During the medical check-up of 35 students of a class, their weights were recorded as follows. Draw a less than type of ogive for the given data: Weight in kg Number of students Less than 38 0 Less than 40 3 Less than 42 5 Less than 44 9 Less than 46 14 Less than 48 28 Less than 50 32 Less than 52 35

Less than type cumulative frequency distribution is, Weight (in kg ) Cumulative frequency f i Points 36 - 38 0 ( 38 , 0 ) 38 - 40 3 ( 40 , 3 ) 40 - 42 5 42 , 5 42 - 44 9 ( 44 , 9 ) 44 - 46 14 ( 46 , 14 ) 46 - 48 28 ( 48 , 28 ) 48 - 50 32 ( 50 , 32 ) 50 - 52 35 ( 52 , 35 ) Here, N = 35 , which is odd. So we need to find N + 1 2 which is 36 2 = 18 . So, the graph we get is, The value N + 1 2 = 18 is marked on y - axis and a line parallel to x - axis is drawn. This line meets the curve at a point. From that point draw a perpendicular to meet x - axis at median 46 . 5 . Hence, the median of the frequency distribution is 46 . 5 .

From the frequency distribution table given below, draw less than ogive: Marks obtained 50 - 60 60 - 70 70 - 80 80 - 90 90 - 100 Frequency 4 8 12 6 10

An ogive graph serves as a graphical representation of the cumulative relative frequency distribution for quantitative variables. Marks obtained Frequency CF 50 – 60 4 4 60 - 70 8 8 + 4 = 12 70 - 80 12 12 + 12 = 24 80 - 90 6 24 + 6 = 30 90 - 100 10 30 + 10 = 40

An insurance policy agent found the following data for distribution of ages of 35 policy holders. Draw a ''less than type'' (below) of ogive for the given data: Age(in years) Number of policy holders Below 20 2 Below 25 6 Below 30 12 Below 35 16 Below 40 20 Below 45 25 Below 50 35

Less than type cumulative frequency distribution is, Age (in years) Cumulative frequency f i Points 15 - 20 2 ( 20 , 2 ) 20 - 25 6 + 2 = 8 ( 25 , 8 ) 25 - 30 8 + 12 = 20 30 , 20 30 - 35 20 + 16 = 36 ( 35 , 36 ) 35 - 40 36 + 20 = 56 ( 40 , 56 ) 40 - 45 56 + 25 = 81 ( 45 , 81 ) 45 - 50 81 + 35 = 116 ( 50 , 116 ) Mark upper class limits on x - axis and cumulative frequency on y - axis. We plot the following points: 20 , 2 , 25 , 8 , 30 , 20 , 35 , 36 , 40 , 56 , 45 , 81 , 50 , 116 Joining the above points by line segments, we get less than type ogive as below.

The following distribution gives the daily income of 50 workers of a factory. Daily income (in Rupees) 250 - 300 300 - 350 350 - 400 400 - 450 450 - 500 Number of workers 12 14 8 6 10 Convert the distribution above to a less than type cumulative frequency distribution, and draw its ogive.

Let's make less than type cumulative frequency table: Daily income (less than type) Cumulative frequency c f Less than 300 12 Less than 350 12 + 14 = 26 Less than 400 26 + 8 = 34 Less than 450 34 + 6 = 40 Less than 500 40 + 10 = 50 Plot points 300 , 12 , 350 , 26 , 400 , 34 , 450 , 40 and 500 , 50 on a graph. The ogive is drawn below:

50 students enter for a school javelin throw competition. The distance (in metre) thrown are recorded below Distance (in m) 0 - 20 20 - 40 40 - 60 60 - 80 80 - 100 Number of students 6 11 17 12 4 Draw a cumulative frequency curve (less than type) and calculate the median distance drawn by using this curve.

Distance (in m) Cumulative frequency 0 0 Less than 20 6 Less than 40 17 Less than 60 34 Less than 80 46 Less than 100 50 To draw less than type ogive we plot the points ( 0 , 0 ) , ( 20 , 6 ) , ( 40 , 17 ) , ( 60 , 34 ) , ( 80 , 46 ) , 100 , 50 , join all these points by free hand. Now, N 2 = 50 2 = 25 Taking Y = 25 on Y-axis and draw a line parallel to X -axis, which meets the curve at point A From point A . we draw a line perpendicular to to X -axis, where this meets that point is the required median i.e., 49 . 4 .

