Draw the less than type and more than type ogives for the data and use them to find the median weight. The weights of tea in 70 packets shown in the following table Weight (in g) Number of packets 200 - 201 13 201 - 202 27 202 - 203 18 203 - 204 10 204 - 205 1 205 - 206 1

For less than type table Here, we observe that, the weight of all 70 packets is more than or equal to 200 . Since, 13 packets lie in the interval 200 - 201 . So, the weight of 70 - 13 = 57 packets is more than or equal to 201 . Continuing in this manner we will get remaining more than or equal to 202 , 203 , 204 , 205 and 206 . (i) Less than type (ii) More than type Less than 200 0 More than or equal to 200 70 Less than 201 13 More than or equal to 201 70 - 13 = 57 Less than 202 40 More than or equal to 202 57 - 27 = 30 Less than 203 58 More than or equal to 203 30 - 18 = 12 Less than 204 68 More than or equal to 204 12 - 10 = 2 Less than 205 69 More than or equal to 205 2 - 1 = 1 Less than 206 70 More than or equal to 206 1 - 1 = 0 To draw the less than type ogive, we plot the points ( 200 , 0 ) , ( 201 , 13 ) , ( 202 , 40 ) , ( 203 , 58 ) ( 204 , 68 ) . (205, 69), (206, 70) on the paper and join them by free hand. To draw the more than type ogive plot the points ( 200 , 70 ) , ( 201 , 57 ) , ( 202 30 ) , ( 203 , 12 ) ( 204 , 2 ) , ( 205 , 1 ) , ( 206 , 0 ) on the the graph paper and join them by free hand. Hence, required median weight = Intersection point of X-axis = 201 . 8 g .

The table below shows the salaries of 280 persons. Salary (in ₹ thousand) Number of persons 5 - 10 49 10 - 15 133 15 - 20 63 20 - 25 15 25 - 30 6 30 - 35 7 40 - 45 2 45 - 50 1 Calculate the median and mode of the data.

First, we construct a cumulative frequency table Salary (in ₹ thousand) Number of persons Cumulative frequency c f 5 - 10 49 = f 1 49 = c f 10 - 15 f m 133 = f 133 + 49 = 182 15 - 20 63 = f 2 182 + 63 = 245 20 - 25 15 245 + 15 = 260 25 - 30 6 260 + 6 = 266 30 - 35 7 266 + 7 = 273 40 - 45 2 277 + 2 = 279 45 - 50 1 279 + 1 = 280 N = 280 ∴ N 2 = 280 2 = 140 (i) Here, median class is 10 - 15 , because 140 lies in it. Lower limit ( i ) = 10 , Frequency ( f ) = 133 , Cumulative frequency c f = 49 and class width ( h ) = 5 ∴ Median = l + N 2 - c f f × h = 10 + 140 - 49 133 × 5 = 10 + 91 × 5 133 = 10 + 455 133 = 10 + 3 . 421 = ₹ 13 . 421 (in thousand) = 13 . 421 × 1000 = ₹ 13421 (ii) Here, the highest frequency is 133 , which lies in the interval 10 - 15 , called modal class. Lower limit ( i ) = 10 , class width ( h ) = 5 , f m = 133 , t 1 = 49 , and f 2 = 63 . ∴ Mode = l + f m - f 1 2 f m - f 1 - f 2 × h = 10 + 133 - 49 2 × 133 - 49 - 63 × 5 = 10 + 84 × 5 266 - 112 = 10 + 84 × 5 154 = 10 + 2 . 727 = ₹ 12 . 727 (in thousand) = 12 . 727 × 1000 = ₹ 12727 Hence, the median and modal salary are ₹ 13421 and ₹ 12727 respectively.

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.

Neha Tyagi and Amit Rastogi Solutions for Chapter: Statistics and Probability, Exercise 4: Exercise

Author:Neha Tyagi & Amit Rastogi

Neha Tyagi Mathematics Solutions for Exercise - Neha Tyagi and Amit Rastogi Solutions for Chapter: Statistics and Probability, Exercise 4: Exercise

Attempt the practice questions on Chapter 13: Statistics and Probability, Exercise 4: Exercise with hints and solutions to strengthen your understanding. NCERT EXEMPLAR PROBLEMS-SOLUTIONS MATHEMATICS solutions are prepared by Experienced Embibe Experts.

Questions from Neha Tyagi and Amit Rastogi Solutions for Chapter: Statistics and Probability, Exercise 4: Exercise with Hints & Solutions

MEDIUM

10th CBSE

IMPORTANT

NCERT EXEMPLAR PROBLEMS-SOLUTIONS MATHEMATICS>Chapter 13 - Statistics and Probability>Exercise>Q 6

Draw the less than type and more than type ogives for the data and use them to find the median weight.

The weights of tea in $12 - 10 = 2$ packets shown in the following table

Weight (in g)	Number of packets
$200 - 201$	$13$
$201 - 202$	$27$
$70$	$18$
$203 - 204$	$10$
$204 - 205$	$1$
$205 - 206$	$1$

MEDIUM

10th CBSE

IMPORTANT

NCERT EXEMPLAR PROBLEMS-SOLUTIONS MATHEMATICS>Chapter 13 - Statistics and Probability>Exercise>Q 7

The table below shows the salaries of $280$ persons.

Salary (in $₹$ thousand)	Number of persons
$5 - 10$	$49$
$10 - 15$	$133$
$15 - 20$	$63$
$20 - 25$	$15$
$25 - 30$	$6$
$30 - 35$	$7$
$40 - 45$	$2$
$45 - 50$	$1$

Calculate the median and mode of the data.

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	$22$	$10$	$8$	$15$	$5$	$6$

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?

Neha Tyagi and Amit Rastogi Solutions for Chapter: Statistics and Probability, Exercise 4: Exercise

Neha Tyagi Mathematics Solutions for Exercise - Neha Tyagi and Amit Rastogi Solutions for Chapter: Statistics and Probability, Exercise 4: Exercise

Questions from Neha Tyagi and Amit Rastogi Solutions for Chapter: Statistics and Probability, Exercise 4: Exercise with Hints & Solutions

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