Statistics - Test Papers

 CBSE Test Paper 01

Chapter 14 Statistics

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1. Mode+23(Mean−Mode)$Mode+\frac{2}{3}\left(Mean-Mode\right)$=. (1)

   1. Mode
   2. Median
   3. Mean
   4. None of these

2. Construction of cumulative frequency table is useful to determine (1)

   1. mean
   2. all the three
   3. median
   4. mode

3. For the following distribution

   Class	Below 10	Below 20	Below 30	Below 40	Below 50	Below 60
   Frequency	3	12	27	57	75	80

   the modal class is (1)

   1. 50 - 60
   2. 40 - 50
   3. 20 - 30
   4. 30 - 40

4. The mean of the first 10 natural numbers is (1)

   1. 4.5
   2. 5
   3. 6
   4. 5.5

5. The marks obtained by 9 students in Mathematics are 59, 46, 30, 23, 27, 44, 52, 40 and 29. The median of the data is (1)

   1. 35
   2. 29
   3. 30
   4. 40

6. Find the mode of the given data 3, 3, 7, 4, 5, 3, 5, 6, 8, 9, 5, 3, 5, 3, 6, 9, 7, 4. (1)

7. If the median of a series exceeds the mean by 3, find by what number the mode exceeds its mean? (1)

8. If the values of mean and median are 26.4 and 27.2, what will be the value of mode? (1)

9. In the following frequency distribution, find the median class. (1)

   Height (in cm)	140 -145	145-150	150-155	155 -160	160 -165	165 -170
   Frequency	5	15	25	30	15	10

10. Find median of the data, using an empirical relation when it is given that Mode = 12.4 and Mean = 10.5. (1)

11. Find the mode of the following distribution. (2)

    Class interval	0-10	10-20	20-30	30-40	40-50	50-60	60-70	70-80
    Frequency	5	8	7	12	28	20	10	10

12. Convert the following data into 'more than type' distribution: (2)

    Class	50 - 55	55 - 60	60 - 65	65 - 70	70 - 75	75 - 80
    Frequency	2	8	12	24	38	16

13. Calculate the mean of the following data, using direct method: (2)

    Class	25 - 35	35 - 45	45 - 55	55 - 65	65 - 75
    Frequency	6	10	8	12	4

14. If the median of the following frequency distribution is 46, find the missing frequencies. (3)

    Variable	10-20	20-30	30-40	40-50	50-60	60-70	70-80	Total
    Frequency	12	30	?	65	?	25	18	229

15. Find median for the following data: (3)

    Wages(in Rs)	Number of workers
    More than 150	Nil
    More than 140	12
    More than 130	27
    More than 120	60
    More than 110	105
    More than 100	124
    More than 90	141
    More than 80	150

16. Draw a pie-chart for the following data of expenditure on various items in a family.

    Item	Education	Food	Rent	Clothing	Others
    Expenditure (in Rs.)	1600	3200	4000	2400	3200

    3

17. Find the mean and mode of the following frequency distribution: (3)

    Classes	0 - 10	10 -20	20 -30	30 -40	40 -50	50 -60	60 -70
    Frequency	3	8	10	15	7	4	3

18. From the following frequency distribution, prepare the 'more than' ogive. (4)

    Score	Number of candidates
    400 - 450	20
    450 - 500	35
    500 - 550	40
    550 - 600	32
    600 - 650	24
    650 - 700	27
    700 - 750	18
    750 - 800	34
    Total	230

    Also, find the median.

19. Find the mean marks of students from the following cumulative frequency distribution: (4)

    Marks	Number of students
    0 and above	80
    10 and above	77
    20 and above	72
    30 and above	65
    40 and above	55
    50 and above	43
    60 and above	28
    70 and above	16
    80 and above	10
    90 and above	8
    100 and above	0

20. The marks obtained by 100 students of a class in an examination are given below:

    Marks	Number of students
    0 - 5	2
    5 - 10	5
    10 - 15	6
    15 - 20	8
    20 - 25	10
    25 - 30	25
    30 - 35	20
    35 - 40	18
    40 - 45	4
    45 - 50	2

    Draw cumulative frequency curves by using (i) 'less than' series and (ii) 'more than' series.
    Hence, find the median. (4)

