# 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

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

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

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

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

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

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

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

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

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

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