Q: Find the mean of the following data, by using the assumed mean method.Class Interval\$0-10\$\$10-20\$\$20-30\$\$30-40\$\$40-50\$Frequency\$7\$\$8\$\$12\$\$13\$\$10\$

Answer:- Class intervalFrequency (\$f_i\$) \$ x_i = \cfrac{\text{lower limit + upper limit}}{2} \$\$ f_ix_i \$ \$0-10\$\$ 7\$ \$5\$ \$35\$ \$10-20\$ \$8\$ \$15\$ \$120\$ \$20-30\$ \$12\$ \$25\$ \$300\$ \$30-40\$ \$13\$ \$35\$ \$455\$ \$40-50\$ \$10\$\$45\$ \$450\$ \$ \Sigma f_i = 50 \$ \$ \Sigma f_ix_i = 1360 \$ Mean = \$ \cfrac{\Sigma f_ix_i}{\Sigma f_i}= \cfrac{1360}{50} = 27.2 \$

Q: The following frequency distribution gives the monthly consumption of electricity of \$68\$ consumers of a locality. Find the median, mean and mode of the data and compare them.Monthly consumption (in units)Number of consumers\$65-85\$\$4\$\$85-105\$\$5\$\$105-125\$\$13\$\$125-145\$\$20\$\$145-165\$\$14\$\$165-185\$\$8\$\$185-205\$\$4\$

\$C.I\$ \$x_i\$ \$f_i\$ \$f_ix_i\$ \$c.f\$ \$65-85\$ \$75\$ \$4\$ \$300\$ \$4\$ \$85-105\$ \$95\$ \$5\$ \$475\$ \$9\$ \$105-125\$ \$115\$ \$13\$ \$1495\$ \$22\$ \$125-145\$ \$135\$ \$20\$ \$2700\$ \$42\$ \$145-165\$ \$155\$ \$14\$ \$2170\$ \$56\$ \$165-185\$ \$175\$ \$8\$ \$1400\$ \$64\$ \$185-205\$ \$195\$ \$4\$ \$780\$ \$68\$ \$\sum f_i=68\$ \$\sum f_ix_i=9320\$ \$\Rightarrow\$ \$Mean=\dfrac{\sum f_ix_i}{\sum f_i}=\dfrac{9320}{68}=137.05\$Now, \$n=68\$ and \$\dfrac{n}{2}=34\$\$\therefore\$ Median class \$=125-145\$We have, \$l=125,c.f=22,f=20\$ and \$h=20\$\$\Rightarrow\$ \$Median=l+\dfrac{\dfrac{n}{2}-c.f}{f}\times h\$ \$=125+\dfrac{34-22}{20}\times 20\$ \$=125+12\$\$\therefore\$ \$Median=137\$Now, modal class \$=125-145\$Then we have \$l=125,f_0=13,f_1=20,f_2=14\$ and \$h=20\$\$\Rightarrow\$ \$Mode=l+\dfrac{f_1-f_0}{2f_1-f_0-f_2}\times h\$ \$=125+\dfrac{20-13}{2\times 20-13-14}\times 20\$ \$=125+10.77\$\$\therefore\$ \$Mode=135.77\$

Q: Find the median of the following data 12, 17, 3, 14, 5, 8, 7, 15

\$12,17,3,14,5,8,7,15\$No' of observations (n) =\$8\$first of all we arrange the observations in an ascending order\$3,5,7,8,12,14,15,17\$Here observe that No' of observations (n) is even soMedian is average of \$\dfrac{n}{2}\$ th term and \$\left(\dfrac{n}{2}+1\right)\$ th termMedian \$=\dfrac{\dfrac{n}{2}\text{th term}+(\dfrac{n}{2}+1)\text{th term}}{2}\$ \$=\dfrac{\dfrac{8}{2}\text{th term}+(\dfrac{8}{2}+1)\text{th term}}{2}\$ \$=\dfrac{4 \text{th term}+5\text{th term}}{2}\$ \$=\dfrac{8+12}{2}\$ \$=\dfrac{20}{2}\$ Median = \$10\$