CBSE Test Paper 01
Chapter 14 Statistics

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Solution

1. 2. Median
      Explanation: Since, 3 Median = Mode + 2 Mean
      ⇒$\Rightarrow$ Median = Mode3+23Mean$\frac{Mode}{3}+\frac{2}{3}Mean$
      ⇒$\Rightarrow$ Median = Mode3+23Mean−23Mode+23Mode$\frac{Mode}{3}+\frac{2}{3}Mean-\frac{2}{3}Mode+\frac{2}{3}Mode$
      ⇒$\Rightarrow$ Median = Mode+23[Mean−Mode]$Mode+\frac{2}{3}[Mean-Mode]$
2. 3. median
      Explanation: A cumulative frequency distribution is the sum of the class and all classes below it in a frequency distribution. Construction of cumulative frequency table is useful to determine Median.
3. 4. 30 – 40
      Explanation: According to the question,

      Class	0 – 10	10 – 20	20 – 30	30 – 40	40 – 50	50 – 60
      Freq	3	9	15	30	18	5

      Here Maximum frequency is 30.
      Therefore, the modal class is 30 – 40.
4. 4. 5.5
      Explanation: The first 10 natural numbers are 1, 2, 3, …………, 10
      ∴$\therefore$ Mean = Sum of first 10 natural numbers10$\frac{\text{Sum of first 10 natural numbers}}{10}$
      = 1+2+3+.......+1010$\frac{1+2+3+.......+10}{10}$
      = 5510$\frac{55}{10}$ = 5.5
5. 4. 40
      Explanation: Arranging the given data in ascending order: 23, 27, 29, 30, 40, 44, 46, 52, 59
      Here n = 9, which is even.
      ∴$\therefore$ Median = (n+12)th${\left(\frac{n+1}{2}\right)}^{th}$
      = (9+12)th term${\left(\frac{9+1}{2}\right)}^{th}\text{ term}$
      = 5th term = 40

6. Value x	3	4	5	6	7	8	9
   Frequency f	5	2	4	2	2	1	2

   We observe that the value 3 has the maximum frequency i.e 5 .
   The mode of data is 3.

7. Given,
   Median=Mean+3$\mathrm{M}\mathrm{e}\mathrm{d}\mathrm{i}\mathrm{a}\mathrm{n}=\mathrm{M}\mathrm{e}\mathrm{a}\mathrm{n}+3$
   Since, Mode=3Median−2Mean$\mathrm{M}\mathrm{o}\mathrm{d}\mathrm{e}=3\mathrm{M}\mathrm{e}\mathrm{d}\mathrm{i}\mathrm{a}\mathrm{n}-2\mathrm{M}\mathrm{e}\mathrm{a}\mathrm{n}$
   =3(Mean+3)−2Mean$=3(\mathrm{M}\mathrm{e}\mathrm{a}\mathrm{n}+3)-2\mathrm{M}\mathrm{e}\mathrm{a}\mathrm{n}$
   =3Mean+9−2mean$=3Mean+9-2mean$
   ⇒$\Rightarrow$ Mode = Mean + 9
   Hence Mode exceeds Mean by 9.

8. We know that
   Mode = 3 median -2 mean
   = 3(27.2) - 2(26.4)
   = 81.6 - 52.8 = 28.8
   Mode = 28.8

9. Height	Frequency	c.f.
   140−145$140-145$	5$5$	5$5$
   145−150$145-150$	15$15$	5+15=20$5+15=20$
   150−155$150-155$	25$25$	25+20=45$25+20=45$
   155−160$155-160$	30$30$	45+30=75$45+30=75$
   160−165$160-165$	15$15$	75+15=90$75+15=90$
   165−170$165-170$	10$10$	90+10=100$90+10=100$
   	∑$\sum$f = 100

   N=100$N=100$
   ⇒$\Rightarrow$N2$\frac{N}{2}$th term = 1002$\frac{100}{2}$ = 50th term
   Hence, Median class is 155 - 160.

10. Mode = 3 median - 2 mean
    Mode = 12.4$12.4$ and mean = 10.5$10.5$
    Median = 13$\frac{1}{3}$Mode + 23$\frac{2}{3}$Mean
    = 13$\frac{1}{3}$(12.4) + 23$\frac{2}{3}$(10.5)
    =12.43+213$=\frac{12.4}{3}+\frac{21}{3}$
    =12.4+213$=\frac{12.4+21}{3}$
    =33.43$=\frac{33.4}{3}$
    =11.13$=11.13$
    So, median is 11.13.