Profit (in ' 000 ) earned by a company during the period 2011-2016 are given below. Draw a simple bar diagram to this data showing profit Year Profit ' 000 2011 15000 2012 18000 2013 20000 2014 16000 2015 13000 2016 17000 Find the year in which the company earned a maximum profit.

Simple bar diagram for profit of earned by company in 2011 - 2016 Year Profit ( ' 000 ) 2011 15000 2012 18000 2013 20000 2014 16000 2015 13000 2016 17000 By using above diagram the company earned maximum profit in year 2013 .

During the medical check-up of 35 students of a class, their weights were recorded as follows: Weight (in kg ) Number of students Less than 38 Less than 40 Less than 42 Less than 44 Less than 46 Less than 48 Less than 50 Less than 52 0 3 5 9 14 28 32 35 Draw a less than type ogive for the given data. Hence, obtain the median weight from the graph and verify the result by using the formula.

Using points ( 38 , 0 ) , ( 40 , 3 ) , ( 42 , 5 ) , ( 44 , 9 ) , ( 46 , 14 ) , ( 48 , 28 ) , ( 50 , 32 ) and ( 52 , 35 ) plot a graph. The less than type ogive is shown below: To find median graphically, We have total frequency, N = 35 ⇒ N 2 = 35 2 = 17 . 5 Plot 17 . 5 on the y - a x i s and draw a horizontal line intersecting the ogive parallel to x - a x i s . Observe, from the graph the horizontal line starting from the cumulative frequency 17 . 5 on the y - axis touches the graph corresponding to 46 . 5 kg (approx) point on the x - axis . Hence, median is 46 . 5 kg . Finding median by formula: Weight ( kg ) Cumulative frequency c f Class Frequency f i Less than 38 0 36 - 38 0 Less than 40 3 38 - 40 3 - 0 = 3 Less than 42 5 40 - 42 5 - 3 = 2 Less than 44 9 42 - 44 9 - 5 = 4 Less than 46 14 44 - 46 14 - 9 = 5 Less than 48 28 46 - 48 28 - 14 = 14 Less than 50 32 48 - 50 32 - 28 = 4 Less than 52 35 50 - 52 35 - 32 = 3 N = 35 For median: We have, total frequency, N = 35 N 2 = 35 2 = 17 . 5 Observe, c f = 28 is just greater than 17 . 5 Thus, median class = 46 - 48 Median is given by Median = L + N 2 - c f f × h Where, L = Lower class limit of median class = 46 N 2 = 17 . 5 c f = Cumulative frequency of the class preceding median class = 14 f = Frequency of the median class = 14 h = Class interval of the median class = 2 Substituting these values in the formula of median, we get, Median = 46 + 17 . 5 - 14 14 × 2 ⇒ Median = 46 + 3 . 5 14 × 2 ⇒ Median = 46 + 0 . 5 ⇒ Median = 46 . 5 Thus, median is 46 . 5 kg . Hence, verified.

The production yield per hectare of wheat of some farms of a village are given in the following table: Production yield (in kg/ha) 40 - 45 45 - 50 50 - 55 55 - 60 60 - 65 65 - 70 70 - 75 75 - 80 80 - 85 Number of farms 1 9 15 18 40 26 16 14 10 Draw a less than type ogive and a more than type ogive for this data

Class interval Less than 45 1 Less than 50 10 Less than 55 25 Less than 60 43 Less than 65 83 Less than 70 109 Less than 75 125 Less than 80 139 Less than 85 149 We plot the points 45 , 1 , 50 , 10 , 55 , 25 , 60 , 43 , 65 , 83 , 35 , 27 , 70 , 109 , 75 , 125 , 80 , 139 and 85 , 149 to get less than ogive More than Series Class interval More than 40 149 More than 45 148 More than 50 139 More than 55 124 More than 60 106 More than 65 66 More than 70 40 More than 75 24 More than 80 10 We plot the points 40 , 149 , 45 , 148 , 50 , 139 , 55 , 124 , 60 , 106 , 65 , 66 , 70 , 40 , 75 , 24 and 80 , 10 to get more than ogive

The following table gives production yield per hectare of wheat of 100 farms of a village. Production yield ( Qui / Hec ) 50 - 55 55 - 60 60 - 65 65 - 70 70 - 75 75 - 80 Number of farmers 2 8 12 24 38 16 Change the distribution to a more than type distribution and draw its ogive.