Q: If \$x_i\$'s are the mid points of the class intervals of grouped data, \$f_1\$'s are the corresponding frequencies and \$\bar x\$ is the mean, then \$\sum (f_i x_i - \bar x)\$ is equal to

Given that, \$x_i\$ are the mid points of the class intervals.\$f_i\$ are the corresponding frequencies.The mean \$\bar x\$ for the grouped data is \$\bar x=\dfrac{\sum f_ix_i}{n}\$\$\implies \bar xn=\sum f_ix_i\$ ------(1)\$\sum (f_ix_i-\bar x)=\sum f_ix_i-\sum \bar x\$\$\implies \sum (f_ix_i-\bar x)=\bar xn-\sum \bar x\$ (from (1))\$\implies \sum (f_ix_i-\bar x)=\bar xn-\bar xn=0\$

Q: The mean and variance of \$7\$ observation are \$8\$ and \$16\$ respectively. If five of the observation are \$2,4, 10, 12\$ and \$14\$. Find the remaining two observations.

Let the remaining two observation be \$x\$ and \$y\$The observation are \$2, 4, 10, 12, 14, x, y\$Mean \$ =\displaystyle \bar{x}=\frac{2+4+10+12+14+x+y}{7}=8 \$ \$ \displaystyle \Rightarrow 56=42+x+y\$ \$ \displaystyle \Rightarrow x+y=14 \$ ......(1)Variance \$=16= \displaystyle \frac{1}{n}\sum_{i=l}^{7}(X_{i}-\bar{X})^{2} \$ \$ \displaystyle 16=\frac{1}{7}\left [ (-6)^{2}+(-4)^{2}+(2)^{2}+(4)^{2}+(6)^{2}+x^{2}+y^{2}-2 \times 8(x+y)+2\times (8)^{2} \right ] \$ \$ \displaystyle 16=\frac{1}{7}\left [ 36+16+4+6+36+x^{2}+y^{2}-16(14)+2(64) \right ] \$ ...........[using (1)] \$ \displaystyle 16=\frac{1}{7}\left [ 108+x^{2}+y^{2}-224+128 \right ] \$ \$ \displaystyle 16=\frac{1}{7}\left [ 12+x^{2}+y^{2} \right ] \$ \$ \displaystyle \Rightarrow x^{2}+y^{2}=112-12=100 \$ \$ \displaystyle x^{2}+y^{2}=100 \$ ..........(2)From (1), we obtain \$ \displaystyle x^{2}+y^{2}+2xy=196 \$ ....(3)From (2) and (3), we obtain\$2xy = 196 - 100\$\$ \displaystyle \Rightarrow 2xy = 96\$ ...........(4)subtracting (4) from (2), we obtain \$ \displaystyle x^{2}+ y^{2}-2xy=100 - 96 \$ \$ \displaystyle \Rightarrow (x-y)^{2} = 4 \$ \$ \displaystyle \Rightarrow x-y=\pm 2 \$ ............(5)Therefore, from (1) and (5) we obtain \$x = 8\$ and \$y = 6\$ when \$x - y = 2\$\$x = 6\$ and \$y = 8\$ when \$x - y = -2\$Thus the remaining observation are \$6\$ and \$8\$

Q: Calculate the mean for the following frequency distribution:Marks\$10-20\$\$20-30\$\$30-40\$\$40-50\$\$50-60\$\$60-70\$\$70-80\$Frequency\$3\$\$7\$\$10\$\$6\$\$8\$\$2\$\$4\$

Marks\$(I)\$ 10-20 20-30 30-40 40-50 50-60 60-70 70-80 \$x_i\$ 15 25 35 45 55 65 75 \$f_i\$ 3 7 10 6 8 2 4 \$f_ix_i\$ 45 175 350 270 440 130 300 \$\bar x=\cfrac{\sum f_ix_i}{\sum f_i}=\cfrac{1710}{40}=42.75\$

