9th Class Mathematics Statistics Mode

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, \[M={{x}_{k}}+\left[ h\times \frac{{{f}_{k}}-{{f}_{k-1}}}{2{{f}_{k}}-{{f}_{k-1}}-{{f}_{k+1}}} \right]\]

\[=40+\left[ 10\times \frac{28-12}{2\times 28\times 12-20} \right]=46.67\]    

 

 

 

  • 17 is also the only prime number which is the sum of 4 consecutive prime 2,3,5 and 7
  • 17 is the lowest number that can be written as \[{{x}^{3}}+{{y}^{2}}\] in 2 distinct ways: \[17={{2}^{3}}+{{3}^{2}}\] and \[17={{1}^{3}}+{{4}^{2}}\]
  • One of the earliest numerical approximation of \[\sqrt{2}\] was found on a Babylonian clay tablet approximately between 1800 B.C. and 1600 B.C.
  • Among all shapes with the same perimeter a circle has the largest area.  

 

 

 

  • Mean \[\overline{X}=\frac{{{x}_{1}}+{{x}_{2}}+....+{{x}_{n}}}{n}=\frac{\sum {{x}_{i}}}{n}\]
  • Mean \[\overline{X}=A+\frac{\sum fidi}{n}\],where symbols have their usual meaning
  • Media \[=l+\left[ h\times \frac{\frac{N}{2}-C}{f} \right]\], where symbols have their usual meaning
  • Mode \[={{x}_{k}}+h\left[ \frac{{{f}_{k}}-{{f}_{k-1}}}{2{{f}_{k}}-{{f}_{k-1}}-{{f}_{k+1}}} \right]\], where symbols have their usual mining  

 

 

 

 

  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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