Download App

Maths 4 Marks Questions

Statistics · Gujarat Board (GSEB) STD 10 (GSEB - English Medium). 68 questions.

Questions · Page 2 of 2

4 Marks Questions

Question 514 Marks
Find the mean of each of the following frequency distributions:
Class interval $0-8$ $8-16$ $16-24$ $24-32$ $32-40$
Frequency $5$ $6$ $4$ $3$ $2$
Answer
Let the assumed mean be $A = 20$ and $h = 8.$
Class interval: Mid value$(x_i):$ Frequency$(f_i):$ $d_i = x_i - A = x_{i }- 20$ $\text{u}_\text{i}=\frac{1}{\text{h}}(\text{d}_\text{i})$
$=\frac{1}{8}(\text{d}_\text{i})$		 $f_iu_i$
$0-8$ $4$ $5$ $-16$ $-2$ $-10$
$8-16$ $12$ $6$ $-8$ $-1$ $-6$
$16-24$ $20$ $4$ $0$ $0$ $0$
$24-32$ $28$ $3$ $8$ $1$ $3$
$32-40$ $36$ $2$ $16$ $2$ $4$
    $\sum\text{f}_\text{i}=20$     $\sum\text{f}_\text{i}\text{u}_\text{i}=-9$
We know that mean, $\overline{\text{X}}=\text{A}+\text{h}\Big(\frac{1}{\text{N}}\sum\text{f}_\text{i}\text{u}_\text{i}\Big)$
Now, we have $\text{N}= \sum\text{f}_\text{i}=20,\sum\text{f}_\text{i}\text{u}_\text{i}=-9,\text{h}=8$ and $A = 20$
Putting the values in the above formula, we get
$\overline{\text{X}}=\text{A}+\text{h}\Big(\frac{1}{\text{N}}\sum\text{f}_\text{i}\text{u}_\text{i}\Big)$
$=20+8\Big(\frac{1}{20}\times(-9)\Big)$
$=20-\frac{72}{20}$
$=20-3.6$
$=16.4$
Hence, the mean is $16.4.$
View full question & answer→
Question 524 Marks
If median of the following data is $50.$ Find the values of $p$ and $q,$ if the sum of all the frequencies is $90.$
Marks: $20-30$ $30-40$ $40-50$ $50-60$ $60-70$ $70-80$ $80-90$
Frequency: $p$ $15$ $25$ $20$ $q$ $8$ $10$
Answer
The given series is in inclusive form. Converting it to exclusive form and preparing the cumulative frequency table, we have
Class interval Frequency $(f_i)$ Cumulative Frequency $(c.f.)$
$20-30$ $p$ $p$
$30-40$ $15$ $p + 15$
$40-50$ $25$ $p + 40$
$50-60$ $20$ $p + 60$
$60-70$ $q$ $p + q + 60$
$70-80$ $8$ $p + q + 68$
$80-90$ $10$ $p + q + 78$
  $78 + p + q = 90$  
Median $= 50$
It lies in the inrerval $50 - 60,$ so the median class is $50 - 60$. now, we have
$\text{l}=50, \text{h}=10, \text{f}=20, \text{F}= \text{p}+40,\text{N}=90$
We know that
$\text{Median}=\text{l}\ +\bigg\{\frac{\frac{\text{N}}{2}-\text{F}}{\text{f}}\bigg\}\times\text{h}$
$50=50+\frac{45-(\text{p}+40)}{20}\times10$
$\Rightarrow 0=\frac{5-\text{p}}{2}$
$\Rightarrow\text{p}=5$
And,
$\text{p}+\text{q}+78=90$
$\Rightarrow\text{p}+\text{q}=12$
$\Rightarrow \text{q}=12-5=7$
View full question & answer→
Question 534 Marks
The distribution below gives the weight of $30$ students in a class. Find the median weight of students:
Weight $(in \ kg)$ $40-45$ $45-50$ $50-55$ $55-60$ $60-65$ $65-70$ $70-75$
No. of students $2$ $3$ $8$ $6$ $6$ $3$ $2$
Answer
We prepare the cumulative frequency table, as given below.