Q: Calculate mean with step Deviation method from the following data.Marks\$0-10\$\$10-20\$\$20-30\$\$30-40\$\$40-50\$No. of student\$4\$\$6\$\$10\$\$20\$\$10\$

Class inreval Mid values \$(x_i)\$ Frequency \$(f_i)\$ \$d_i=x_i-A\$ \$u_i=\frac{d_i}{h}\$ \$F_iu_i\$ 0-10 5 4 -20 -2 -8 10-20 15 6 -10 -1 -6 20-30 25=A 10 0 0 0 30-40 35 20 10 1 20 40-50 45 10 20 2 20 N=50 Total=26 \$h=10,A=25,N=50, \sum f_iu_i=26\$Mean \$(\overline x)=A+h[\frac{1}{N} \sum f_iu_i]\$put the values \$Mean=30.2\$

Question 1

Find the mean of the following data, by using the assumed mean method.Class Interval$0-10$$10-20$$20-30$$30-40$$40-50$Frequency$7$$8$$12$$13$$10$

Accepted Answer

Answer:- Class intervalFrequency ($f_i$) $ x_i = \cfrac{\text{lower limit + upper limit}}{2} $$ f_ix_i $   $0-10$$ 7$ $5$ $35$ $10-20$ $8$ $15$ $120$ $20-30$ $12$ $25$ $300$ $30-40$ $13$ $35$ $455$ $40-50$ $10$$45$  $450$  $ \Sigma f_i = 50 $ $ \Sigma f_ix_i = 1360 $ Mean = $ \cfrac{\Sigma f_ix_i}{\Sigma f_i}= \cfrac{1360}{50} = 27.2 $

Question 2

The following frequency distribution gives the monthly consumption of electricity of $68$ consumers of a locality. Find the median, mean and mode of the data and compare them.Monthly consumption (in units)Number of consumers$65-85$$4$$85-105$$5$$105-125$$13$$125-145$$20$$145-165$$14$$165-185$$8$$185-205$$4$

Accepted Answer

$C.I$ $x_i$ $f_i$ $f_ix_i$ $c.f$ $65-85$ $75$ $4$ $300$ $4$ $85-105$ $95$ $5$ $475$ $9$ $105-125$ $115$ $13$ $1495$ $22$ $125-145$ $135$ $20$ $2700$ $42$ $145-165$ $155$ $14$ $2170$ $56$ $165-185$ $175$ $8$ $1400$ $64$ $185-205$ $195$ $4$ $780$ $68$   $\sum f_i=68$ $\sum f_ix_i=9320$  $\Rightarrow$  $Mean=\dfrac{\sum f_ix_i}{\sum f_i}=\dfrac{9320}{68}=137.05$Now, $n=68$ and $\dfrac{n}{2}=34$$\therefore$  Median class $=125-145$We have, $l=125,c.f=22,f=20$ and $h=20$$\Rightarrow$  $Median=l+\dfrac{\dfrac{n}{2}-c.f}{f}\times h$                     $=125+\dfrac{34-22}{20}\times 20$                     $=125+12$$\therefore$  $Median=137$Now, modal class $=125-145$Then we have $l=125,f_0=13,f_1=20,f_2=14$ and $h=20$$\Rightarrow$  $Mode=l+\dfrac{f_1-f_0}{2f_1-f_0-f_2}\times h$                 $=125+\dfrac{20-13}{2\times 20-13-14}\times 20$                 $=125+10.77$$\therefore$  $Mode=135.77$

Question 3

Find the mean marks of students from the following cumulative frequency distribution:MarksNumber of studentso and above8010 and above7720 and above7230 and above6540 and above5550 and above4360 and above2870 and above1680 and above1090 and above8100 and above 0

