Q1: Find the median of the following data by making a â€˜less than giveâ€™

Marks | 0-10 | 10-20 | 20-30 | 30-40 | 40-50 | 50-60 | 60-70 | 70-80 | 80-90 | 90-100 |

No. of students | 5 | 3 | 4 | 3 | 3 | 4 | 7 | 9 | 7 | 8 |

Sol:

The frequency distribution table of less than type is given as follows:

Maks (upper class limits) | Cumulative frequency (cf) |

Less than 10 | 5 |

Less than 20 | 5+3=8 |

Less than 30 | 8+4=12 |

Less than 40 | 12+3=15 |

Less than 50 | 15+3 = 18 |

Less than 60 | 18+4 = 22 |

Less than 70 | 22+7 =29 |

Less than 80 | 29+9 =38 |

Less than 90 | 38+7 =45 |

Less than 100 | 45+8 =53 |

Taking upper class limits of class intervals on x-axis and their respective frequencies on y-axis, its ogive can be drawn as follows:

Here, N=53,

\(\Rightarrow \frac{N}{2}= 26.5\)

Mark the point A whose ordinate is 26.5 and its x-coordinate is 66.4.

The median of the data is 66.4.

Q2: The given distribution shows the number of wickets taken by the bowlers in one-day international cricket matches:

No. of wickets | Less than 15 | Less than 30 | Less than 45 | Less than 60 | Less than 75 | Less than 90 | Less than 105 | Less than 120 |

No. of bowlers | 2 | 5 | 9 | 17 | 39 | 54 | 70 | 80 |

Sol:

Taking upper class limits of class intervals on x-axis and their respective frequencies on y-axis, its ogive can be drawn as follows:

Here, N=80,

\(\Rightarrow \frac{N}{2}= 40 \)

Mark the point A whose ordinate is 40 and its x-coordinate is 76.

The median of the data is 76.

Q3: Draw a â€˜more thanâ€™ ogive for the data given below which gives the marks of 100 students.

Marks | 0-10 | 10-20 | 20-30 | 30-40 | 40-50 | 50-60 | 60-70 | 70-80 |

No. of students | 4 | 6 | 10 | 10 | 25 | 22 | 18 | 5 |

Sol:

Taking upper class limits of class intervals on x-axis and their respective frequencies on y-axis, its ogive can be drawn as follows:

Marks (lower class limits) | Cumulative frequency (cf) |

More than 0 | 96+4 = 100 |

More than 10 | 90+6 = 96 |

More than 20 | 80+10=90 |

More than 30 | 70+10=80 |

More than 40 | 45+25=70 |

More than 50 | 23+22=45 |

More than 60 | 18+5=23 |

More than 70 | 5 |

Taking the lower class limits on x-axis and their respective cumulative frequencies on y-axis, its ogive can be obtained as follows:

Q4: The height of 50 girls of class X of a school are recorded as follows:

Height ( in cm) | 135-140 | 140-145 | 145-150 | 150-155 | 155-160 | 160-165 |

No. of girls | 5 | 8 | 9 | 12 | 14 | 2 |

Draw a â€˜more than typeâ€™ ogive for the above data.

Sol:

The frequency distribution table of more than type is as follows:

Height (in cm) (lower class limits) | Cumulative frequency (cf) |

More than 135 | 45+5=50 |

More than 140 | 37+8=45 |

More than 145 | 28+9=37 |

More than 150 | 16+12=28 |

More than 155 | 14+2=16 |

More than 160 | 2 |

Taking the lower class limits on x-axis and their respective cumulative frequencies on y-axis, its ogive can be obtained as follows:

Q5: The monthly consumption of electricity ( in units) of some families of a locality is given in the following frequency distribution:

Monthly consumption (in units) | 140-160 | 160-180 | 180-200 | 200-220 | 220-240 | 240-260 | 260-280 |

No. of families | 3 | 8 | 15 | 40 | 50 | 30 | 10 |

Sol:

The frequency distribution table of more than type is as follows:

Height (in cm) (lower class limits) | Cumulative frequency (cf) |

More than 140 | 3+153=156 |

More than 160 | 8+145=153 |

More than 180 | 15+130=145 |

More than 200 | 40+90=130 |

More than 220 | 50+40=90 |

More than 240 | 30+10=40 |

More than 260 | 10 |

Taking the lower class limits on x-axis and their respective cumulative frequencies on y-axis, its ogive can be obtained as follows:

Q6: The following table gives the production yield per hectare of wheat of 100 farms of a village.

Production yield (kg/ha) | 50-55 | 55-60 | 60-65 | 65-70 | 70-75 | 75-80 |

No. of farms | 2 | 8 | 12 | 24 | 38 | 16 |

Change the distribution to â€˜a more than typeâ€™ distribution and draw its ogive. Using ogive, find the median of the given data.

Sol:

The frequency distribution table of more than type is as follows:

Production yield (kg/ha)(lower class limits) | Cumulative frequency (cf) |

More than 50 | 2+98=100 |

More than 55 | 8+90=98 |

More than 60 | 12+78=90 |

More than 65 | 24+54=78 |

More than 70 | 38+16=54 |

More than 75 | 16 |

Here, N=100,

\(\Rightarrow \frac{N}{2}= 50 \)

Mark the point A whose ordinate is 50 and its x-coordinate is 70.5.