weight $( in \ kg)$ No. of students: $(f_i)$ cumulative frequency $(e.f.)$
$40-45$ $2$ $2$
$45-50$ $3$ $5$
$50-55$ $8$ $13$
$55-60$ $6$ $19$
$60-65$ $6$ $25$
$65-70$ $3$ $28$
$70-75$ $2$ $30$
  $N = 30$  
We have, $N = 30$
So, $\frac{\text{N}}{2}=15$
Now, the cumulative frequency just greater than $15$ is $19$ and the
corresponding class is $55-65$
Therefore, $55-60$ is the median class.
Here, $\text{l}=55,\text{f}=6,\text{F}=13$ and $\text{h}=5$
We know that
$\text{Median}=\text{l}+\bigg\{\frac{\frac{\text{N}}{2}-\text{F}}{\text{f}}\bigg\}\times\text{h}$
$=55+\Big\{ \frac{15-13}{6}\Big\}\times5$
$=55+\frac{2\times5}{6}$
$=55+1.667$
$=56.667$
Hence, the median weight of students is $56.67\ kg.$
View full question & answer→
Question 544 Marks
Find the missing frequencies and the median for the following distribution if the mean is $1.46.$
No. of accidents $0$ $1$ $2$ $3$ $4$ $5$ Total
Frequency $($No. of days$)$ $46$ $?$ $?$ $25$ $10$ $5$ $200$
Answer
Mean $= 1.46, N = 200$
Let $p_1$ and $p_{2 }$ be the missing frequencies
No. of accidents $(x)$ Frequency No. of days $(f)$ $c.f.$ $f$
$0$ $46$ $46$ $0$
$1$ $p_1$ $46 + p_1$ $p_1$
$2$ $p_2$ $46 + p_1 + p_2$ $2p_2$
$3$ $25$ $71 + p_{1 }+ p_2$ $75$
$4$ $10$ $81 + p_1 + p_2$ $40$
$5$ $5$ $86 + p_1 + p_2$ $25$
Total $200$   $140 + p_1 + p_2$
$\therefore86+\text{p}_1+\text{p}=200$
$\Rightarrow \text{p}_1+\text{p}_2=200-86=114$
$\therefore \text{p}_1=114-\text{p}_2\ ......(1)$
Now, $\text{Mean}=\frac{\sum\text{f}\ \text{x}}{\sum\text{f}}=\frac{140+\text{p}_1+2\text{p}_2}{200}$
$\Rightarrow1.46=\frac{140+\text{p}_1+2\text{p}_2}{200}$
$\Rightarrow292=140+\text{p}_1+2\text{p}_2$
$\Rightarrow\text{p}_1+2\text{p}_2=292-140=152$
$114-\text{p}_2+2\text{p}_2=152$
$\Rightarrow\text{p}_2=152-114=38$
$\therefore \text{p}_1=114- \text{p}_2=114-38=76$
Hence missing frequencies are $76$ and $38$
$\text{Median}=\text{Here}\frac{\text{N}}{2}=\frac{200}{2}=100$
$\therefore cf$ of $2^{nd}$ class is $46 + 76 = 122$
View full question & answer→
Question 554 Marks
Candidate of four schools appear in a mathematics test. The data were as follows:
Schools No. of Candidates Average Score
i 60 75
ii 48 80
iii Not available 55
iv 40 50
If the average score of the candidates of all the four schools is 66, find the number of candidates that appeared from school III.
