A survey regarding the heights (in cm ) of 51 girls of class X of a school was conducted and the following data was obtained: Height (in cm ) Number of Girls Less than 140 4 Less than 145 11 Less than 150 29 Less than 155 40 Less than 160 46 Less than 165 51 Find the median height.

Here, we are given less than cumulative frequency table. First, we prepare the frequency table: Height (in cm ) Class interval Number of Girls c . f . Frequency f i Less than 140 135 - 140 4 4 Less than 145 140 - 145 11 11 - 4 = 7 Less than 150 145 - 150 29 29 - 11 = 18 Less than 155 150 - 155 40 40 - 29 = 11 Less than 160 155 - 160 46 46 - 40 = 6 Less than 165 160 - 165 51 51 - 46 = 5 We have, N = 51 So, N 2 = 51 2 = 25 . 5 The cumulative frequency just greater than 25 . 5 is 29 , which corresponds to the class 145 - 150 . Thus, the median class is 145 - 150 . Here, l = 145 (the lower limit of the median class), f = 18 (frequency of the median class), h = 5 (width of each class), N = 51 (number of observations), and C = 11 (cumulative frequency of class preceding the median class) ∴ Median = l + N 2 - C f × h = 145 + 25 . 5 - 11 18 × 5 = 145 + 14 . 5 18 × 5 = 145 + 4 . 03 = 149 . 03 approx Hence, the median is 149 . 03 .

The following distribution gives daily income of 50 workers of a factory: Daily income (in ₹ ) 100 - 120 120 - 140 140 - 160 160 - 180 180 - 200 Number of workers 12 14 8 6 10 Convert the distribution given above to a less than type, cumulative frequency distribution and draw its ogive.

The cumulative frequency distribution table by less than type is as below: Daily income (in ₹ ) Number of workers f i Daily income (in ₹ ) Cumulative frequency c . f . 100 - 120 12 Less than 120 12 120 - 140 14 Less than 140 14 + 12 = 26 140 - 160 8 Less than 160 8 + 26 = 34 160 - 180 6 Less than 180 6 + 34 = 40 180 - 200 10 Less than 200 10 + 40 = 50 Mark upper class limits on x - axis and cumulative frequency on y - axis. We plot the points 120 , 12 , 140 , 26 , 160 . 34 , 180 , 40 , and 200 , 50 . Joining the above points by line segments, the cumulative frequency polygon is as shown below:

Find the mean, mode and median of the following data: Class-interval 0 - 10 10 - 20 20 - 30 30 - 40 40 - 50 50 - 60 60 - 70 Frequency 3 8 10 15 7 4 3

We prepare the following frequency distribution table to find mean, mode and median: Class interval Frequency f i Cumulative frequency c . f . Class Mark x i Deviations d i = x i - 35 u i = d i h = x i - a 10 f i u i 0 - 10 3 3 0 + 10 2 = 10 2 = 5 5 - 35 = - 30 - 30 10 = - 3 - 9 10 - 20 8 3 + 8 = 11 15 - 20 - 2 - 16 20 - 30 10 11 + 10 = 21 25 - 10 - 1 - 10 30 - 40 15 21 + 15 = 36 35 = a 0 0 0 40 - 50 7 36 + 7 = 43 45 10 1 7 50 - 60 4 43 + 4 = 47 55 20 2 8 60 - 70 3 47 + 3 = 50 65 30 3 9 Total ∑ f i = 50 ∑ f i u i = - 11 (i) Let the assumed mean a = 35 and h = 10 , then we have, Mean x ¯ = a + Σ f i u i Σ f i × h = 35 + - 11 50 × 10 = 35 - 2 . 2 = 32 . 8 Hence, the mean is 32 . 8 . (ii) In the given frequency distribution, the class 30 - 40 has maximum frequency. So, 30 - 40 is the modal class. Here, l = 30 , f 1 = 15 , f 0 = 10 , f 2 = 7 , h = 10 , Mode = l + f 1 - f 0 2 f 1 - f 0 - f 2 × h = 30 + 15 - 10 2 × 15 - 10 - 7 × 10 = 30 + 5 30 - 17 × 10 = 30 + 5 13 × 10 = 30 + 3 . 85 = 33 . 85 approx . Hence, the mode is 33 . 85 . (iii) We have, N = 50 So, N 2 = 50 2 = 25 . The cumulative frequency just greater than 25 is 36 , which corresponds to the class 30 - 40 . Thus, the median class is 30 - 40 . Here, l = 30 , f = 15 , h = 10 , N = 50 , and C = 21 ∴ Median = l + N 2 - C f × h = 30 + 25 - 21 15 × 10 = 30 + 4 15 × 10 = 30 + 8 3 = 30 + 2 . 67 = 32 . 67 approx Hence, the median is 32 . 67 . Hence, the mean, mode and median of the given data are 32 . 8 , 33 . 85 and 32 . 67 respectively.

