The annual rainfall record of a city for 66 days is given in the following table. Rainfall (in cm) 0 - 10 10 - 20 20 - 30 30 - 40 40 - 50 50 - 60 Number of days 22 10 8 15 5 6 Calculate the median rainfall using ogives (or more than type and less than type).

We observe that, the annual rainfall record of a city less than 0 is 0 . Similarly, less than 10 include the annual rainfall record of a city from 0 as well as the annual rainfall record of a city from 0 - 10 . So, the total annual rainfall record of a city for less than 10 cm is 0 + 22 = 22 days. Continuing in this manner, we will get remaining less than 20 , 30 , 40 , 50 and 60 . Also, we observe that annual rainfall record of a city for 66 days is more than or equal to 0 cm . Since, 22 days lie in the interval 0 - 10 . So, annual rainfall record for 66 - 22 = 44 days is more than or equal to 10 cm . Continuing in this manner we will get remaining more than or equal to 20 , 30 , 40 , 50 and 60 . Now, we construct a table for less than and more than types. (i) Less than type (ii) More than type Rainfall (in cm Number of days Rainfall (in cm Number of days Less than 0 0 More than or equal to 0 66 Less than 10 0 + 22 = 22 More than or equal to 10 66 - 22 = 44 Less than 20 22 + 10 = 32 More than or equal to 20 44 - 10 = 34 Less than 30 32 + 8 = 40 More than or equal to 30 34 - 8 = 26 Less than 40 40 + 15 = 55 More than or equal to 40 26 - 15 = 11 Less than 50 55 + 5 = 60 More than or equal to 50 11 - 5 = 6 Less than 60 60 + 6 = 66 More than or equal to 60 6 - 6 = 0 To draw less than type ogive we plot the points 0 , 0 , 10 , 22 , 20 , 32 , 30 , 40 , 40 , 55 , 50 , 60 , 60 , 66 on the paper and join them by free hand. To draw more than type ogive we plot the points 0 , 66 , 10 , 44 , 20 , 34 , 30 , 26 , 40 , 11 , 50 , 6 , 60 , 0 on the graph paper and join them by free hand. Total number of days n = 66 Now, n 2 = 33 Firstly, we plot a line parallel to x - axis at intersection point of both ogives, which further intersect at 0 , 33 on y - axis. Now, we draw a line perpendicular to x - axis at intersection point of both ogives, which further intersect at 21 . 25 , 0 on x - axis. Which is the required median using ogives. Hence, median rainfall = 21 . 25 cm

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 .

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 Construct a cumulative frequency table.

Distance (in m) Number of students f i Cumulative frequency c f 0 - 20 6 6 20 - 40 11 17 40 - 60 17 34 60 - 80 12 46 80 - 100 4 50

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 .

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 Calculate the median distance by using the formula for median.

Now, N 2 = 50 2 = 25 Which lies is the interval 40 - 60 . ∴ l = 40 , h = 20 , c f = 17 and f = 17 ∴ Median = l + N 2 - c f f × h = 40 + 25 - 17 17 × 20 = 40 + 8 × 20 17 = 40 + 9 . 41 = 49 . 41

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 Are the median distance calculated in (ii) by drawing ogive and (iii) by cumulative frequency same?

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 . Now, N 2 = 50 2 = 25 Which lies is the interval 40 - 60 . ∴ l = 40 , h = 20 , c f = 17 and f = 17 ∴ Median = l + N 2 - c f f × h = 40 + 25 - 17 17 × 20 = 40 + 8 × 20 17 = 40 + 9 . 41 = 49 . 41 Median obtained by drawing an ogive is 49 . 4 and by calculating the cumulative frequency is 49 . 41 Therefore, Yes, median distance calculated by parts (ii) and (iii) are same.

MEDIUM

10th CBSE

IMPORTANT

Earn 100

Size of agriculture holdings in a survey of $200$ families is given in the following table.

Size of Agricultural Holdings (in hec) Number of families

$0 - 5$ $10$

$5 - 10$ $15$

$10 - 15$ $30$

$15 - 20$ $80$

$20 - 25$ $40$

$25 - 30$ $20$

$30 - 35$ $5$

Compute median and mode size of the holdings.

Important Questions on Statistics and Probability

MEDIUM

10th CBSE

IMPORTANT

NCERT EXEMPLAR PROBLEMS-SOLUTIONS MATHEMATICS>Chapter 13 - Statistics and Probability>Exercise>Q 12

The annual rainfall record of a city for $66$ days is given in the following table.

Rainfall (in cm)	$0 - 10$	$10 - 20$	$20 - 30$	$30 - 40$	$40 - 50$	$50 - 60$
Number of days	$= 15 + (\frac{80 - 30}{2 \times 80 - 30 - 40}) \times 5$	$10$	$8$	$15$	$5$	$6$

Calculate the median rainfall using ogives (or more than type and less than type).

HARD

10th CBSE

IMPORTANT

NCERT EXEMPLAR PROBLEMS-SOLUTIONS MATHEMATICS>Chapter 13 - Statistics and Probability>Exercise>Q 13

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.

EASY

10th CBSE

IMPORTANT

NCERT EXEMPLAR PROBLEMS-SOLUTIONS MATHEMATICS>Chapter 13 - Statistics and Probability>Exercise>Q 14

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	$6$	$11$	$17$	$12$	$4$

Construct a cumulative frequency table.

MEDIUM

10th CBSE

IMPORTANT

NCERT EXEMPLAR PROBLEMS-SOLUTIONS MATHEMATICS>Chapter 13 - Statistics and Probability>Exercise>Q 14

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.

MEDIUM

10th CBSE

IMPORTANT

NCERT EXEMPLAR PROBLEMS-SOLUTIONS MATHEMATICS>Chapter 13 - Statistics and Probability>Exercise>Q 14

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	$6$	$11$	$17$	$12$	$4$

Calculate the median distance by using the formula for median.

MEDIUM

10th CBSE

IMPORTANT

NCERT EXEMPLAR PROBLEMS-SOLUTIONS MATHEMATICS>Chapter 13 - Statistics and Probability>Exercise>Q 14

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	$6$	$11$	$17$	$12$	$4$

Are the median distance calculated in (ii) by drawing ogive and (iii) by cumulative frequency same?

Size of Agricultural Holdings (in hec)	Number of families
$0 - 5$	$10$
$5 - 10$	$15$
$10 - 15$	$30$
$15 - 20$	$80$
$20 - 25$	$40$
$25 - 30$	$20$
$30 - 35$	$5$

Size of agriculture holdings in a survey of 200 families is given in the following table. Size of Agricultural Holdings (in hec) Number of families 0-5 10 5-10 15 10-15 30 15-20 80 20-25 40 25-30 20 30-35 5 Compute median and mode size of the holdings.

Important Questions on Statistics and Probability

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