Answer
Let unknown quantity = p and mean = 66
Schools No. of Candidates (f) Average Score (x) Total Score (fx)
i 60 75 4500
ii 48 80 3840
iii p 55 55p
iv 40 50 2000
Total 148 + p   10340 + 55p
$\therefore$ Average $=\frac{\sum\text{fx}}{\sum\text{f}}$
$\Rightarrow66=\frac{10340+55\text{P}}{148+\text{P}}$
$\Rightarrow66(148+\text{p})=10340+55\text{p}$
$\Rightarrow9768+66\text{p}=10340+55\text{p}$
$\Rightarrow66\text{p}-55\text{p}=10340-9768$
$\Rightarrow11\text{p}=572$
$\Rightarrow\text{p}=\frac{572}{11}=52$
$\therefore$ Number of candidates for school III = 52
View full question & answer→
Question 564 Marks
The following distribution gives the state$-$wise teacher$-$student ratio in higher secondary schools of India. Find the mode and mean of this data. Interpret, the two measures:
Number of students per Teacher Number of States $U.T.$
$15-20$ $3$
$20-25$ $8$
$25-30$ $9$
$30-35$ $10$
$35-40$ $3$
$40-45$ $0$
$45-50$ $0$
$50-55$ $2$
Answer
Let assumed mean $(A) = 32.5$
Number of students per Teacher Number of States Class Mark $(f)$ $d_1 = x - A A = 32.5$ $f_1d_1$
$15-20$ $3$ $17.5$ $-15$ $-45$
$20-25$ $8$ $22.5$ $-10$ $-80$
$25-30$ $9$ $27.5$ $-5$ $-45$
$30-35$ $10$ $32.5 - A$ $0$ $0$
$35-40$ $3$ $37.5$ $5$ $15$
$40-45$ $0$ $42.5$ $15$ $0$
$45-50$ $0$ $47.5$ $15$ $0$
$50-55$ $2$ $52.5$ $20$ $40$
Total $35$     $-115$
We see that the class $30-35$ has the maximum frequency
$\therefore$ It is the modal class
Here $I = 30, f = 10, f_1 = 9, f_2 = 3, h = 5$
$\therefore\text{Mode}=\text{l}+\frac{\text{f}-\text{f}_1}{2\text{f}-\text{f}_1-\text{f}_2}\times\text{h}$
$=30+\frac{10-9}{2\times10-9-3}\times5=30+\frac{1}{20-12}\times5$
$=30+\frac{5}{8}=30+0.625=30.625$
$\text{Mean}=\text{A}+\frac{\sum\text{f}_\text{i}\text{d}_\text{i}}{\sum\text{f}_\text{i}}$
$=32.5-\frac{115}{35}=32.5-\frac{23}{7}$
$= 32.5 - 3.28 = 29.22 = 29.2$
View full question & answer→
Question 574 Marks
The following table gives the literacy rate $($in percentage$)$ of $35$ cities. Find the mean literacy rate.
Literacy rate $(in \ \%)$ $45-55$ $55-65$ $65-75$ $75-85$ $85-95$
Number of cities $3$ $10$ $11$ $8$ $3$
Answer
We may find class marks by using the relation
$\text{x}_\text{i}=\frac{\text{Upper class limit+Lower class limit}}{2}$
Class size $(h)$ for this data $= 10$
Now taking $70$ as assumed mean $(a)$ wrong Calculate $d_i, u_i,$ and $f_i, u_{i }$ as follows,
Library rate $(in \ r_i)$ Number of cities $(f_i)$ $x_i$ $d_i = x_i - 70$ $\text{x}_\text{i}=\frac{\text{d}_\text{i}}{10}$ $f_iu_i$
$45-55$ $3$ $50$ $-20$ $-2$ $-6$
$55-65$ $10$ $60$ $-10$ $-1$ $-10$
$65-75$ $11$ $70$ $0$ $0$ $0$
$75-85$ $8$ $80$ $10$ $1$ $8$
$85-95$ $3$ $90$ $20$ $2$ $6$
Total $35$       $-2$
Now we may observe that
$\sum\text{f}_\text{i}=35$
$\sum\text{f}_\text{i}\text{u}_\text{i}=-2$
$\text{Mean}\ \bar{\text{(x)}}=\text{a}+\Big(\frac{{{\sum\text{f}_\text{i}\text{u}_\text{i}}}}{{\sum\text{u}_\text{i}}}\Big)\times\text{h}$
$=70+\Big(\frac{-2}{35}\Big)10$
$=70\frac{-4}{7}$
$=70-0.57=69.43$
So, mean literacy rate is $69.437.$
View full question & answer→
Question 584 Marks
Calculate the median from the following data:
Marks below
10
20
30
40
