Find the arithmetic mean of the following frequency distribution: Class interval 25 - 29 30 - 34 35 - 39 40 - 44 45 - 49 50 - 54 55 - 59 Frequency 14 22 16 6 5 3 4

Here, the class interval is non-overlapping. ∴ Class size h = 29 - 25 + 30 - 29 = 4 + 1 = 5 Class interval Frequency f i Class Mark x i Deviations d i = x i - 42 u i = d i h = x i - 42 5 f i u i 25 - 29 14 25 + 29 2 = 54 2 = 27 27 - 42 = - 15 - 15 5 = - 3 - 42 30 - 34 22 32 - 10 - 2 - 44 35 - 39 16 37 - 5 - 1 - 16 40 - 44 6 42 = a 0 0 0 45 - 49 5 47 5 1 5 50 - 54 3 52 10 2 6 55 - 59 4 57 15 3 12 ∑ f i = 70 ∑ f i u i = - 79 Let, the assumed mean a = 42 and class size h = 5 , then we have, Mean x ¯ = a + Σ f i u i Σ f i × h = 42 + ( - 79 ) 70 × 5 = 42 - 79 14 = 42 - 5 . 64 = 36 . 36 approx Hence, the mean is 36 . 36 .

Find the mean age (in years) from the following frequency distribution: Age (in years) 15 - 19 20 - 24 25 - 29 30 - 34 35 - 39 40 - 44 45 - 49 Frequency 3 13 21 15 5 4 2

Here, the class interval is non-overlapping. ∴ Class size h = 19 - 15 + 20 - 19 = 4 + 1 = 5 Age (in years) Frequency f i Class Mark x i Deviations d i = x i - 32 u i = d i h = x i - 32 5 f i u i 15 - 19 3 15 + 19 2 = 34 2 = 17 17 - 32 = - 15 - 15 5 = - 3 - 9 20 - 24 13 22 - 10 - 2 - 26 25 - 29 21 27 - 5 - 1 - 21 30 - 34 15 32 = a 0 0 0 35 - 39 5 37 5 1 5 40 - 44 4 42 10 2 8 45 - 49 2 47 15 3 6 ∑ f i = 63 ∑ f i u i = - 37 Let, the assumed mean a = 32 and class size h = 5 , then we have, Mean x ¯ = a + Σ f i u i Σ f i × h = 32 + ( - 37 ) 63 × 5 = 32 - 185 63 = 32 – 2 . 94 = 29 . 06 approx Hence, the mean age is 29 . 06 years.

Find the arithmetic mean of the following frequency distribution: Marked below 10 20 30 40 50 60 70 80 90 100 No. of students 12 22 35 50 70 86 97 104 109 115

The given data may be written as: Class interval 0 - 10 10 - 20 20 - 30 30 - 40 40 - 50 50 - 60 60 - 70 70 - 80 80 - 90 90 - 100 No. of students 12 22 - 12 = 10 35 - 22 = 13 50 - 35 = 15 70 - 50 = 20 86 - 70 = 16 97 - 86 = 11 104 - 97 = 7 109 - 104 = 5 115 - 109 = 6 We prepare the table as below: Class interval Frequency f i Class Mark x i Deviations d i = x i - 55 u i = d i h = x i - 55 10 f i u i 0 - 10 12 0 + 10 2 = 10 2 = 5 5 - 55 = - 50 - 50 10 = - 5 - 60 10 - 20 10 15 - 40 - 4 - 40 20 - 30 13 25 - 30 - 3 - 39 30 - 40 15 35 - 20 - 2 - 30 40 - 50 20 45 - 10 - 1 - 20 50 - 60 16 55 = a 0 0 0 60 - 70 11 65 10 1 11 70 - 80 7 75 20 2 14 80 - 90 5 85 30 3 15 90 - 100 6 95 40 4 24 ∑ f i = 115 ∑ f i u i = - 125 Let, the assumed mean a = 55 and class size h = 10 , then we have, Mean x ¯ = a + Σ f i u i Σ f i × h = 55 + - 125 115 × 10 = 55 - 10 . 87 = 44 . 13 approx Hence, the arithmetic mean is 44 . 13 .

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 .

MEDIUM

10th CBSE

IMPORTANT

Earn 100

Find the mean, by Step Deviation Method.

Class interval $0 - 50$ $50 - 100$ $100 - 150$ $150 - 200$ $200 - 250$ $250 - 300$

Frequency $17$ $35$ $43$ $40$ $21$ $24$

Important Questions on Statistics

MEDIUM

10th CBSE

IMPORTANT

Modern's abc+ of Mathematics for Class 10>Chapter 14 - Statistics>EXERCISE>Q 17

Find the arithmetic mean of the following frequency distribution:

Class interval	$25 - 29$	$30 - 34$	$35 - 39$	$40 - 44$	$45 - 49$	$50 - 54$	$55 - 59$
Frequency	$14$	$22$	$16$	$6$	$5$	$3$	$4$

MEDIUM

10th CBSE

IMPORTANT

Modern's abc+ of Mathematics for Class 10>Chapter 14 - Statistics>EXERCISE>Q 18

Find the mean age (in years) from the following frequency distribution:

Age (in years)	$15 - 19$	$20 - 24$	$25 - 29$	$30 - 34$	$35 - 39$	$40 - 44$	$45 - 49$
Frequency	$3$	$13$	$21$	$15$	$5$	$4$	$2$

HARD

10th CBSE

IMPORTANT

Modern's abc+ of Mathematics for Class 10>Chapter 14 - Statistics>EXERCISE>Q 19

Find the arithmetic mean of the following frequency distribution:

Marked below	$10$	$20$	$30$	$40$	$50$	$60$	$70$	$80$	$90$	$100$
No. of students	$12$	$22$	$35$	$50$	$70$	$86$	$97$	$104$	$109$	$115$

MEDIUM

10th CBSE

IMPORTANT

Modern's abc+ of Mathematics for Class 10>Chapter 14 - Statistics>EXERCISE>Q 20

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.

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

Find the mode of the following frequency distribution:

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

EASY

10th CBSE

IMPORTANT

Modern's abc+ of Mathematics for Class 10>Chapter 14 - Statistics>EXERCISE>Q 2

Find the mode of the following frequency distribution:

Class interval	$10 - 15$	$15 - 20$	$20 - 25$	$25 - 30$	$30 - 35$	$35 - 40$
Frequency	$30$	$45$	$75$	$35$	$25$	$15$

Class interval	$0 - 50$	$50 - 100$	$100 - 150$	$150 - 200$	$200 - 250$	$250 - 300$
Frequency	$17$	$35$	$43$	$40$	$21$	$24$

Find the mean, by Step Deviation Method. Class interval 0-50 50-100 100-150 150-200 200-250 250-300 Frequency 17 35 43 40 21 24

Important Questions on Statistics

Learn and Prepare for any exam you want !

Find the mean, by Step Deviation Method.

Class interval $0 - 50$ $50 - 100$ $100 - 150$ $150 - 200$ $200 - 250$ $250 - 300$

Frequency $17$ $35$ $43$ $40$ $21$ $24$