The annual profits earned by 30 shops of a shopping complex in a locality give rise to the following distribution: Profit (in lakhs ₹ ) Number of shops (Frequency) More than or equal to 5 30 More than or equal to 10 28 More than or equal to 15 16 More than or equal to 20 14 More than or equal to 25 10 More than or equal to 30 7 More than or equal to 35 3 Draw the ogives for the data above. Hence, obtain the median profit.

We draw the co-ordinate axes, with lower-class limits on x - axis and number of shops (cumulative frequency) on y - axis. We plot the points 5 , 30 , 10 , 28 , 15 , 16 , 20 , 14 , 25 , 10 , 30 , 7 and 35 , 3 . Joining the above points by line segments, the cumulative frequency polygon is as shown below: To locate the median point on the graph, take N 2 = 30 2 = 15 on the y - axis. From this point, draw a line parallel to the x - axis cutting the curve at a point P . From P , draw a perpendicular to the x - axis. The point of intersection of this perpendicular with the x - axis at M which gives the median of the data. Here, the median profit is ₹ 17 . 5 lakhs.

J P Mohindru and Bharat Mohindru Solutions for Chapter: Statistics, Exercise 8: CHAPTER TEST

Author:J P Mohindru & Bharat Mohindru

J P Mohindru Mathematics Solutions for Exercise - J P Mohindru and Bharat Mohindru Solutions for Chapter: Statistics, Exercise 8: CHAPTER TEST

Attempt the free practice questions on Chapter 14: Statistics, Exercise 8: CHAPTER TEST with hints and solutions to strengthen your understanding. Modern's abc+ of Mathematics for Class 10 solutions are prepared by Experienced Embibe Experts.

Questions from J P Mohindru and Bharat Mohindru Solutions for Chapter: Statistics, Exercise 8: CHAPTER TEST with Hints & Solutions

EASY

10th CBSE

IMPORTANT

Modern's abc+ of Mathematics for Class 10>Chapter 14 - Statistics>CHAPTER TEST>Q 4

Find the mode of the following frequency distribution.

Class interval	$0 - 20$	$20 - 40$	$40 - 60$	$60 - 80$
Frequency	$15$	$6$	$18$	$10$

MEDIUM

10th CBSE

IMPORTANT

Modern's abc+ of Mathematics for Class 10>Chapter 14 - Statistics>CHAPTER TEST>Q 5

A survey regarding the heights (in $cm$ ) of $51$ girls of class X of a school was conducted and the following data was obtained:

Height (in $cm$ )	Number of Girls
Less than $140$	$4$
Less than $145$	$11$
Less than $150$	$29$
Less than $155$	$40$
Less than $160$	$46$
Less than $165$	$51$

Find the median height.

