(1)$: $24 + y = 39 \implies y = 15$.
Thus,$x = 24$ and $y = 15$.
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View Solution18
Medium
The mode of the following frequency distribution is $64$ and the total frequency is $200.$ Find the missing frequencies $x$ and $y.$
| Class | $20-30$ | $30-40$ | $40-50$ | $50-60$ | $60-70$ | $70-80$ | $80-90$ |
| Frequency | $8$ | $12$ | $27$ | $x$ | $55$ | $37$ | $y$ |
Solution
(X=43, Y=18) Given total frequency $N = 200.$
Sum of frequencies: $8 + 12 + 27 + x + 55 + 37 + y = 200 \implies 139 + x + y = 200 \implies x + y = 61$ (Equation $1$).
Since the mode is $64,$ the modal class is $60-70.$
Formula for mode: $\text{Mode} = l + \left( \frac{f_1 - f_0}{2f_1 - f_0 - f_2} \right) \times h.$
Here,$l = 60, f_1 = 55, f_0 = x, f_2 = 37, h = 10.$
$64 = 60 + \left( \frac{55 - x}{2(55) - x - 37} \right) \times 10.$
$4 = \left( \frac{55 - x}{110 - x - 37} \right) \times 10 \implies 4 = \frac{550 - 10x}{73 - x}.$
$292 - 4x = 550 - 10x \implies 6x = 258 \implies x = 43.$
Substituting $x = 43$ in Equation $1$: $43 + y = 61 \implies y = 18.$
Thus,$x = 43$ and $y = 18.$
Sum of frequencies: $8 + 12 + 27 + x + 55 + 37 + y = 200 \implies 139 + x + y = 200 \implies x + y = 61$ (Equation $1$).
Since the mode is $64,$ the modal class is $60-70.$
Formula for mode: $\text{Mode} = l + \left( \frac{f_1 - f_0}{2f_1 - f_0 - f_2} \right) \times h.$
Here,$l = 60, f_1 = 55, f_0 = x, f_2 = 37, h = 10.$
$64 = 60 + \left( \frac{55 - x}{2(55) - x - 37} \right) \times 10.$
$4 = \left( \frac{55 - x}{110 - x - 37} \right) \times 10 \implies 4 = \frac{550 - 10x}{73 - x}.$
$292 - 4x = 550 - 10x \implies 6x = 258 \implies x = 43.$
Substituting $x = 43$ in Equation $1$: $43 + y = 61 \implies y = 18.$
Thus,$x = 43$ and $y = 18.$
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View Solution19
EasyMCQ
In the formula $Z = l + \left( \frac{f_{1} - f_{0}}{2f_{1} - f_{0} - f_{2}} \right) \times c$ for the mode,$l = \ldots \ldots \ldots$
A
lower limit of the modal class
B
frequency of the modal class
C
frequency of the class preceding the modal class
D
frequency of the class succeeding the modal class
Solution
(A) In the formula for calculating the mode of grouped data,$Z = l + \left( \frac{f_{1} - f_{0}}{2f_{1} - f_{0} - f_{2}} \right) \times c$,the term $l$ represents the lower limit of the modal class. The modal class is the class interval with the highest frequency.
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View Solution20
EasyMCQ
In the formula $Z = l + \left( \frac{f_{1} - f_{0}}{2f_{1} - f_{0} - f_{2}} \right) \times c$ for the mode,$f_{0} = \ldots \ldots \ldots$
A
frequency of the modal class
B
frequency of the class preceding the modal class
C
class length
D
frequency of the class succeeding the modal class
Solution
(B) In the formula for the mode,$Z = l + \left( \frac{f_{1} - f_{0}}{2f_{1} - f_{0} - f_{2}} \right) \times c$,the terms are defined as follows:
$l$ is the lower limit of the modal class.
$f_{1}$ is the frequency of the 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.
$c$ is the class size or class length.
Therefore,$f_{0}$ represents the frequency of the class preceding the modal class.
$l$ is the lower limit of the modal class.
$f_{1}$ is the frequency of the 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.
$c$ is the class size or class length.
Therefore,$f_{0}$ represents the frequency of the class preceding the modal class.