Accepted Answer

Here we have, the cumulative frequency distribution. So, first we convert it into an ordinary frequency distribution. we observe that are 80 students getting marks greater than or equal to 0 and 77 students have secured 10 and more marks. Therefore, the number of students getting marks between 0 and 10 is 80-77= 3.Similarly, the number of students getting marks between 10 and 20 is 77-72= 5 and so on. Thus, we obtain the following frequency distribution.Marks Number of students0-10310-20520-30730-401040-501250-601560-70 1270-80680-90290-1008Now, we compute mean arithmetic mean by taking 55 as the assumed mean.Computative of Mean Marks$(x_i)$Mid-value $(f_i)$Frequency $u_i\frac{x_i-55}{10}$$f_iu_i$0-1053-5-1510-20155-4-2020-30257-3-2130-403510-2-2040-504512-1-2050-6055150060-70651211270-8075621280-908523690-100958432Total                  $\sum f_i=80$$\sum f_iu_i$= -26 We have,N= $sum f_i= 80, \sum f_iu_i= - 26$, A= 55 and h= 10 $\therefore \overline {X} = A + h[\frac{1}{N} \sum f_iu_i]$$\Rightarrow \overline {X} = A+ h[ \frac{1}{N} \sum f_iu_i]$$\Rightarrow \overline {X} = 55+ 10 \times \frac{-26}{80}= 55- 3.25 = 51. 75 $Marks.

Question 4

Find the median of the following data 12, 17, 3, 14, 5, 8, 7, 15

Accepted Answer

$12,17,3,14,5,8,7,15$No' of observations (n) =$8$first of all we arrange the observations in an ascending order$3,5,7,8,12,14,15,17$Here observe that No' of observations (n) is even soMedian is average of $\dfrac{n}{2}$ th term and $\left(\dfrac{n}{2}+1\right)$ th termMedian $=\dfrac{\dfrac{n}{2}\text{th term}+(\dfrac{n}{2}+1)\text{th term}}{2}$              $=\dfrac{\dfrac{8}{2}\text{th term}+(\dfrac{8}{2}+1)\text{th term}}{2}$               $=\dfrac{4 \text{th term}+5\text{th term}}{2}$               $=\dfrac{8+12}{2}$               $=\dfrac{20}{2}$               Median = $10$

Question 5

If $x_i$'s are the mid points of the class intervals of grouped data, $f_1$'s are the corresponding frequencies and $\bar x$ is the mean, then $\sum (f_i x_i - \bar x)$ is equal to

Accepted Answer

Given that, $x_i$ are the mid points of the class intervals.$f_i$ are the corresponding frequencies.The mean $\bar x$ for the grouped data is $\bar x=\dfrac{\sum f_ix_i}{n}$$\implies \bar xn=\sum f_ix_i$ ------(1)$\sum (f_ix_i-\bar x)=\sum f_ix_i-\sum \bar x$$\implies \sum (f_ix_i-\bar x)=\bar xn-\sum \bar x$ (from (1))$\implies \sum (f_ix_i-\bar x)=\bar xn-\bar xn=0$

Question 6

The mean and variance of $7$ observation are $8$ and $16$ respectively. If five of the observation are $2,4, 10, 12$ and $14$. Find the remaining two observations.

