Find the mode of the following frequency distribution: Monthly salary (in ₹ ) 0 - 5000 5000 - 10000 10000 - 15000 15000 - 20000 20000 - 25000 25000 - 30000 Number of employees 90 150 100 80 70 10

In the given frequency distribution class 5000 - 10000 has maximum frequency. So, 5000 - 10000 is the modal class. Here, l = lower limit of the modal class = 5000 , f 1 = frequency of modal class = 150 , f 0 = frequency of the class preceding the modal class = 90 , f 2 = frequency of the class succeeding the modal class = 100 , h = size of the modal class interval = 5000 , Mode = l + f 1 - f 0 2 f 1 - f 0 - f 2 × h = 5000 + 150 - 90 2 × 150 - 90 - 100 × 5000 = 5000 + 60 300 - 190 × 5000 = 5000 + 60 110 × 5000 = 5000 + 2727 . 27 = 7727 . 27 approx Hence, the mode is ₹ 7727 . 27 .

Given below is the frequency distribution of the heights of players in a school: Heights (in cm ) 160 - 162 163 - 165 166 - 168 169 - 171 172 - 174 Number of students: 15 118 142 127 18 Find the average height of maximum number of students.

The given frequency distribution is not continuous. We convert it into continuous by subtracting the conversion factor from the lower limit and adding it to the upper limit of each class interval. Here, the conversion factor = 163 - 162 2 = 1 2 = 0 . 5 . Thus, we have the table, Heights (in cm ) Class interval Number of students f i 160 - 162 159 . 5 - 162 . 5 15 163 - 165 162 . 5 - 165 . 5 118 = f 0 166 - 168 165 . 5 - 168 . 5 = Modal class 142 = f 1 169 - 171 168 . 5 - 171 . 5 127 = f 2 172 - 174 171 . 5 - 174 . 5 18 Therefore, the class interval 165 . 5 - 168 . 5 has maximum frequency. So, 165 . 5 - 168 . 5 is the modal class. Here, l = lower limit of the modal class = 165 . 5 , f 1 = frequency of modal class = 142 , f 0 = frequency of the class preceding the modal class = 118 , f 2 = frequency of the class succeeding the modal class = 127 , h = size of the modal class interval = 3 , Mode = l + f 1 - f 0 2 f 1 - f 0 - f 2 × h = 165 . 5 + 142 - 118 2 × 142 - 118 - 127 × 3 = 165 . 5 + 24 284 - 245 × 3 = 165 . 5 + 24 39 × 3 = 165 . 5 + 1 . 85 = 167 . 35 approx Hence, the average height of the maximum number of students is 167 . 35 cm .

Given below is the distribution of cases of a certain disease admitted during a year in a particular hospital: Age (in years) 6 - 15 16 - 25 26 - 35 36 - 45 46 - 55 56 - 65 Number of cases 6 11 21 23 14 5 Find the average age for which maximum cases occurred.

The given frequency distribution is not continuous. We convert it into continuous by subtracting the conversion factor from the lower limit and adding it to the upper limit of each class interval. Here, the conversion factor = 16 - 15 2 = 1 2 = 0 . 5 Thus, we have the table, Heights (in cm ) Class interval Number of students f i 6 - 15 5 . 5 - 15 . 5 6 16 - 25 15 . 5 - 25 . 5 11 26 - 35 25 . 5 - 35 . 5 21 = f 0 36 - 45 35 . 5 - 45 . 5 = Modal class 23 = f 1 36 - 55 45 . 5 - 55 . 5 14 = f 2 56 - 65 55 . 5 - 65 . 5 5 Therefore, the class interval 35 . 5 - 45 . 5 has maximum frequency. So, 35 . 5 - 45 . 5 is the modal class. Here, l = lower limit of the modal class = 35 . 5 , f 1 = frequency of modal class = 23 , f 0 = frequency of the class preceding the modal class = 21 , f 2 = frequency of the class succeeding the modal class = 14 , h = size of the modal class interval = 10 , Mode = l + f 1 - f 0 2 f 1 - f 0 - f 2 × h = 35 . 5 + 23 - 21 2 × 23 - 21 - 14 × 10 = 35 . 5 + 2 46 - 35 × 10 = 35 . 5 + 2 11 × 10 = 35 . 5 + 1 . 82 = 37 . 32 approx Hence, the average age for which maximum cases occurred is 37 . 32 years.