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View Solution21
EasyMCQ
In the formula $Z = l + \left( \frac{f_{1} - f_{0}}{2f_{1} - f_{0} - f_{2}} \right) \times c$ for the mode,$f_{1} = \ldots \ldots \ldots$
A
frequency of the class preceding the modal class
B
class length
C
frequency of the modal class
D
frequency of the class succeeding the modal class
Solution
(C) In the formula for the mode of grouped data,$Z = l + \left( \frac{f_{1} - f_{0}}{2f_{1} - f_{0} - f_{2}} \right) \times c$,the term $f_{1}$ represents the frequency of the modal class. Here,$l$ is the lower limit of the 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,and $c$ is the class size (or class length).
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View Solution22
EasyMCQ
In the formula $Z = l + \left( \frac{f_{1} - f_{0}}{2f_{1} - f_{0} - f_{2}} \right) \times c$ for the mode,$f_{2} = \ldots \ldots \ldots$
A
frequency of the modal class
B
lower limit of the modal class
C
frequency of the class preceding the modal class
D
frequency of the class succeeding the modal class
Solution
(D) In the formula for calculating the mode of grouped data,the variables are defined as follows:
$Z$ is the mode.
$l$ is the lower limit of the modal class.
$f_{1}$ is the frequency of the 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.
$c$ is the size of the class interval.
Therefore,$f_{2}$ represents the frequency of the class succeeding the modal class.
$Z$ is the mode.
$l$ is the lower limit of the modal class.
$f_{1}$ is the frequency of the 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.
$c$ is the size of the class interval.
Therefore,$f_{2}$ represents the frequency of the class succeeding the modal class.
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View Solution23
EasyMCQ
In the formula $Z = l + \left( \frac{f_{1} - f_{0}}{2f_{1} - f_{0} - f_{2}} \right) \times c$ for the mode,$c = \dots$
A
class length
B
frequency of the modal class
C
frequency of the class preceding the modal class
D
frequency of the class succeeding the modal class
Solution
(A) In the formula for the mode of grouped data,$Z = l + \left( \frac{f_{1} - f_{0}}{2f_{1} - f_{0} - f_{2}} \right) \times c$,the term $c$ represents the class size or class length of the modal class. The other terms are defined as follows: $l$ is the lower limit of the modal class,$f_{1}$ is the frequency of the modal class,$f_{0}$ is the frequency of the class preceding the modal class,and $f_{2}$ is the frequency of the class succeeding the modal class.
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View Solution24
EasyMCQ
For the following data,the modal class is .... .
| Class | $0-10$ | $10-20$ | $20-30$ | $30-40$ | $40-50$ |
| Frequency | $7$ | $16$ | $25$ | $32$ | $20$ |
A
$10-20$
B
$20-30$
C
$30-40$
D
$40-50$
Solution
(C) The modal class is the class interval with the highest frequency.
From the given table,the frequencies are:
$0-10: 7$
$10-20: 16$
$20-30: 25$
$30-40: 32$
$40-50: 20$
The maximum frequency is $32$,which corresponds to the class interval $30-40$.
Therefore,the modal class is $30-40$.
From the given table,the frequencies are:
$0-10: 7$
$10-20: 16$
$20-30: 25$
$30-40: 32$
$40-50: 20$
The maximum frequency is $32$,which corresponds to the class interval $30-40$.
Therefore,the modal class is $30-40$.
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View Solution25
EasyMCQ
The formula to find the mode of a grouped data is $\ldots \ldots \ldots . . .$
A
$Z=l+\frac{(\frac{n}{2}-cf)}{f} \times c$
B
$Z=l+\left(\frac{f_{1}-f_{0}}{2 f_{1}-f_{0}-f_{2}}\right) \times h$
C
$Z=3M - 2\bar{x}$
D
$Z= A +\frac{\sum f_{i} u_{i}}{\sum f_{i}} \times h$
Solution
(B) The mode of a grouped data is calculated using the formula:
$Z = l + \left( \frac{f_1 - f_0}{2f_1 - f_0 - f_2} \right) \times h$
Where:
$l$ = lower limit of the modal class
$h$ = size of the class interval
$f_1$ = frequency of the modal class
$f_0$ = frequency of the class preceding the modal class
$f_2$ = frequency of the class succeeding the modal class.
$Z = l + \left( \frac{f_1 - f_0}{2f_1 - f_0 - f_2} \right) \times h$
Where:
$l$ = lower limit of the modal class
$h$ = size of the class interval
$f_1$ = frequency of the modal class
$f_0$ = frequency of the class preceding the modal class
$f_2$ = frequency of the class succeeding the modal class.
