Mode
Category : 9th Class
Mode
Mode is the value that occurs the most of the time in a data or mode is a way of capturing important information about a random variable or a population in a single quantity. The mode is generally different from the mean and median.
Model Class
In a frequency distribution, the class having maximum frequency is called modal class.
Formula for Mode
Mode can be calculated by a formula, which is given below:
Mode \[={{x}_{k}}+h\left[ \frac{{{f}_{k}}-{{f}_{k-1}}}{2{{f}_{k}}-{{f}_{k-1}}-{{f}_{k+1}}} \right]\]
where \[{{x}_{k}}\]= lower limit of the modal class interval.
\[{{f}_{k}}\]= frequency of the modal class
\[{{f}_{k-1}}\]= frequency of the class preceding the modal class
\[{{f}_{k+1}}\]= frequency of the class succeeding the modal class
h = width of the class interval
Find the mode for the following frequency distribution:
| Class Interval | Frequency | Class Interval | Frequency |
| 0 - 10 | 5 | 40 - 50 | 28 |
| 10 -20 | 8 | 50 - 60 | 20 |
| 20 - 30 | 7 | 60 - 70 | 10 |
| 30 - 40 | 12 | 70 - 80 | 10 |
(a) 46.67
(b) 45.2
(c) 42.2
(d) 43.2
(e) None of these
Answer: (a)
Explanation:
From table 40-50 is modal class
\[\therefore \]\[{{x}_{k}}=40,h=10,{{f}_{k}}=28,{{f}_{k-1}}=12,{{d}_{k+1}}=20\]
then by using formula, \[M={{x}_{k}}+\left[ h\times \frac{{{f}_{k}}-{{f}_{k-1}}}{2{{f}_{k}}-{{f}_{k-1}}-{{f}_{k+1}}} \right]\]
\[=40+\left[ 10\times \frac{28-12}{2\times 28\times 12-20} \right]=46.67\]
The arithmetic Mean of the following frequency distribution is 25. Determine the value of P.
| Class Interval | Frequency | Class Interval | Frequency |
| 0 - 10 | 5 | 30 - 40 | P |
| 10 - 20 | 18 | 40 - 50 | 6 |
| 20 - 30 | 15 |
(a) P=16
(b) P=15
(c) P=14
(d) P=16
(e) None of these
Answer: (e)
Explanation:
| Class Interval | Frequency\[({{f}_{i}})\] | Class mark \[({{x}_{i}})\] | \[{{f}_{i}}{{x}_{i}}\] |
| 0-10 | 5 | 5 | 25 |
| 10-20 | 18 | 15 | 270 |
| 20-30 | 15 | 25 | 375 |
| 30-40 | P | 35 | 35P |
| 40-50 | 645 | 270 | |
| \[\sum {{f}_{i}}=(44+P)\] | \[\sum {{f}_{i}}{{x}_{i}}=(940+35P)\] |
\[\therefore \]Mean \[x=\frac{\sum ({{F}_{1}}\times {{x}_{1}})}{\sum {{F}_{1}}}\]\[\Rightarrow \]\[\frac{(940+35\,P)}{(44+P)}=25\]
\[\Rightarrow \]\[(940+35P)=25(44+P)\]
\[\Rightarrow \]\[(35P-25P)=(1100-940)\]
\[\Rightarrow \]\[10\,P=160\] \[\Rightarrow \]\[P=16\]
If the Mean of the following frequency distribution is 54 then find the value of P.
| Class Interval | Frequency | Class Interval | Frequency |
| 0 - 20 | 7 | 60 - 80 | 9 |
| 20 - 40 | P | 80 - 100 | 13 |
| 40 - 60 | 10 |
(a) P=8
(b) P=9
(c) P=10
(d) P=11
(e) None of these
Answer: (d)
Explanation:
| Class Interval | Frequency\[({{f}_{i}})\] | Class mark \[({{x}_{i}})\] | \[{{f}_{i}}{{x}_{i}}\] |
| 0-20 | 7 | 10 | 70 |
| 20-40 | P | 30 | 30P |
| 40-60 | 10 | 50 | 500 |
| 60-80 | 9 | 70 | 630 |
| 80-100 | 13 | 90 | 1170 |
| \[\sum {{f}_{i}}=(39+P)\] | \[\sum {{f}_{i}}{{x}_{i}}=(2370+30P)\] |
\[\therefore \]Mean, \[x=\frac{\sum ({{f}_{i}}\times {{x}_{i}})}{\sum {{f}_{i}}}\]
\[\Rightarrow \]\[\frac{2370+30P}{(39+P)}=54\] \[\Rightarrow \]\[(2370+30P)=54(39+P)\]
\[\Rightarrow \]\[24P=(2370-2106)=264\] \[\Rightarrow \]\[P=1\]
The mean of the following frequency distribution is 62.8 and the sum of all frequencies is 50. Compute the missing frequencies \[{{f}_{1}}\]and\[{{f}_{2}}\].