Let's make more than type cumulative frequency table: Production yield (more than type) Number of farmers Cumulative frequency c f More than 50 2 2 + 8 + 12 + 24 + 38 + 16 = 100 More than 55 8 100 - 2 = 98 More than 60 12 98 - 8 = 90 More than 65 24 90 - 12 = 78 More than 70 38 78 - 24 = 54 More than 75 16 54 - 38 = 16 Plot points 50 , 100 , 55 , 98 , 60 , 90 , 65 , 78 , 70 , 54 and 75 , 16 on a graph. The more than type ogive is sown below:

The given distribution shows the number of wickets taken by the bowlers in one-day international cricket matches: Number 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 Number of bowlers 2 5 9 17 39 54 70 80 Draw a ‘less than type’ ogive from the above data. Find the median.

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 ⇒ N 2 = 40 . Mark the point A whose ordinate is 40 and its x -coordinate is 76 . Thus, median of the data is 76 .

The distribution of heights (in cm ) of 96 children is given below Height (in cm ) Number of Children 124 - 128 5 128 - 132 8 132 - 136 17 136 - 140 24 140 - 144 16 144 - 148 12 148 - 152 6 152 - 156 4 156 - 160 3 160 - 164 1 Draw a less than type cumulative frequency curve for this data and use it to compute median height of the children.

Height (in cm ) Number of Children Less than 124 0 Less than 128 5 Less than 132 13 Less than 136 30 Less than 140 54 Less than 144 70 Less than 148 82 Less than 152 88 Less than 156 92 Less than 160 95 Less than 164 96 To draw the less than type ogive, we plot the points ( 124 , 0 ) , ( 128 , 5 ) , ( 132 , 13 ) , ( 136 , 30 ) , ( 140 , 54 ) , ( 144 , 70 ) , ( 148 , 82 ) , ( 152 , 88 ) , ( 156 , 92 ) , ( 160 , 95 ) , ( 164 , 96 ) and join all these point by free hand . Here, N 2 = 96 2 = 48 We take, y = 48 in y - coordinate and draw a line parallel to x - axis, meets the curve at A and draw a perpendicular line from point A to the x - axis and this line meets the x - axis at the point which is the median i.e., median = 141 . 17 .

The following is the frequency distribution of duration for 100 calls made on a mobile phones. Duration (in s) Number of calls 95 - 125 14 125 - 155 22 155 - 185 28 185 - 215 21 215 - 245 15 Calculate the average duration (in sec) of a call and also find the median from a cumulative frequency curve.

First, we calculate class marks as follows. Duration (in s) Number of calls f i Class marks x i u i = x i - a h f i u i 95 - 125 14 110 - 2 - 28 125 - 155 22 140 - 1 - 22 155 - 185 28 a = 170 0 0 185 - 215 21 200 1 21 215 - 245 15 230 2 30 ∑ f i = 100 ∑ f i u i = 1 Here, (assumed mean) a = 170 and (class width) h = 30 By step deviation method, Average x ¯ = a + ∑ f i u i ∑ f i × h = 170 + 1 100 × 30 = 170 + 0 . 3 = 170 . 3 Hence, average duration is 170 . 3 s . For calculating median from a cumulative frequency curve. We prepare less than type or more than type ogive. We observe that, number of calls in less than 95 s is 0 . Similarly, in less than 125 include the number of the calls in less than 95 as well as the number of calls from 95 - 125 s. So, the total number of calls less than 125 s is 0 + 14 = 14 . Continuing in this manner, we will get remaining in less than 155 , 185 , 215 and 245 s . Now, we construct a table for less than ogive (cumulative frequency curve). Less than type Duration (in s) Number of calls Less than 95 0 Less than 125 0 + 14 = 14 Less than 155 14 + 22 = 36 Less than 185 36 + 28 = 64 Less than 215 64 + 21 = 85 Less than 245 85 + 15 = 100 To draw less than type ogive we plot them the points 95 , 0 , 125 , 14 , 155 , 36 , 185 , 64 , 215 , 85 , 245 , 100 on the paper and join them by free hand. ∵ Total number of calls n = 100 ∴ n 2 = 100 2 = 50 Now, point 50 taking y - axis draw a line parallel to x - axis meet at a point P and draw a perpendicular line from P to the x - axis, the intersection point of x - axis is the median. Hence, required median is 170 .