The median of the data is 70.5

Q7: The table below shows the weekly expenditures on food of some household in a locality:

Weekly introduction (in Rs.) | No. of household |

100-200 | 5 |

200-300 | 6 |

300-400 | 11 |

400-500 | 13 |

500-600 | 5 |

600-700 | 4 |

700-800 | 3 |

800-900 | 2 |

Draw a â€˜less than type ogiveâ€™ and a â€˜more than type ogiveâ€™ for this distribution.

Sol:

The frequency distribution table of less than type is as follows:

Weekly expenditure ( in Rs)
(Upper class limits) |
Cumulative frequency (cf) |

Less than 200 | 5 |

Less than 300 | 5+6=11 |

Less than 400 | 11+11=22 |

Less than 500 | 22+13=35 |

Less than 600 | 35+5=40 |

Less than 700 | 40+4=44 |

Less than 800 | 44+3=47 |

Less than 900 | 47+2=49 |

Now,

The frequency distribution table of more than type is as follows:

Weekly expenditure (in Rs) ( lower class limits) | Cumulative frequencies (cf) |

More than 100 | 44+5=49 |

More than 200 | 38+6=44 |

More than 300 | 27+11=38 |

More than 400 | 14+13=27 |

More than 500 | 9+5=14 |

More than 600 | 5+4=9 |

More than 700 | 2+3=5 |

More than 800 | 2 |

Q8: From the following frequency distribution , prepare the â€˜more than ogiveâ€™.

Score | No. 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.

Sol:

From the given table, we may prepare the â€˜more thanâ€™ frequency table as shown below:

Score | Number of candidates |

More than 750 | 34 |

More than 700 | 52 |

More than 650 | 79 |

More than 600 | 103 |

More than 550 | 135 |

More than 500 | 175 |

More than 450 | 210 |

More than 400 | 230 |

We plot the points A(750,34), B(700,50), C(650,79), D(600,103), E(550,135), F(500,175), G(450,210) and H(400,230).

Join AB,BC,CD,DE,EF,FG,GH and HA with a free hand to get the curve representing the â€˜more than typesâ€™ series.

Here N= 230

\(\Rightarrow \frac{N}{2}= 115\)

From P(0,115), draw PQ meeting the curve at Q.

Draw QM meeting at M.

Clearly, QM = 590 units

Hence, median = 590 units

Q9: The marks obtained by 100 students of a class in an examination are given below:

Marks | No. 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) â€˜Less thanâ€™ series and

(ii) â€˜More thanâ€™ series

Hence find the median.

Sol:

(i) From the given table, we may prepare the â€˜less thanâ€™ frequency table as shown below:

Marks | No. 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 |

We plot the points A(5,2), B(10,7), C(15,13), D(20,21), E(25,31), F(30,56), G(35,76), H(40,94), I(45,98), and J(50,100).

Join AB,BC,CD,DE,EF,FG,GH,HI,IJ and JA with a free hand to get the curve representing the â€˜less than typesâ€™ series.

(ii) More than series:

Marks | No. of students |

More than 0 | 100 |

More than 5 | 98 |

More than 10 | 93 |

More than 15 | 87 |

More than 20 | 79 |

More than 25 | 69 |

More than 30 | 44 |

More than 35 | 24 |

More than 40 | 6 |

More than 45 | 2 |

Now on the same graph paper, we plot the points (0,100), (5,98), (10,94), (15,76), (20,56), (25,31), (30,21), (35,13), (40,6) and (45,2).

Join, with a free hand to get the â€˜more than typeâ€™ series.

The two curves intersect at point L. Draw \(LM \perp OX\)

Hence, Median = 29.5

Q10: From the following data, draw two types of cumulative frequency curves and determine the median

Height ( in cm) | Frequency |

140-144 | 3 |

144-148 | 9 |

148-152 | 24 |

152-156 | 31 |

156-160 | 42 |

160-164 | 64 |

164-168 | 75 |

168-172 | 82 |

172-176 | 86 |

176-180 | 34 |

Sol:

(i) Less than series:

Marks | No. of students |

Less than 144 | 3 |

Less than 148 | 12 |

Less than 152 | 36 |

Less than 156 | 67 |

Less than 160 | 109 |

Less than 164 | 173 |

Less than 168 | 248 |

Less than 172 | 330 |

Less than 176 | 416 |

Less than 180 | 450 |

We plot the points A(144,3), B(148,12), C(152,36), D(156,67), E(160,109), F(164,173), G(168,248), H(172,330), I(176,416), and J(180,450).

Join AB,BC,CD,DE,EF,FG,GH,HI,IJ and JA with a free hand to get the curve representing the â€˜less than typesâ€™ series.

(ii) More than series:

Marks | No. of students |

More than 140 | 450 |

More than 144 | 447 |

More than 148 | 438 |

More than 152 | 414 |

More than 156 | 383 |

More than 160 | 341 |

More than 164 | 277 |

More than 168 | 202 |

More than 172 | 120 |

More than 176 | 34 |

Now on the same graph paper, we plot the points Â \(A_{1}(140,4510) ,B_{1}(144,447), C_{1}(148,447), D_{1}(152,414), E_{1}(156,383), F_{1}(160,341),G_{1}(164,277), H_{1}(168,202),I_{1}(172,120) and J_{1}(176,34)\)

Join, \(A_{1}B_{1}, B_{1}C_{1}, C_{1}D_{1},D_{1}E_{1},E_{1}F_{1},F_{1}G_{1},G_{1}H_{1},H_{1}I_{1},I_{1}J_{1}\)

The two curves intersect at point L. Draw \(LM \perp OX\)

Hence, Median = 166 cm.