Accepted Answer

Let the remaining two observation be $x$ and $y$The observation are $2, 4, 10, 12, 14, x, y$Mean  $ =\displaystyle \bar{x}=\frac{2+4+10+12+14+x+y}{7}=8 $  $ \displaystyle \Rightarrow 56=42+x+y$  $ \displaystyle \Rightarrow x+y=14 $        ......(1)Variance $=16=   \displaystyle \frac{1}{n}\sum_{i=l}^{7}(X_{i}-\bar{X})^{2} $ &#10; $ \displaystyle 16=\frac{1}{7}\left [ &#10;(-6)^{2}+(-4)^{2}+(2)^{2}+(4)^{2}+(6)^{2}+x^{2}+y^{2}-2 \times &#10;8(x+y)+2\times (8)^{2} \right ] $  $ \displaystyle 16=\frac{1}{7}\left [ 36+16+4+6+36+x^{2}+y^{2}-16(14)+2(64) \right ] $   ...........[using (1)]  $ \displaystyle 16=\frac{1}{7}\left [ 108+x^{2}+y^{2}-224+128 \right ] $  $ \displaystyle 16=\frac{1}{7}\left [ 12+x^{2}+y^{2} \right ] $  $ \displaystyle \Rightarrow x^{2}+y^{2}=112-12=100  $  $ \displaystyle x^{2}+y^{2}=100  $       ..........(2)From (1), we obtain   $ \displaystyle x^{2}+y^{2}+2xy=196  $         ....(3)From (2) and (3), we obtain$2xy = 196 - 100$$ \displaystyle \Rightarrow  2xy = 96$ ...........(4)subtracting (4) from (2), we obtain   $ \displaystyle x^{2}+ y^{2}-2xy=100 - 96  $  $ \displaystyle \Rightarrow (x-y)^{2} = 4  $  $ \displaystyle \Rightarrow x-y=\pm 2  $   ............(5)Therefore, from (1) and (5) we obtain $x = 8$ and $y = 6$ when $x - y = 2$$x = 6$ and $y = 8$ when $x - y = -2$Thus the remaining observation are $6$ and $8$

Question 7

Calculate the mean for the following frequency distribution:Marks$10-20$$20-30$$30-40$$40-50$$50-60$$60-70$$70-80$Frequency$3$$7$$10$$6$$8$$2$$4$

Accepted Answer

&#10; &#10;  &#10;  Marks$(I)$&#10;  &#10;  &#10;  10-20&#10;  &#10;  &#10;  20-30&#10;  &#10;  &#10;  30-40&#10;  &#10;  &#10;  40-50&#10;  &#10;  &#10;  50-60&#10;  &#10;  &#10;  60-70&#10;  &#10;  &#10;  70-80&#10;  &#10; &#10; &#10;  &#10;  $x_i$&#10;  &#10;  &#10;  15&#10;  &#10;  &#10;  25&#10;  &#10;  &#10;  35&#10;  &#10;  &#10;  45&#10;  &#10;  &#10;  55&#10;  &#10;  &#10;  65&#10;  &#10;  &#10;  75&#10;  &#10; &#10; &#10;  &#10;  $f_i$&#10;  &#10;  &#10;  3&#10;  &#10;  &#10;  7&#10;  &#10;  &#10;  10&#10;  &#10;  &#10;  6&#10;  &#10;  &#10;  8&#10;  &#10;  &#10;  2&#10;  &#10;  &#10;  4&#10;  &#10; &#10; &#10;  &#10;  $f_ix_i$&#10;  &#10;  &#10;  45&#10;  &#10;  &#10;  175&#10;  &#10;  &#10;  350&#10;  &#10;  &#10;  270&#10;  &#10;  &#10;  440&#10;  &#10;  &#10;  130&#10;  &#10;  &#10;  300&#10;  &#10; &#10;&#10;&#10; $\bar x=\cfrac{\sum f_ix_i}{\sum f_i}=\cfrac{1710}{40}=42.75$

Question 8

Calculate mean with step Deviation method from the following data.Marks$0-10$$10-20$$20-30$$30-40$$40-50$No. of student$4$$6$$10$$20$$10$