In More than Ogive curve cumulative frequencies are marked against the lower limit of the respective classes

MEDIUM

Mathematics>Statistics and Probability>Statistics>Graphical Representation of Cumulative Frequency Distribution

During the medical check-up of $35$ students of a class, their weights were recorded as follows. Draw a less than type of ogive for the given data:

Weight in kg	Number of students
Less than $38$	$0$
Less than $40$	$3$
Less than $42$	$5$
Less than $44$	$9$
Less than $46$	$14$
Less than $48$	$28$
Less than $50$	$32$
Less than $52$	$35$

MEDIUM

Mathematics>Statistics and Probability>Statistics>Graphical Representation of Cumulative Frequency Distribution

From the frequency distribution table given below, draw less than ogive:

Marks obtained	$50 - 60$	$60 - 70$	$70 - 80$	$80 - 90$	$90 - 100$
Frequency	$4$	$8$	$12$	$6$	$10$

EASY

Mathematics>Statistics and Probability>Statistics>Graphical Representation of Cumulative Frequency Distribution

What type of Ogive is given
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MEDIUM

Mathematics>Statistics and Probability>Statistics>Graphical Representation of Cumulative Frequency Distribution

An insurance policy agent found the following data for distribution of ages of $35$ policy holders. Draw a ''less than type'' (below) of ogive for the given data:

Age(in years)	Number of policy holders
Below $20$	$2$
Below $25$	$6$
Below $30$	$12$
Below $35$	$16$
Below $40$	$20$
Below $45$	$25$
Below $50$	$35$

MEDIUM

Mathematics>Statistics and Probability>Statistics>Graphical Representation of Cumulative Frequency Distribution

Find the median of the data from the above graph.

Question Image

MEDIUM

Mathematics>Statistics and Probability>Statistics>Graphical Representation of Cumulative Frequency Distribution

State the type for curve A and curve $B$

Question Image

EASY

Mathematics>Statistics and Probability>Statistics>Graphical Representation of Cumulative Frequency Distribution

Ogive is graphical representation of _____ in a frequency distribution.

HARD

Mathematics>Statistics and Probability>Statistics>Graphical Representation of Cumulative Frequency Distribution

The following distribution gives the daily income of $50$ workers of a factory.

Daily income (in Rupees)	$250 - 300$	$300 - 350$	$350 - 400$	$400 - 450$	$450 - 500$
Number of workers	$12$	$14$	$8$	$6$	$10$

Convert the distribution above to a less than type cumulative frequency distribution, and draw its ogive.

MEDIUM

Mathematics>Statistics and Probability>Statistics>Graphical Representation of Cumulative Frequency Distribution

50 students enter for a school javelin throw competition. The distance (in metre) thrown are recorded below

Distance (in m)	$0 - 20$	$20 - 40$	$40 - 60$	$60 - 80$	$80 - 100$
Number of students	$6$	$11$	$17$	$12$	$4$

Draw a cumulative frequency curve (less than type) and calculate the median distance drawn by using this curve.