The following table shows the age distribution of cases of a certain disease admitted during a year in a particular hospital: Age (in years) 5 - 14 15 - 24 25 - 34 35 - 44 45 - 54 55 - 64 Number of cases 6 11 21 23 14 5 Find the average age for which maximum cases occurred.

The given frequency distribution is not continuous. We convert it into continuous by subtracting the conversion factor from the lower limit and adding it to the upper limit of each class interval. Here, the conversion factor = 15 - 14 2 = 1 2 = 0 . 5 Thus, we have the table, Heights (in cm ) Class interval Number of students f i 5 - 14 4 . 5 - 14 . 5 6 15 - 24 14 . 5 - 24 . 5 11 25 - 34 24 . 5 - 34 . 5 21 = f 0 35 - 44 34 . 5 - 44 . 5 = Modal class 23 = f 1 35 - 54 44 . 5 - 54 . 5 14 = f 2 55 - 64 54 . 5 - 64 . 5 5 Therefore, the class interval 34 . 5 - 44 . 5 has maximum frequency. So, 34 . 5 - 44 . 5 is the modal class. Here, l = lower limit of the modal class = 34 . 5 , f 1 = frequency of modal class = 23 , f 0 = frequency of the class preceding the modal class = 21 , f 2 = frequency of the class succeeding the modal class = 14 , h = size of the modal class interval = 10 , Mode = l + f 1 - f 0 2 f 1 - f 0 - f 2 × h = 34 . 5 + 23 - 21 2 × 23 - 21 - 14 × 10 = 34 . 5 + 2 46 - 35 × 10 = 34 . 5 + 2 11 × 10 = 34 . 5 + 1 . 81 = 36 . 31 approx Hence, the average age for which maximum cases occurred is 36 . 31 years.

The following table shows the ages of the patients admitted in a hospital during a year: Age (in years) 5 - 15 15 - 25 25 - 35 35 - 45 45 - 55 55 - 65 Number of patients 6 11 21 23 14 5 Find the mode and the mean of the data given above. Compare and interpret the two measures of central tendency.

In the given frequency distribution class 35 - 45 has maximum frequency. So, 35 - 45 is the modal class. Here, l = lower limit of the modal class = 35 , f 1 = frequency of modal class = 23 , f 0 = frequency of the class preceding the modal class = 21 , f 2 = frequency of the class succeeding the modal class = 14 , h = size of the modal class interval = 10 , Mode = l + f 1 - f 0 2 f 1 - f 0 - f 2 × h = 35 + 23 - 21 2 × 23 - 21 - 14 × 10 = 35 + 2 46 - 35 × 10 = 35 + 2 11 × 10 = 35 + 1 . 82 = 36 . 82 approx . The given frequency distribution table: Age (in years) Number of patients f i Class Mark x i Deviations d i = x i - 30 u i = d i h = x i - a 10 f i u i 5 - 15 6 5 + 15 2 = 20 2 = 10 10 - 30 = - 20 - 20 10 = - 2 - 12 15 - 25 11 20 - 10 - 1 - 11 25 - 35 21 30 = a 0 0 0 35 - 45 23 40 10 1 23 45 - 55 14 50 20 2 28 55 - 65 5 60 30 3 15 ∑ f i = 80 ∑ f i u i = 43 Let a = 30 be the assumed mean and class size h = 10 , then we have, Mean ( x ¯ ) = a + Σ f i u i Σ f i × h = 30 + 43 80 × 10 = 30 + 5 . 375 = 35 . 375 . Hence, the maximum number of patients admitted in the hospital of the age 36 . 82 years (approx), which on an average the age of a patient is 35 . 375 years.