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View Solution26
EasyMCQ
In the formula $Z = l + \left( \frac{f_{1} - f_{0}}{2f_{1} - f_{0} - f_{2}} \right) \times c$,$f_{1}$ represents:
A
lower limit of the modal class
B
frequency of the modal class
C
frequency of the class preceding the modal class
D
frequency of the class succeeding the modal class
Solution
(B) In the formula for the mode of grouped data,$Z = l + \left( \frac{f_{1} - f_{0}}{2f_{1} - f_{0} - f_{2}} \right) \times c$,the variables are defined as follows:
$l$: lower limit of the modal class.
$f_{1}$: frequency of the modal class.
$f_{0}$: frequency of the class preceding the modal class.
$f_{2}$: frequency of the class succeeding the modal class.
$c$: size of the class interval.
Therefore,$f_{1}$ represents the frequency of the modal class.
$l$: lower limit of the modal class.
$f_{1}$: frequency of the modal class.
$f_{0}$: frequency of the class preceding the modal class.
$f_{2}$: frequency of the class succeeding the modal class.
$c$: size of the class interval.
Therefore,$f_{1}$ represents the frequency of the modal class.
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View Solution27
EasyMCQ
The mode of the observations $4, 5, 6, 3, 4, 3, 3, 2, 3, 5$ is $\ldots \ldots \ldots \ldots .$
A
$3$
B
$6$
C
$4$
D
$5$
Solution
(A) The mode is defined as the value that appears most frequently in a data set.
First,let us list the frequency of each observation:
$2$ appears $1$ time.
$3$ appears $4$ times.
$4$ appears $2$ times.
$5$ appears $2$ times.
$6$ appears $1$ time.
Since the observation $3$ has the highest frequency of $4$,the mode of the given data is $3$.
First,let us list the frequency of each observation:
$2$ appears $1$ time.
$3$ appears $4$ times.
$4$ appears $2$ times.
$5$ appears $2$ times.
$6$ appears $1$ time.
Since the observation $3$ has the highest frequency of $4$,the mode of the given data is $3$.
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View Solution28
EasyMCQ
For the calculations of mode of a grouped data,the maximum frequency of a class is denoted by $\ldots \ldots \ldots . .$
A
$f_0$
B
$f_1$
C
$f_2$
D
$cf$
Solution
(B) In the formula for calculating the mode of grouped data,the modal class is the class with the highest frequency. This maximum frequency is denoted by $f_1$. The frequency of the class preceding the modal class is denoted by $f_0$,and the frequency of the class succeeding the modal class is denoted by $f_2$.
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View Solution29
EasyMCQ
If the modal class of a frequency distribution is $70-85,$ then in the formula for mode,$l=$ ..........
A
$15$
B
$77.5$
C
$70$
D
$85$
Solution
(C) In the formula for calculating the mode of a grouped frequency distribution,$l$ represents the lower limit of the modal class.
Given the modal class is $70-85$.
The lower limit of this class is $70$.
Therefore,$l = 70$.
Given the modal class is $70-85$.
The lower limit of this class is $70$.
Therefore,$l = 70$.
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View Solution30
EasyMCQ
The modal class of the following frequency distribution is ........
| Class | $4-8$ | $8-12$ | $12-16$ | $16-20$ | $20-24$ | $24-28$ |
| Frequency | $9$ | $8$ | $12$ | $7$ | $15$ | $1$ |
A
$12-16$
B
$20-24$
C
$16-20$
D
$24-28$
Solution
(B) The modal class is defined as the class interval with the highest frequency in a given frequency distribution.
By observing the provided table,we compare the frequencies:
$9, 8, 12, 7, 15, 1$.
The maximum frequency is $15$.
The class interval corresponding to the frequency $15$ is $20-24$.
Therefore,the modal class is $20-24$.
By observing the provided table,we compare the frequencies:
$9, 8, 12, 7, 15, 1$.
The maximum frequency is $15$.
The class interval corresponding to the frequency $15$ is $20-24$.
Therefore,the modal class is $20-24$.