Accepted Answer

&#10; &#10;  &#10;  Class inreval &#10;  &#10;  &#10;  Mid values $(x_i)$&#10;  &#10;  &#10;  Frequency $(f_i)$&#10;  &#10;  &#10;  $d_i=x_i-A$&#10;  &#10;  &#10;  $u_i=\frac{d_i}{h}$&#10;  &#10;  &#10;  $F_iu_i$&#10;  &#10; &#10; &#10;  &#10;   &#10;  &#10;  &#10;   &#10;  &#10;  &#10;   &#10;  &#10;  &#10;   &#10;  &#10;  &#10;   &#10;  &#10;  &#10;   &#10;  &#10; &#10; &#10;  &#10;  0-10&#10;  &#10;  &#10;  5&#10;  &#10;  &#10;  4&#10;  &#10;  &#10;  -20&#10;  &#10;  &#10;  -2&#10;  &#10;  &#10;  -8&#10;  &#10; &#10; &#10;  &#10;  10-20&#10;  &#10;  &#10;  15&#10;  &#10;  &#10;  6&#10;  &#10;  &#10;  -10&#10;  &#10;  &#10;  -1&#10;  &#10;  &#10;  -6&#10;  &#10; &#10; &#10;  &#10;  20-30&#10;  &#10;  &#10;  25=A&#10;  &#10;  &#10;  10&#10;  &#10;  &#10;  0&#10;  &#10;  &#10;  0&#10;  &#10;  &#10;  0&#10;  &#10; &#10; &#10;  &#10;  30-40&#10;  &#10;  &#10;  35&#10;  &#10;  &#10;  20&#10;  &#10;  &#10;  10&#10;  &#10;  &#10;  1&#10;  &#10;  &#10;  20&#10;  &#10; &#10; &#10;  &#10;  40-50&#10;  &#10;  &#10;  45&#10;  &#10;  &#10;  10&#10;  &#10;  &#10;  20&#10;  &#10;  &#10;  2&#10;  &#10;  &#10;  20&#10;  &#10; &#10; &#10;  &#10;   &#10;  &#10;  &#10;   &#10;  &#10;  &#10;  N=50&#10;  &#10;  &#10;   &#10;  &#10;  &#10;   &#10;  &#10;  &#10;  Total=26&#10;  &#10; &#10;$h=10,A=25,N=50, \sum f_iu_i=26$Mean $(\overline x)=A+h[\frac{1}{N} \sum f_iu_i]$put the values $Mean=30.2$

Question 9

Marks0-10 10-20 20-30 30-40 40-50 50-60 60-70 70- 8080-90 90-100  Frequency3 5 6 7 8  9 10 12 64 Find the mean mark using step deviation method:

Accepted Answer

XMid Point (X) Frequency (F) i=class interval width A=45 Assumed mean u=$\dfrac{x-A}{i}$ fu  0-105 3 10 45 -4 -12 10-20 15 5 10 45 -3 -1520-30 25 6 10 45 -2 -12 30-40 35 7 10 45 -1 -7 40-50 45=A8 10 45 0 0 50-60 55 9 10 45 1 9 60-70 65 10 10 45 2 20 70-80 75 12 10 45 3 36 80-90 85 6 10 45 4 24 90-100  954 10 45 5 20   $\Sigma f=70$    $\Sigma fu=63$ The formula used for arithmetic mean of grouped data by step deviation method is, $\overline {X} =A + \dfrac{\sum fu}{\sum f} \times i$ $A =$ Assumed mean of the given data$\sum$ = Summation of the frequencies given in the grouped data$\sum fu$ = Summation of the frequencies and deviation of a given mean data$u = \dfrac{(x - A)}{i}$$i =$ Class interval width$\overline {X}$ = arithmetic mean$\overline {X} =45 +\dfrac{63}{70}\times 10$$= 45 + 9$$= 54$

Question 10

The mean of the following frequency distribution is $25.2$. Find the missing frequency $x$$C.I.$$0-10$$10-20$$20-30$$30-40$$40-50$frequency$8$$X$$10$$11$$9$

Accepted Answer

Solution -  C.I0-10 10-20 20-30 30-40 40-50  Frequency8 X 10 11 9.  Midvalue5 15 25 35 45 Mean = 25.2$ \frac{8\times 5+15x+10\times 25+11\times 35+9\times 45}{8+x+10+11+9} = 25.2$$ 1080+15x = 957.6+25.2x$$ x= \frac{122.4}{10.2}$$ x = 12.$