J P Mohindru and Bharat Mohindru Solutions for Chapter: Statistics, Exercise 2: EXERCISE

Author:J P Mohindru & Bharat Mohindru

J P Mohindru Mathematics Solutions for Exercise - J P Mohindru and Bharat Mohindru Solutions for Chapter: Statistics, Exercise 2: EXERCISE

Attempt the free practice questions on Chapter 14: Statistics, Exercise 2: EXERCISE 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 2: EXERCISE with Hints & Solutions

EASY

10th CBSE

IMPORTANT

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

Find the mode of the following frequency distribution:

Monthly salary (in $₹$ )	$0 - 5000$	$5000 - 10000$	$10000 - 15000$	$15000 - 20000$	$20000 - 25000$	$25000 - 30000$
Number of employees	$90$	$150$	$100$	$80$	$70$	$10$

MEDIUM

10th CBSE

IMPORTANT

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

Given below is the frequency distribution of the heights of players in a school:

Heights (in $cm$ )	$160 - 162$	$163 - 165$	$166 - 168$	$169 - 171$	$172 - 174$
Number of students:	$15$	$118$	$142$	$127$	$18$

Find the average height of maximum number of students.

MEDIUM

10th CBSE

IMPORTANT

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

Given below is the distribution of cases of a certain disease admitted during a year in a particular hospital:

Age (in years)	$6 - 15$	$16 - 25$	$26 - 35$	$36 - 45$	$46 - 55$	$56 - 65$
Number of cases	$6$	$11$	$21$	$23$	$14$	$5$

Find the average age for which maximum cases occurred.

MEDIUM

10th CBSE

IMPORTANT

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

The mode of the following data is $36$ . Find the missing frequency in it.

Class interval	$0 - 10$	$10 - 20$	$20 - 30$	$30 - 40$	$40 - 50$	$50 - 60$	$60 - 70$
Frequency	$8$	$10$	_____	$16$	$12$	$6$	$7$

MEDIUM

10th CBSE

IMPORTANT

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

Find the missing frequency $(x)$ of the following data if its mode is $240$ .

Expenditure	$0 - 100$	$100 - 200$	$200 - 300$	$300 - 400$	$400 - 500$
Number of families	$140$	$230$	$270$	$x$	$150$

MEDIUM

10th CBSE

IMPORTANT

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

The following table shows the age distribution of cases of a certain disease admitted during a year in a particular hospital:

Age (in years)	$5 - 14$	$15 - 24$	$25 - 34$	$35 - 44$	$45 - 54$	$55 - 64$
Number of cases	$6$	$11$	$21$	$23$	$14$	$5$

Find the average age for which maximum cases occurred.

MEDIUM

10th CBSE

IMPORTANT

Modern's abc+ of Mathematics for Class 10>Chapter 14 - Statistics>EXERCISE>Q 12

The following table shows the ages of the patients admitted in a hospital during a year:

Age (in years)	$5 - 15$	$15 - 25$	$25 - 35$	$35 - 45$	$45 - 55$	$55 - 65$
Number of patients	$6$	$11$	$21$	$23$	$14$	$5$

Find the mode and the mean of the data given above. Compare and interpret the two measures of central tendency.

MEDIUM

10th CBSE

IMPORTANT

Modern's abc+ of Mathematics for Class 10>Chapter 14 - Statistics>EXERCISE>Q 13

The following frequency distribution gives monthly consumption of electricity of $68$ consumers of a locality. Find the mean and mode of the data and compare them:

Monthly consumption	$65 - 85$	$85 - 105$	$105 - 125$	$125 - 145$	$145 - 165$	$165 - 185$	$185 - 205$
Number of consumers	$4$	$5$	$13$	$20$	$14$	$8$	$4$

J P Mohindru and Bharat Mohindru Solutions for Chapter: Statistics, Exercise 2: EXERCISE

J P Mohindru Mathematics Solutions for Exercise - J P Mohindru and Bharat Mohindru Solutions for Chapter: Statistics, Exercise 2: EXERCISE

Questions from J P Mohindru and Bharat Mohindru Solutions for Chapter: Statistics, Exercise 2: EXERCISE with Hints & Solutions

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