11. Class interval	0-10	10-20	20-30	30-40	40-50	50-60	60-70	70-80
    Frequency	5	8	7	12	28	20	10	10

    Here the maximum frequency is 28 then the corresponding class 40 - 52 is the modal class
    l = 40, h = 50 - 40 = 10, f = 28, f1 = 12, f2 = 20
    Mode =l+f−f12f−f1−f2×h$=l+\frac{f-f_1}{2f-f_1-f_2}\times h$
    =40+28−122×28−12−20×10$=40+\frac{28-12}{2\times 28-12-20}\times 10$
    =40+16024$=40+\frac{160}{24}$
    = 40 + 6.67
    = 46.67

12. Class	Frequency	Cumulative Frequency
    More than 50	2	98 + 2 = 100
    More than 55	8	90 + 8 = 98
    More than 60	12	78 + 12 = 90
    More than 65	24	54 + 24 = 78
    More than 70	38	16 + 38 = 54
    More than 75	16	16

13. Class Interval	Frequency	Class mark xi$x_i$	fixi$f_i{x}_i$
    25 - 35	6	30	180
    35 - 45	10	40	400
    45 - 55	8	50	400
    55 - 65	12	60	720
    65 - 75	4	70	280
    	Σfi=40$\mathrm{\Sigma }{f}_i=40$		Σ(fixi)=1980$\mathrm{\Sigma }\left(f_i{x}_i\right)=1980$

    from table ,
    Σfi=40$\mathrm{\Sigma }{f}_i=40$ , Σ(fixi)=1980$\mathrm{\Sigma }\left(f_i{x}_i\right)=1980$
    we know that,
    mean = ΣfixiΣfi$\frac{\mathrm{\Sigma }{f}_i{x}_i}{\mathrm{\Sigma }{f}_i}$
    =198040$=\frac{1980}{40}$
    =49.5$=49.5$

14. Let the frequency of the class 30 - 40 be f1 and that of the class 50 - 60 be f2. The total frequency is 229.
    ⇒$\Rightarrow$ 12 + 30 + f1 + 65 + f2 + 25 + 18 =229
    ⇒$\Rightarrow$ f1 + f2 =79
    It is given that the median is 46
    Clearly, 46 lies in the class 40 - 50. So, 40 - 50 is the median class.
    ∴$\therefore$ l=40,h=10,f=65$l=40,h=10,f=65$ and
    F=12+30+f1$F=12+30+f_1$
    =42+f1,$=42+f_1,$
    N=229
    Median=l+N2−Ff×h$Median=l+\frac{\frac{N}{2}-F}{f}\times h$
    46=40+2292−(42+f1)65×10$46=40+\frac{\frac{229}{2}-\left(42+f_1\right)}{65}\times 10$
    46=40+145−2f113$46=40+\frac{145-2f_1}{13}$
    6=145−2f113⇒2f1=67⇒f1=33.5 or 34( say )$6=\frac{145-2f_1}{13}\Rightarrow 2f_1=67\Rightarrow f_1=33.5\text{ or }34(\text{ say })$
    Since f1+f2=79$f_1+f_2=79$ ,
    f2=79−34$f_2=79-34$
    =45
    Hence, f1 = 34 and f2 =45

15. C.I.	f	c.f.
    80 - 90	9	9
    90 - 100	17	26
    100 - 110	19	45
    110 - 120	45	90
    120 - 130	33	123
    130 - 140	15	138
    140 - 150	12	150

    n=150$n=150$⇒n2=75$\Rightarrow \frac{n}{2}=75$
    Median Class =110−120$=110-120$
    l=110,f=45,c.f.=45,h=10$l=110,f=45,c.f.=45,h=10$
    we know that, Median = l+n2−cff×h$l+\frac{\frac{n}{2}-cf}{f}\times h$
    =110+75−4545×10$=110+\frac{75-45}{45}\times 10$
    =116.67$=116.67$

16. Item	Expenditure (Ei)	Central angle = [Ei14400×360]$\left[\frac{E_i}{14400}\times 360\right]$
    Education	1600	[160014400×360]=40∘$\left[\frac{1600}{14400}\times 360\right]={40}^{\circ }$
    Food	3200	[320014400×360]=80∘$\left[\frac{3200}{14400}\times 360\right]={80}^{\circ }$
    Rent	4000	[400014400×360]=100∘$\left[\frac{4000}{14400}\times 360\right]={100}^{\circ }$
    Clothing	2400	[240014400×360]=60∘$\left[\frac{2400}{14400}\times 360\right]={60}^{\circ }$
    Others	3200	[320014400×360]=80∘$\left[\frac{3200}{14400}\times 360\right]={80}^{\circ }$

    

17. Class interval	xi$x_i$	fi$f_i$	fixi$f_i{x}_i$
    0−10$0-10$	5	3	15$15$
    10−20$10-20$	15	8	120$120$
    20−30$20-30$	25	10	250$250$
    30−40$30-40$	35	15	525$525$
    40−50$40-50$	45	7	315$315$
    50−60$50-60$	55	4	220$220$
    60−70$60-70$	65	3	195$195$
    		Σfi=50$\mathrm{\Sigma }{f}_i=50$	Σfixi=1640$\mathrm{\Sigma }{f}_i{x}_i=1640$