| Class Interval | 0-20 | 20-40 | 40-60 | 60-80 | 80-100 | 100-120 | Total |
| Frequency | 5 | \[{{f}_{1}}\] | 10 | \[{{f}_{2}}\] | 7 | 8 | 50 |
(a) 8 & 12
(b) 8 & 13
(c) 14 & 12
(d) 15 & 8
(e) None of these
Answer: (a)
Explanation:
\[5+{{f}_{1}}+10+{{F}_{2}}+7+8=50\] \[\Rightarrow \]\[{{F}_{2}}=(20-{{F}_{1}})\]
Now we may prepare the table given below:
| Class Interval | Frequency\[({{f}_{i}})\] | Class mark \[({{x}_{i}})\] | \[{{f}_{i}}{{x}_{i}}\] |
| 0-20 | 5 | 10 | 50 |
| 20-40 | \[{{F}_{1}}\] | 30 | 30\[{{F}_{1}}\] |
| 40-60 | 10 | 50 | 500 |
| 60-80 | \[20-{{F}_{1}}\] | 70 | 1400-70F |
| 80-100 | 7 | 90 | 630 |
| 100-120 | 8 | 110 | 880 |
| \[\sum {{f}_{i}}=50\] | \[\sum {{f}_{i}}{{x}_{i}}=(3460-40{{F}_{1}})\] |
\[\therefore \]Mean, \[x=\frac{\sum ({{f}_{i}}\times {{x}_{i}})}{\sum {{f}_{i}}}=\frac{(3640-40{{f}_{1}})}{50}\]
But mean = 62.8 (given) \[\therefore \]\[\frac{3400-40{{F}_{1}}}{50}=62.8\]\[\Rightarrow \]\[3460-40{{F}_{1}}=3140\]
\[\Rightarrow \]\[40{{F}_{1}}=320\] \[\Rightarrow \]\[{{F}_{1}}=8\]
Thus\[{{F}_{1}}=8\,\And \,{{F}_{2}}=(20-8)=12\]
Find the arithmetic mean of the following frequency distribution:
| Class Interval | 25-29 | 30-34 | 35-39 | 40-44 | 45-49 | 50-54 | 55-59 |
| Frequency | 14 | 22 | 16 | 6 | 5 | 3 | 4 |
(a) 36.46
(b) 36.36
(c) 48.56
(d) 99.95
(e) None of these
Answer: (b)
Explanation:
The given series is in inclusive series. Making it exclusive series, we get
| Class Interval | Frequency\[({{f}_{i}})\] | Class Mark \[({{x}_{i}})\] | \[{{u}_{i}}=\frac{{{x}_{i}}-A}{h}\] \[({{x}_{i}})=\frac{{{x}_{i}}-42}{5}\] | \[{{f}_{i}}{{u}_{i}}\] |
| 24.5-29.5 | 14 | 27 | -3 | -42 |
| 29.5-34.5 | 22 | 32 | -2 | -44 |
| 34.5-39.5 | 16 | 37 | -1 | -16 |
| 39.5-44.5 | 6 | 42=A | 0 | 0 |
| 44.5-49.5 | 5 | 47 | 1 | 5 |
| 49.5-54.5 | 4 | 52 | 2 | 6 |
| 54.5-59.5 | 3 | 57 | 3 | 12 |
| \[\sum {{f}_{i}}=70\] | \[\sum {{f}_{i}}{{u}_{i}}=-70\] |
Thus \[A=42,h=5,\sum {{F}_{i}}\And \sum ({{F}_{i}}\times {{u}_{i}})=79\]
Mean, \[\overline{X}=A+\left[ h\times \frac{\sum ({{F}_{i}}\times {{u}_{i}})}{\sum {{F}_{i}}} \right]=42\left[ 5\times \frac{(-79)}{70} \right]\] \[=(42-5.64)=36.36\]
Find the mean age (in years) from the following frequency distribution:
| Age(in year) | 15-19 | 20-24 | 25-29 | 30-34 | 35-39 | 40-44 | 45-49 | Total |
| Frequency | 3 | 13 | 21 | 15 | 5 | 4 | 2 | 63 |
(a) 29.06
(b) 26.96
(c) 42.96
(d) 95.99
(e) None of these
Answer: (a)
Explanation:
The given series is in inclusive. Making it an exclusive series. We get:
| Class Interval | Frequency\[({{f}_{i}})\] | Class Mark \[({{x}_{i}})\] | \[{{u}_{i}}=\frac{{{x}_{i}}-A}{h}\] \[({{x}_{i}})=\frac{{{x}_{i}}-32}{5}\] | \[{{f}_{i}}{{u}_{i}}\] |
| 14.5-19.5 | 3 | 17 | -3 | -9 |
| 19.5-24.5 | 13 | 22 | -2 | -26 |
| 24.5-29.5 | 21 | 27 | -1 | -21 |
| 29.5-34-5 | 15 | 32=A | 0 | 0 |
| 34.5-39.5 | 5 | 37 | 1 | 5 |
| 39.5-44.5 | 4 | 42 | 2 | 8 |
| 44.5-49.5 | 2 | 47 | 3 | 6 |
| \[\sum {{f}_{i}}=63\] | \[\sum {{f}_{i}}{{u}_{i}}=19-56=-37\] |
\[=-37\]
Thus\[A=32,h={{5}_{1}}\,\sum {{F}_{i}}=63\,\And \,\sum ({{F}_{i}}\times {{u}_{i}})=-37\]
Mean, \[\overline{X}=A+\left[ h\times \frac{\sum ({{F}_{i}}\times {{u}_{1}})}{\sum {{F}_{i}}} \right]\]
\[=32+\left[ 5\times \left( \frac{-37}{63} \right) \right]=(32-2.936)\] \[=32-2.94)=29.06\]
Find the median for the following frequency distribution:
| Height(In cm) | Frequency | Height (In cm) | Frequency |
| 160 - 162 | 15 | 169 - 171 | 118 |
| 163 - 165 | 117 | 171 - 174 | 14 |
| 166 - 168 | 136 |
(a) 168cm
(b) 167cm
(c) 189cm
(d) 198cm
(e) None of these
Answer: (b)
Explanation:
The given series is in inclusive form. Converting it into exclusive form and preparing the cumulative frequency table, we get
| Class Interval | Frequency \[({{f}_{i}})\] | C F |
| 159.5-162.5 | 15 | 15 |
| 162.5-165.5 | 117 | 132 |
| 165.5-168.5 | 136 | 268 |
| 168.5-171.5 | 118 | 368 |
| 171.5-174.5 | 14 | 400 |
| \[N=\sum {{f}_{i}}=400\] |
\[N=400\] \[\Rightarrow \] \[(N/2)=200\]
The cumulative frequency Just greater than 200 in 268 and the corresponding class is 165.5 - 168.5
Thus the median class is 165.5 - 168.5
\[\therefore \]L = 165.5, h = 3 F = 136, C = c.f of preceding class = 132 & (N/2) = 200
Median, \[{{M}_{c}}=l+\left[ n\times \frac{\left( \frac{N}{2}-C \right)}{F} \right]\]
\[=165.5+\left[ 3\times \frac{(200-132)}{136} \right]=165.5+\left( \frac{3\times 68}{136} \right)=\]\[(165.5+1.5)=167\]
The following is the distribution of IQ of 100 students, find the Median IQ.
| IQ | 75-84 | 85-94 | 95-104 | 105-114 | 114-124 |
| Frequency | 8 | 11 | 26 | 31 | 18 |
(a) 100.1
(b) 106.1
(c) 146.1
(d) 149.7
(e) None of these
Answer: (b)
Explanation:
The following series is in inclusive form. Converting it into exclusive form and preparing the cumulative frequency table, we get:
| Class Interval | Frequency \[({{f}_{i}})\] | C F |
| 74.5-84.5 | 8 | 8 |
| 84.5-94.5 | 11 | 19 |
| 94.5-104.5 | 26 | 45 |
| 104.5-114.5 | 31 | 76 |
| 114.5-124.5 | 18 | 94 |
| 124.5-134.5 | 4 | 98 |
| 134.5-144.5 | 2 | 100 |
| \[N=\sum {{f}_{i}}=100\] |
Now N=100 \[\Rightarrow \]\[(N/2)=50\]
The cumulative frequency just greater than 50 is 70 and the corresponding class interval is 104.5 - 114.5.