EASY

Mathematics>Statistics and Probability>Statistics>Graphical Representation of Cumulative Frequency Distribution

In a frequency distribution, ogives are graphical representation of

MEDIUM

Mathematics>Statistics and Probability>Statistics>Graphical Representation of Cumulative Frequency Distribution

Profit (in $' 000$ ) earned by a company during the period 2011-2016 are given
below. Draw a simple bar diagram to this data showing profit

Year	Profit $(' 000)$
$2011$	$15000$
$2012$	$18000$
$2013$	$20000$
$2014$	$16000$
$2015$	$13000$
$2016$	$17000$

Find the year in which the company earned a maximum profit.

HARD

Mathematics>Statistics and Probability>Statistics>Graphical Representation of Cumulative Frequency Distribution

During the medical check-up of $35$ students of a class, their weights were recorded as follows:

Weight (in kg )	Number of students
Less than $38$ Less than $40$ Less than $42$ Less than $44$ Less than $46$ Less than $48$ Less than $50$ Less than $52$	$\begin{array}{l} 0 \\ 3 \\ 5 \\ 9 \\ 14 \\ 28 \\ 32 \\ 35 \end{array}$

Draw a less than type ogive for the given data. Hence, obtain the median weight from the graph and verify the result by using the formula.

EASY

Mathematics>Statistics and Probability>Statistics>Graphical Representation of Cumulative Frequency Distribution

The production yield per hectare of wheat of some farms of a village are given in the following table:

Production yield (in kg/ha)	$40 - 45$	$45 - 50$	$50 - 55$	$55 - 60$	$60 - 65$	$65 - 70$	$70 - 75$	$75 - 80$	$80 - 85$
Number of farms	$1$	$9$	$15$	$18$	$40$	$26$	$16$	$14$	$10$

Draw a less than type ogive and a more than type ogive for this data

HARD

Mathematics>Statistics and Probability>Statistics>Graphical Representation of Cumulative Frequency Distribution

The following table gives production yield per hectare of wheat of $100$ farms of a village.

Production yield $(Qui / Hec)$	$50 - 55$	$55 - 60$	$60 - 65$	$65 - 70$	$70 - 75$	$75 - 80$
Number of farmers	$2$	$8$	$12$	$24$	$38$	$16$

Change the distribution to a more than type distribution and draw its ogive.

EASY

Mathematics>Statistics and Probability>Statistics>Graphical Representation of Cumulative Frequency Distribution

The median of a given frequency distribution is found graphically with the help of

MEDIUM

Mathematics>Statistics and Probability>Statistics>Graphical Representation of Cumulative Frequency Distribution

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

Number 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 $120$
Number of bowlers	$2$	$5$	$9$	$17$	$39$	$54$	$70$	$80$

Draw a ‘less than type’ ogive from the above data. Find the median.

MEDIUM

Mathematics>Statistics and Probability>Statistics>Graphical Representation of Cumulative Frequency Distribution

The median of a given frequency distribution is found graphically with the help of

MEDIUM

Mathematics>Statistics and Probability>Statistics>Graphical Representation of Cumulative Frequency Distribution

The distribution of heights (in $cm$ ) of $96$ children is given below

Height (in $cm$ )	Number of Children
$124 - 128$	$5$
$128 - 132$	$8$
$132 - 136$	$17$
$136 - 140$	$24$
$140 - 144$	$16$
$144 - 148$	$12$
$148 - 152$	$6$
$152 - 156$	$4$
$156 - 160$	$3$
$160 - 164$	$1$

Draw a less than type cumulative frequency curve for this data and use it to compute median height of the children.

HARD

Mathematics>Statistics and Probability>Statistics>Graphical Representation of Cumulative Frequency Distribution

The following is the frequency distribution of duration for $100$ calls made on a mobile phones.

Duration (in s)	Number of calls
$95 - 125$	$14$
$125 - 155$	$22$
$155 - 185$	$28$
$185 - 215$	$21$
$215 - 245$	$15$

Calculate the average duration (in sec) of a call and also find the median from a cumulative frequency curve.

MEDIUM

Mathematics>Statistics and Probability>Statistics>Graphical Representation of Cumulative Frequency Distribution

The abscissa of the point of intersection of less than type and of the more than types cumulative frequency curves of a grouped data gives its

In More than Ogive curve, cumulative frequencies are marked against the lower limit of the respective classes.

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