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View Solution31
EasyMCQ
For the calculation of mode of the following frequency distribution,$f_{1} = \dots \dots \dots \dots \dots \dots$
| Class | $0-100$ | $100-200$ | $200-300$ | $300-400$ | $400-500$ | $500-600$ |
| Frequency | $12$ | $18$ | $27$ | $20$ | $17$ | $6$ |
A
$18$
B
$20$
C
$27$
D
$17$
Solution
(C) In the formula for calculating the mode of grouped data,$f_{1}$ represents the frequency of the modal class.
The modal class is the class interval with the highest frequency.
Looking at the given frequency distribution:
- The frequencies are $12, 18, 27, 20, 17, 6$.
- The maximum frequency is $27$,which corresponds to the class interval $200-300$.
Therefore,the frequency of the modal class is $f_{1} = 27$.
The modal class is the class interval with the highest frequency.
Looking at the given frequency distribution:
- The frequencies are $12, 18, 27, 20, 17, 6$.
- The maximum frequency is $27$,which corresponds to the class interval $200-300$.
Therefore,the frequency of the modal class is $f_{1} = 27$.
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View Solution32
EasyMCQ
For the calculation of mode of the following frequency distribution,$f_{0} = \dots \dots \dots \dots \dots$
| Class | $1-3$ | $3-5$ | $5-7$ | $7-9$ | $9-11$ |
| Frequency | $6$ | $3$ | $8$ | $2$ | $1$ |
A
$3$
B
$8$
C
$2$
D
$6$
Solution
(A) In the given frequency distribution,the maximum frequency is $8$,which corresponds to the class interval $5-7$.
Therefore,the modal class is $5-7$.
In the formula for mode,$f_{1}$ is the frequency of the modal class,$f_{0}$ is the frequency of the class preceding the modal class,and $f_{2}$ is the frequency of the class succeeding the modal class.
Here,the modal class is $5-7$,so its frequency $f_{1} = 8$.
The class preceding the modal class is $3-5$,and its frequency is $f_{0} = 3$.
Thus,$f_{0} = 3$.
Therefore,the modal class is $5-7$.
In the formula for mode,$f_{1}$ is the frequency of the modal class,$f_{0}$ is the frequency of the class preceding the modal class,and $f_{2}$ is the frequency of the class succeeding the modal class.
Here,the modal class is $5-7$,so its frequency $f_{1} = 8$.
The class preceding the modal class is $3-5$,and its frequency is $f_{0} = 3$.
Thus,$f_{0} = 3$.
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View Solution33
EasyMCQ
The modal class of a frequency distribution is $20-24$. Then,in the calculations of mode,$l=$ ..........
A
$20$
B
$22$
C
$24$
D
$25$
Solution
(A) The modal class is the class interval with the highest frequency.
Given that the modal class is $20-24$.
In the formula for calculating the mode,$l$ represents the lower limit of the modal class.
Therefore,$l = 20$.
Given that the modal class is $20-24$.
In the formula for calculating the mode,$l$ represents the lower limit of the modal class.
Therefore,$l = 20$.
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View Solution34
MediumMCQ
In a frequency distribution,the modal class is $70-85$ with frequency $25$. The frequencies of the classes succeeding and preceding the modal class are $20$ and $8$ respectively. Then,the values of $f_{0}, f_{1}$ and $f_{2}$ are respectively.
A
$25, 20$ and $8$
B
$25, 8$ and $20$
C
$8, 20$ and $25$
D
$8, 25$ and $20$
Solution
(D) In the formula for the mode of a grouped frequency distribution,$f_{1}$ represents the frequency of the modal class,$f_{0}$ represents the frequency of the class preceding the modal class,and $f_{2}$ represents the frequency of the class succeeding the modal class.
Given that the modal class is $70-85$ with frequency $25$,we have $f_{1} = 25$.
The frequency of the class preceding the modal class is $8$,so $f_{0} = 8$.
The frequency of the class succeeding the modal class is $20$,so $f_{2} = 20$.
Therefore,the values of $f_{0}, f_{1}$ and $f_{2}$ are $8, 25$ and $20$ respectively.
Given that the modal class is $70-85$ with frequency $25$,we have $f_{1} = 25$.
The frequency of the class preceding the modal class is $8$,so $f_{0} = 8$.
The frequency of the class succeeding the modal class is $20$,so $f_{2} = 20$.
Therefore,the values of $f_{0}, f_{1}$ and $f_{2}$ are $8, 25$ and $20$ respectively.
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