\[\therefore \]L = 104.5, h = 10 f = 31, C = C.F of preceding class = 45 & (N/2) = 50
Median, \[{{M}_{c}}]=L\left[ h\times \frac{\left( \frac{N}{2}-C \right)}{F} \right]\]
\[=104.5+\left[ 10\times \frac{(50-45)31}{{}} \right]=\left( 104.5+\frac{50}{3} \right)=\] \[(104.5+1.6)=106.1\]
Calculate the median for the following date:
Marks Obtained No. of students
Below 10 6
Below 20 15
Below 30 29
Below 40 41
Below 50 60
Below 60 70
(a) 30
(b) 31
(c) 35
(d) 33
(e) None of these
Answer: (c)
Explanation:
From the given table, we may get back frequency and cumulative frequency as shown below;
| Class Interval | Frequency \[({{f}_{i}})\] | C F |
| 0-10 | 6 | 6 |
| 10-20 | 9 | 15 |
| 20-30 | 14 | 29 |
| 30-40 | 12 | 41 |
| 40-50 | 19 | 60 |
| 50-60 | 10 | 70 |
| \[N=\sum {{f}_{i}}=70\] |
N = 70 \[\Rightarrow \]\[\left( \frac{N}{2} \right)=35\]
The cumulative frequency just greater than 35 is 41 and the corresponding class is 30-40
Thus the median class is 30-40
\[\ell ~=\text{3}0,\text{ h}=\text{1}0,\text{ f}=\text{12},\text{ C}=\text{C}.\text{F}\] of preceding class \[=\text{29 }\And \left( \frac{N}{2} \right)=35\]
Median, \[{{M}_{c}}=\ell +\left[ \frac{\frac{N}{2}-Cf}{f} \right]\times h=30+\frac{35-29}{12}\times 10=35\]
Find the missing frequencies in the following frequency distribution table, if N = 100 and median is 32.
| Marks | 0-10 | 10-20 | 20-30 | 30-40 | 40-50 | 50-60 | Total |
| No. of Students | 10 | ? | 35 | 30 | ? | 10 | 100 |
(a) 12 & 16
(b) 9 & 16
(c) 14 & 9
(d) 15 & 16
(e) None of these
Answer: (b)
Explanation:
Let\[{{F}_{1}}\]&\[{{F}_{2}}\] be the frequencies of class intervals 10-20 & 40-50 respectively.
Then \[10+{{f}_{1}}+25+30+{{f}_{2}}+10=100\]\[\Rightarrow \]\[{{f}_{1}}+{{f}_{2}}=25\]
Median is 32, which is in 30-40, so, the median class is 30-40
\[\therefore \]\[\text{L}=\text{3}0,\text{ h}=\text{1}0\text{ f}=\text{3}0,\text{ N}=\text{1}00\text{ }\And \text{ C}=\text{1}0+{{\text{f}}_{1}}+\]\[\text{25}=({{f}_{1}}+\text{35)}\]
Now median, \[Me=L+\left[ h\times \frac{\left( \frac{N}{2}-C \right)}{F} \right]\]
\[\Rightarrow \]\[30+\frac{(15-{{f}_{1}})}{3}=32\]\[\Rightarrow \]\[(15-{{f}_{1}})=6\] \[\Rightarrow \]\[{{f}_{1}}=9\]
\[\therefore \,\,{{f}_{1}}=9\And {{f}_{2}}=(25-9)=16\]
\[{{f}_{1}}=9\And {{f}_{2}}=16\]
The mode of the following series is 36, find the missing frequency in it.
| Class Interval | 0-10 | 10-20 | 20-30 | 30-40 | 40-50 | 50-60 | 60-70 |
| Frequency | 8 | 10 | .... | 16 | 12 | 6 | 7 |
(a) 9
(b) 10
(c) 12
(d) 15
(e) None of these
Answer: (b)
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