50
60
70
80
No. of students
15
35
60
84
96
127
198
250
Answer
Marks below
No. of students
Class interval
Frequency
Cumulative frequency
10
15
0-10
15
15
20
35
10-20
20
35
30
60
20-30
25
84
40
84
30-40
24
84
50
96
40-50
12
96(f)
60
127
50-60
37(f)
127
70
198
60-70
52
198
80
250
70-80
52
250
 
 
 
N = 250
  
We have N = 250
$\frac{\text{N}}{2}=\frac{250}{2}=125$
The cumulative frequency just greater than $\frac{\text{N}}{2}$ is 127 then mediam class is 50-60 such that
$\text{l}=50,\text{f}=31,\text{F}=96,\text{h}=60-50=10$
$\text{Mediam}= \text{L}+\frac{\frac{\text{N}}{2}-\text{F}}{\text{f}}\times\text{h}$
$=50+\frac{125-96}{31}\times10$
$=50+\frac{29\times10}{31}$
$=\frac{155+290}{31}$
$=\frac{445}{31}$
$=59.35$
View full question & answer→
Question 594 Marks
The following is the distribution of height of students of a certain class in a certain city:
Height (in cms)
160-162
163-165
166-168
16-171
172-174
No. of students
15
118
142
127
18
Find the median height.
Answer
Class interval (inclusive)
Class interval (inclusive)
Class interval Frequency
Cumulative frequency
160-162
159.2-162.5
15
15
163-164
162.5-165.5
118
133 (f)
166-168
165.5-168.5
142 (f)
275
169-171
168.5-168.5
127
402
172-174
171.5-174.5
18
420
 
 
N = 420
  
We have
N = 420
$\frac{\text{N}}{2}=\frac{420}{2}=210$
The cumulativefrequency just greater than $\frac{\text{N}}{2}$ is 275 then 165.5-168.5 is the
median class such, that
$\text{l}=165.5,\text{f}=142,\text{F}=133$ and $\text{h}=168.5-105.5=3$
$\text{Mean}= \text{l}+\frac{\frac{\text{N}}{2}-\text{F}}{\text{f}}\times\text{h}$
$=165.5+\frac{210-133}{142}\times3$
$=165.5+1.626$
$=167.126$
View full question & answer→
Question 604 Marks
Find the missing frequencies in the following distribution, if the sum of the frequencies is $120$ and the mean is $50.$
Class: $0-20$ $20-40$ $40-60$ $60-80$ $80-100$
Frequency: $17$ $f_1$ $32$ $f_2$ $19$
Answer
it is given that mean of the data is $50.$
Let the assumed mean $A = 50$ and $h = 10.$
Marks Mid$-$value$(x_i)$ Frequency $\text{u}_\text{i}=\frac{\text{x}_\text{i}-50}{20}$ $f_iu_i$
$0-20$ $10$ $17$ $-2$ $-34$
$20-40$ $30$ $f_1$ $-1$ $-f1$
$40-60$ $50$ $32$ $0$ $0$
$60-80$ $70$ $f_2$ $1$ $f_2$
$80-100$ $90$ $19$ $2$ $38$
    $N = 68 + f_1 +f_2$   $\sum\text{f}_\text{i}\text{u}_\text{i}=4-\text{f}_1+\text{f}_2$
We know that mean, $\bar{\text{x}}=\text{a}+\text{h}\Big(\frac{1}{\text{N}}\sum\text{f}_\text{i}\text{u}_\text{i}\Big)$
Now, we have $\text{N}=68+\text{f}_1+ \text{f}_2, \ \text{h}=20,\text{A=50,}$
$\sum\text{f}_\text{i}\text{u}_\text{i}=4-\text{f}_1+\text{f}_2$
$50=50+20\Big(\frac{4-\text{f}_1+\text{f}_2}{68+\text{f}_1+\text{f}_2}\Big)$
$\Rightarrow \text{f}_1-\text{f}_2=4$
$\Rightarrow \text{f}_1=4+\text{f}_2\ \dots(\text{i})$
Now, $\text{N}=68+\text{f}_1+\text{f}_2=120$
$\text{f}_1+\text{f}_2=120-68=52$
$\Rightarrow4+\text{f}_2+\text{f}_2=52$ $\big[\text{Using(i)}\big]$
$\Rightarrow\text{f}_2=24$
So, $\text{f}_1=4+24=28$
View full question & answer→
Question 614 Marks
The weekly observation on cost of living index in a certain city for the year $2004 – 2005$ are given below. Compute the weekly cost of living index.
