The following table shows the marks scored by 140 students in an examination of a certain paper. Marks 0 - 10 10 - 20 20 - 30 30 - 40 40 - 50 Number of students 20 24 40 36 20 Calculate the average marks by using all the three methods: Direct Method, Assumed Mean Method and Step Deviation Method.

From the given data, we can plot the below tables: Direct Method: Class interval Frequency f i Class Mark x i f i x i 0 - 10 20 0 + 10 2 = 10 2 = 5 100 10 - 20 24 15 360 20 - 30 40 25 1000 30 - 40 36 35 1260 40 - 50 20 45 900 ∑ f i = 140 ∑ f i x i = 3620 Mean x ¯ = Σ f i x i Σ f i = 3620 140 = 25 . 857 approx Thus, the mean is 25 . 857 . Assumed Mean Method: Class interval Frequency f i Class Mark x i Deviations d i = x i - 25 f i d i 0 - 10 20 0 + 10 2 = 10 2 = 5 5 - 25 = - 20 - 400 10 - 20 24 15 - 10 - 240 20 - 30 40 25 = a 0 0 30 - 40 36 35 10 360 40 - 50 20 45 20 400 ∑ f i = 140 ∑ f i d i = 120 Let, a = 25 be the assumed mean, then we have, Mean x ¯ = a + Σ f i d i Σ f i = 25 + 120 140 = 25 + 0 . 857 = 25 . 857 approx Thus, the mean is 25 . 857 . Step Deviation Method: Class interval Frequency f i Class Mark x i Deviations d i = x i - 25 u i = d i h = x i - 25 10 f i u i 0 - 10 20 0 + 10 2 = 10 2 = 5 5 - 25 = - 20 - 20 10 = - 2 - 40 10 - 20 24 15 - 10 - 1 - 24 20 - 30 40 25 = a 0 0 0 30 - 40 36 35 10 1 36 40 - 50 20 45 20 2 40 ∑ f i = 140 ∑ f i u i = 12 Let a = 25 be the assumed mean and class size h = 10 , then we have, Mean x ¯ = a + Σ f i u i Σ f i × h = 25 + 12 140 × 10 = 25 + 6 7 = 25 + 0 . 857 = 25 . 857 approx Thus, the average marks are 25 . 857 .

Find the mode of the following frequency distribution: Class interval 10 - 15 15 - 20 20 - 25 25 - 30 30 - 35 35 - 40 Frequency 30 45 75 35 25 15

In the given frequency distribution class 20 - 25 has maximum frequency. So, 20 - 25 is the modal class Class interval Frequency 10 - 15 30 15 - 20 45 = f 0 20 - 25 75 = f 1 25 - 30 35 = f 2 30 - 35 25 35 - 40 15 Here, l = lower limit of the modal class = 20 , f 1 = frequency of modal class = 75 , f 0 = frequency of the class preceding the modal class = 45 , f 2 = frequency of the class succeeding the modal class = 35 , h = size of the modal class interval = 5 , Mode = l + f 1 - f 0 2 f 1 - f 0 - f 2 × h = 20 + 75 - 45 2 × 75 - 45 - 35 × 5 = 20 + 30 150 - 80 × 5 = 20 + 30 70 × 5 = 20 + 2 . 14 = 22 . 14 approx Hence, the mode is 22 . 14 .

Find the mode of the following frequency distribution: Class interval 0 - 4 4 - 8 8 - 12 12 - 16 16 - 20 20 - 24 24 - 28 28 - 32 32 - 36 36 - 40 Frequency 0 7 9 17 12 10 6 3 1 0

In the given frequency distribution class 12 - 16 has maximum frequency So, 12 - 16 is the modal class. Here, l = lower limit of the modal class = 12 , f 1 = frequency of modal class = 17 , f 0 = frequency of the class preceding the modal class = 9 , f 2 = frequency of the class succeeding the modal class = 12 , h = size of the modal class interval = 4 Mode = l + f 1 - f 0 2 f 1 - f 0 - f 2 × h = 12 + 17 - 9 2 × 17 - 9 - 12 × 4 = 12 + 8 34 - 21 × 4 = 12 + 8 13 × 4 = 12 + 2 . 46 = 14 . 46 ( approx. ) Hence, the mode is 14 . 46 .

Find the mode of the following frequency distribution: Age (in years) 0 - 5 5 - 10 10 - 15 15 - 20 20 - 25 25 - 30 30 - 35 Number of patients 6 11 18 24 17 13 5

In the given frequency distribution class 15 - 20 has maximum frequency. So, 15 - 20 is the modal class. Here, l = lower limit of the modal class = 15 , f 1 = frequency of modal class = 24 , f 0 = frequency of the class preceding the modal class = 18 , f 2 = frequency of the class succeeding the modal class = 17 , h = size of the modal class interval = 5 , Mode = l + f 1 - f 0 2 f 1 - f 0 - f 2 × h = 15 + 24 - 18 2 × 24 - 18 - 17 × 5 = 15 + 6 48 - 35 × 5 = 15 + 6 13 × 5 = 15 + 2 . 3 = 17 . 3 approx Hence, the mode is 17 . 3 years.

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 .

MEDIUM

10th CBSE

IMPORTANT

Earn 100

Consider the following distribution of daily wages of $50$ workers of a factory.

Daily wages (in $₹$ ) $100 - 120$ $120 - 140$ $140 - 160$ $160 - 180$ $180 - 200$

No. of workers $12$ $14$ $8$ $6$ $10$

Find the mean daily wage (in $₹$ ) of the workers of the factory by using an appropriate method.

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Important Questions on Statistics

HARD

10th CBSE

IMPORTANT

Modern's abc+ of Mathematics for Class 10>Chapter 14 - Statistics>EXERCISE>Q 21

The following table shows the marks scored by $140$ students in an examination of a certain paper.

Marks	$0 - 10$	$10 - 20$	$20 - 30$	$30 - 40$	$40 - 50$
Number of students	$20$	$24$	$40$	$36$	$20$

Calculate the average marks by using all the three methods:

Direct Method, Assumed Mean Method and Step Deviation Method.

HARD

10th CBSE

IMPORTANT

Modern's abc+ of Mathematics for Class 10>Chapter 14 - Statistics>EXERCISE>Q 22

A frequency distribution of the life times of $400$ T.V picture tubes tested in a tube company is given below. Find the average life of the tube (in $hours$ ).

Life time (in $hours$ )	$300 - 399$	$400 - 499$	$500 - 599$	$600 - 699$	$700 - 799$
Frequency	$14$	$46$	$58$	$76$	$68$

Life time (in $hours$ )	$800 - 899$	$900 - 999$	$1000 - 1099$	$1100 - 1199$
Frequency	$62$	$48$	$22$	$6$

EASY

10th CBSE

IMPORTANT

Modern's abc+ of Mathematics for Class 10>Chapter 14 - Statistics>EXERCISE>Q 1