    Mean = ∑fixi∑fi=164050$\frac{\sum f_i{x}_i}{\sum f_i}=\frac{1640}{50}$
    Mean = 32.8
    For Mode, Modal class = 30 - 40
    and l=30,f1=15,f2=7,f0=10,h=10$l=30,f_1=15,f_2=7,f_0=10,h=10$
    Mode = l+f1−f02f1−f0−f2×h$l+\frac{f_1-f_0}{2f_1-f_0-f_2}\times h$
    =30+15−102(15)−10−7×10$=30+\frac{15-10}{2(15)-10-7}\times 10$
    =30+15−1030−10−7×10$=30+\frac{15-10}{30-10-7}\times 10$
    =30+530−17×10$=30+\frac{5}{30-17}\times 10$
    = 30+513×10$30+\frac{5}{13}\times 10$
    = 30+5013$30+\frac{50}{13}$
    = 30+3.85$30+3.85$
    = 33.85$33.85$
    Mean of given data is 32.8 and mode is 33.85.
18. More than series:

    Score	Number of candidates
    More than 400	230
    More than 450	210
    More than 500	175
    More than 550	135
    More than 600	103
    More than 650	79
    More than 700	52
    More than 750	34

    plot the points  (400, 230), (450, 210), (500, 175), (550, 135), (600, 103), (650, 79), (700, 52), (750, 34).
    
    N=230$N=230$ ⇒N2=115$\Rightarrow \frac{N}{2}=115$
    Take a point A(0, 115) on the y-axis and draw AP||x-axis meeting the curve at P, Draw PM ⊥$\mathrm{\perp }$x-axis intersecting x-axis at M
    OM=590$OM=590$
    Hence, median =590$=590$
19. Here we have, the cumulative frequency distribution.
    So, first we convert it into an ordinary frequency distribution.
    We observe that there 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.

    Marks	Mid-value (xi)	Frequency (fi)	ui=xi−5510$u_i=\frac{x_i-55}{10}$	fiui
    0-10	5	3	-5	-15
    10-20	15	5	-4	-20
    20-30	25	7	-3	-21
    30-40	35	10	-2	-20
    40-50	45	12	-1	-12
    50-60	55	15	0	0
    60-70	65	12	1	12
    70-80	75	6	2	12
    80-90	85	2	3	6
    90-100	95	8	4	32
    Total		Σfi=80$\mathrm{\Sigma }{f}_i=80$		Σfiui=−26$\mathrm{\Sigma }{f}_i{u}_i=-26$

    Let assumed mean (a) = 55.
    We have,
    N=Σfi=80,Σfiui=−26,$N=\mathrm{\Sigma }{f}_i=80,\mathrm{\Sigma }{f}_i{u}_i=-26,$ a = 55 and h = 10
    ∴X¯¯¯¯=a+hΣfiuiN$\therefore \overline{X}=a+h\frac{\mathrm{\Sigma }{f}_i{u}_i}{N}$
    ⇒X¯¯¯¯=55+10×−2680$\Rightarrow \overline{X}=55+10\times \frac{-26}{80}$
    =55−3.25=51.75$=55-3.25=51.75$
    Therefore, the mean number of marks is 51.75$51.75$
20. 1. Less than series:

       Marks	Number of students
       Less than 5	2
       Less than 10	7
       Less than 15	13
       Less than 20	21
       Less than 25	31
       Less than 30	56
       Less than 35	76
       Less than 40	94
       Less than 45	98
       Less than 50	100

       Plot the points (5, 2), (10, 7), (15, 13), (20, 21), (25, 31), (30, 56), (35, 76), (40, 94), (45, 98) and (50, 100).
       Join these points free hand to get the "less than" cumulative curve.
    2. 'more than' series:

       Marks	Number of students
       More than 45	2
       More than 40	6
       More than 35	24
       More than 30	44
       More than 25	69
       More than 20	79
       More than 15	87
       More than 10	93
       More than 5	98
       More than 0	100

       Now, on the same graph paper as above, we plot the point (0, 100), (5, 98), (10, 93), (15, 87), (20, 79), (25, 69), (30, 44), (35, 24) , (40, 6) and (45, 2)
       
       N = 100⇒N2=50$\Rightarrow \frac{N}{2}=50$
       Two curves intersect at Point P(28, 50)
       Hence, median = 28