Cost of living Index Number of Students Cost of living Index Number of Students
$1400-1500$ $5$ $1700-1800$ $9$
$1500-1600$ $10$ $1800-1900$ $6$
$1600-1700$ $20$ $1900-2000$ $2$
Answer
Let the assume mean $(A) = 1650$
Class interval Mid$-$value $x_i$ $d_i = x_{i }= - A =x_i - 1650$ $\text{u}_\text{i}=\frac{\text{x}_\text{i}-15}{6}$ Frequency $f_{i }$ $f_iu_i$
$1400-1500$ $1450$ $-200$ $-2$ $5$ $-10$
$1500-1600$ $1550$ $-100$ $-1$ $10$ $-10$
$1600-1700$ $1650$ $0$ $0$ $20$ $0$
$1700-1800$ $1750$ $100$ $1$ $9$ $9$
$1800-1900$ $1850$ $200$ $2$ $6$ $12$
$1900-2000$ $1950$ $300$ $3$ $2$ $6$
        $N = 52$ $\sum\text{f}_\text{i}\text{u}_\text{i}=7$
We have
$A = 1650, h = 100$
$\text{Mean}=\text{A}+\text{h}\times\frac{\sum\text{f}_\text{i}\text{u}_\text{i}}{\text{N}}$
$= 1650+100\times\frac{7}{52}$
$=1650+\frac{700}{52}$
$=1650+13.46$
$=1663.46$
View full question & answer→
Question 624 Marks
The following tables gives the distribution of total household expenditure $($in rupees$)$ of manual workers in a city.
Expenditure $($in rupees$) (x_i)$ Frequency$(f_i)$ Expenditure $($in ruppes$)(x_i)$ Frequency$(f_i)$
$100-150$ $24$ $300-350$ $30$
$150-200$ $40$ $350-400$ $22$
$200-250$ $33$ $400-450$ $16$
$250-300$ $28$ $450-500$ $7$
Find the average expenditure $($in rupees$)$ per household.
Answer
Let the assumed mean $(A) = 275.$
Class interval Mid value $(x_i)$ $d_i = x_{i }- 275$ $\text{u}_\text{i}=\frac{\text{x}_\text{i}-275}{50}$ Frequency $f_i$ $f_iu_i$
$100-150$ $125$ $-150$ $-3$ $24$ $-72$
$150-200$ $175$ $-100$ $-2$ $40$ $-80$
$200-250$ $225$ $-50$ $-1$ $33$ $-33$
$250-300$ $275$ $0$ $0$ $28$ $0$
$300-350$ $325$ $50$ $1$ $30$ $30$
$350-400$ $375$ $100$ $2$ $22$ $44$
$400-450$ $425$ $150$ $3$ $16$ $48$
$450-500$ $475$ $200$ $4$ $7$ $28$
        $N = 200$ $\sum\text{f}_\text{i}\text{u}_\text{i}=-35$
We have
$\text{A}=275,\text{h}=50$
$\text{Mean}=\text{A}+\text{h}\times \frac{\sum\text{f}_1\text{u}_1}{\text{N}}$
$= 275+50\times\frac{-35}{200}$
$=275-8.75$
$=266.25$
View full question & answer→
Question 634 Marks
If the mean of the following data is 20.6. Find the value of p.