MEDIUM

10th CBSE

IMPORTANT

Modern's abc+ of Mathematics for Class 10>Chapter 14 - Statistics>CHAPTER TEST>Q 6

To find the concentration of ${SO}_{2}$ in the air (in parts per million, i.e., $ppm$ ), the data was collected for $30$ localities in a certain city and is presented below:

Concentration of ${SO}_{2}$ (in $ppm$ )	Frequency
$0.00 - 0.04$	$4$
$0.04 - 0.08$	$9$
$0.08 - 0.12$	$9$
$0.12 - 0.16$	$2$
$0.16 - 0.20$	$4$
$0.20 - 0.24$	$2$

Note: This question given in the book seems to have errors and the modified question should be as given below:

To find the concentration of ${SO}_{2}$ in the air (in parts per million, i.e., $ppm$ ), the data was collected for $30$ localities in a certain city and is presented below:

Concentration of ${SO}_{2}$ (in $ppm$ )	Frequency
$0.00 - 0.04$	$4$
$0.04 - 0.08$	$9$
$0.08 - 0.12$	$9$
$0.12 - 0.16$	$2$
$0.16 - 0.20$	$4$
$0.20 - 0.24$	$2$

Find the mean concentration of ${SO}_{2}$ in the air.

MEDIUM

10th CBSE

IMPORTANT

Modern's abc+ of Mathematics for Class 10>Chapter 14 - Statistics>CHAPTER TEST>Q 7

If the median of the following distribution given below is $28.5$ , find the values of $x$ and $y$ :

Class interval	$0 - 10$	$10 - 20$	$20 - 30$	$30 - 40$	$40 - 50$	$50 - 60$	Total
Frequency	$5$	$x$	$20$	$15$	$y$	$5$	$6$

Note: This question given in the book seems to have errors and the modified question should be as given below:

If the median of the following distribution given below is $28.5$ , find the values of $x$ and $y$ :

Class interval	$0 - 10$	$10 - 20$	$20 - 30$	$30 - 40$	$40 - 50$	$50 - 60$	Total
Frequency	$5$	$x$	$20$	$15$	$y$	$5$	$60$

MEDIUM

10th CBSE

IMPORTANT

Modern's abc+ of Mathematics for Class 10>Chapter 14 - Statistics>CHAPTER TEST>Q 8

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

Daily income (in $₹$ )	$100 - 120$	$120 - 140$	$140 - 160$	$160 - 180$	$180 - 200$
Number of workers	$12$	$14$	$8$	$6$	$10$

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

MEDIUM

10th CBSE

IMPORTANT

Modern's abc+ of Mathematics for Class 10>Chapter 14 - Statistics>CHAPTER TEST>Q 9

The following distribution gives the state wise teacher-student ratio in higher secondary school of India. Find the mode and mean of the data. Interpret the two measures.

No. of students per teacher	$15 - 20$	$20 - 25$	$25 - 30$	$30 - 35$	$35 - 40$	$40 - 45$	$45 - 50$	$50 - 55$
Number of states/ UT	$3$	$8$	$9$	$10$	$3$	$0$	$0$	$2$

HARD

10th CBSE

IMPORTANT

Modern's abc+ of Mathematics for Class 10>Chapter 14 - Statistics>CHAPTER TEST>Q 10

Find the mean, mode and median of the following data:

Class-interval	$0 - 10$	$10 - 20$	$20 - 30$	$30 - 40$	$40 - 50$	$50 - 60$	$60 - 70$
Frequency	$3$	$8$	$10$	$15$	$7$	$4$	$3$

MEDIUM

10th CBSE

IMPORTANT

Modern's abc+ of Mathematics for Class 10>Chapter 14 - Statistics>CHAPTER TEST>Q 11

The annual profits earned by $30$ shops of a shopping complex in a locality give rise to the following distribution:

Profit (in lakhs $₹$ )	Number of shops (Frequency)
More than or equal to $5$	$30$
More than or equal to $10$	$28$
More than or equal to $15$	$16$
More than or equal to $20$	$14$
More than or equal to $25$	$10$
More than or equal to $30$	$7$
More than or equal to $35$	$3$

Draw the ogives for the data above. Hence, obtain the median profit.

J P Mohindru and Bharat Mohindru Solutions for Chapter: Statistics, Exercise 8: CHAPTER TEST

J P Mohindru Mathematics Solutions for Exercise - J P Mohindru and Bharat Mohindru Solutions for Chapter: Statistics, Exercise 8: CHAPTER TEST

Questions from J P Mohindru and Bharat Mohindru Solutions for Chapter: Statistics, Exercise 8: CHAPTER TEST with Hints & Solutions

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