x 10 15 p 25 35
y 3 10 25 7 5
Answer
Mean = 20.6
x f fx
10 3 30
15 10 150
p 25 25p
25 7 175
35 5 175
Total 50 $\sum\text{fx}=530 + 25\text{p}$
$\bar{\text{X}}=20.6$
Mean $\bar{\text{X}}=\frac{\sum\text{fx}}{\sum\text{f}}=\frac{530+25p}{50}$
$\Rightarrow\frac{530+25\text{p}}{50}=20.6$
$\Rightarrow530+25\text{p}=20.6\times50$
$\Rightarrow530+25\text{p}=1030$
$\Rightarrow25\text{p}=1030-530=500$
$\therefore\text{p}=\frac{500}{25}=20$
View full question & answer→
Question 644 Marks
Find the missing frequency (p) for the following distribution whose mean is 7.68.
x 3 5 7 9 11 13
f 6 8 15 P 8 4
Answer
x f fx
3 6 18
5 8 40
7 15 105
9 P 9P
11 8 88
13 4 52
  N = P + 41 $\sum\text{fx}=9\text{P}+303$
Given 
$\Rightarrow\text{Mean}=7.68$
$\Rightarrow\frac{\sum\text{fx}}{\text{N}}=68$
$\Rightarrow\frac{9\text{P}+303}{\text{P}+41}=7.68$
$\Rightarrow9\text{P}+303=\text{P}(7.68)+314.88$
$\Rightarrow9\text{P}-7.68\text{P}=314.88-303$
$\Rightarrow1.32\text{P}=11.88$
$\Rightarrow\text{P}=\frac{11.88}{1.32}$
$\Rightarrow\text{P}=9.$
View full question & answer→
Question 654 Marks
A survey regarding the height (in cm) of 51 girls of class X of a school was conducted and the following data was obtained.
Hight in cm
Number of girls
Less than 140
4
Less than 145
11
Less than 150
29
Less than 155
40
Less than 160
46
Less than 165
51
Find the median height.
Answer
Height (in cm)
No. of girls (c.f.)
F
135-140
4
4
140-145
11
7
145-150
29
18
150-155
40
11
155-160
46
6
160-165
51
5
 
 
51
Here $\frac{\text{N}}{2}=\frac{51}{2}$ = 25.5 or 26 which lies in the class 145-150
$\text{I}=145,\text{F}=11,\text{f}=18,\text{h}=5$
$\text{Median}=\text{l}+\frac{\frac{\text{N}}{2}-\text{F}}{\text{f}}\times\text{h}$
$=145+\frac{25.5-11}{18} \times5$
$=145+\frac{14.5}{18} \times5=145+\frac{72.5}{18}$
$=145+4.03=149.03$
View full question & answer→
Question 664 Marks
The following table gives the frequency distribution of married women by age at marriage:
Age (in years)
Frequency
15-19
53
20-24
140
25-29
98
30-34
32
35-39
12
40-44
9
45-49
5
50-54
3
55-59
3
60 and above
2
Calculate the median and interpret the results.
Answer
Writing classes in exclusive form,
Age (in years)
Class intervals in exclusive from
Frequency (f)
e.f.
15-19
14.5-19.5
53
53
20-24
19.5-24.5
140
193
25-29
24.5-29.5
98
291
30-34
29.5-34.5
32
323
35-39
34.5-39.5
12
335
40-44
39.5-44.5
9
344
45-49
44.5-49.5
5
349
50-54
49.5-54.5
3
352
55-59
54.5-59.5
3
355
60 and above
59.5-above
2
357
Here N = 357
$\therefore \frac{\text{N}}{2}=\frac{357}{2}=178.5=179 $ which lies in the class 20-24
$\therefore \text{l}=20,\text{F}=53,\text{f}=140,\text{h}=5$
$\therefore \text{Median}=\text{l}+\frac{\frac{\text{N}}{2}-\text{F}}{\text{f}}\times\text{h}=20+\frac{179-53}{140}\times5$
$=20+\frac{126}{140}\times5=20+\frac{126}{28}$
$=20+4.5=24.5$
View full question & answer→
Question 674 Marks
The table below shows the daily expenditure on food of $25$ households in a locality:
Daily expenditure $( in \  Rs.)$ $100-150$ $150-200$ $200-250$ $250-300$ $300-350$
Number of households $4$ $5$ $12$ $2$ $3$
Answer
We may calcute class marks $(x_i)$ for each interval by using the relatio
$\text{x}_\text{i}=\frac{\text{Upper class limit+lover class limit}}{2}$
Class size $= 50$
Now, taking $225$ as assumed mean can we may calculated $d_i, u_i, f_i, u_i$ as follows
Daily expenditure $(in \ Rs.)$ $f_i$ $x_i$ $d_i = x_i - 225$ $\text{u}_\text{i}=\frac{\text{x}_\text{i}-225}{\text{h}}$ $f_iu_i$
$100-150$ $4$ $125$ $-100$ $-2$ $-8$
$150-200$ $5$ $15$ $-50$ $-1$ $-5$
$200-250$ $12$ $255$ $0$ $0$ $0$
$250-300$ $2$ $275$ $50$ $1$ $2$
$300-350$ $2$ $325$ $100$ $2$ $4$
          $-7$
Now we may observe that
$\sum\text{f}_\text{i}=25$
$\sum\text{f}_\text{i}\text{x}_\text{i}=-7$
Mean $\bar{\text{(x)}}=\text{a}+\Big(\frac{{\sum\text{f}_\text{i}\text{u}_\text{i}}}{\sum\text{f}}\Big)\times\text{h}$
$=225+\Big(\frac{-7}{25}\Big)\times50$
$=225-14=211$
So, mean daily expenditure on food is $Rs. 211$
View full question & answer→
Question 684 Marks
An incomplete distribution is given as follows:
Variable $0-10$ $10-20$ $20-30$ $30-40$ $40-50$ $50-60$ $60-70$
Frequency $10$ $20$ $?$ $40$ $?$ $25$ $15$
You are given that the median value is $35$ and the sum of all the frequencies is $170.$ Using the median formula, fill up the missing frequencies.
Answer
$\text{Median}=25$ and $\sum\text{f}=\text{N}=170$
Let $p_1$ and $p_2$ be two missing frequencies
Variable Frequency $c.f.$
$0-10$ $10$ $10$
$10-20$ $20$ $30$
$20-30$ $p_1$ $30 + p_1$
$30-40$ $40$ $70 + p_1$
$40-50$ $p_2$ $70 + p_1 + p_2$
$50-60$ $25$ $95 + p_1 + p_2$
$60-70$ $15$ $110 + p_1 + p_2$
$\therefore110+\text{p}_1+\text{p}_2=170$
$\Rightarrow\text{p}_1+\text{p}_2=170-110=60$
Here $\text{N}=170,\frac{\text{N}}{2}=\frac{170}{2}=85$
$\therefore$ Median $= 35$ which lies in the class $30 - 40$
Here $\text{I}=30,\text{f}=40,\text{F}=30+\text{p},$ and $\text{h}=10$
$\text{Median}=\text{l}+\frac{\frac{\text{N}}{2}-\text{F}}{\text{f}}\times\text{h}$
$\Rightarrow35=30+\frac{85-(30+\text{p}_1)}{40}\times10$
$\Rightarrow35-30=\frac{85-30-\text{p}_1}{4}$
$\Rightarrow5=\frac{85-30-\text{p}_1}{4}$
$20=55-\text{p}_1$
$\Rightarrow\text{p}_1=55-20=35$
But $\text{p}_1+\text{p}_2=60 $
$\therefore\text{ p}_2=60-\text{p}_1=60-35=25 $
Hence missing frequencies are $35$ and $25$
View full question & answer→
Generate Paper FreeAll Statistics topics