**Question 1: Find the mean of the following set of numbers:**

**(i) 6, 9, 11, 12 and 7**

**(ii) 11, 14, 23, 26, 10, 12, 18 and 6**

**Solution 1:**

(i) x̄ = (x

_{1}+ x

_{2}+...+ x

_{n})/5

Here n = 5

∴ x̄ = (6.9 + 11 + 12 + 7)/5 = 45/5 = 9

(ii) x̄ = (x

Here n = 8

∴ x̄ = (11 + 14 + 23 + 26 + 10 + 12 + 18 + 6)/8 = 120/8 = 15

x̄ = (x

∴ x̄ = (60 + 67 + 52 + 76 + 50 + 51 + 74 + 45 + 56)/9 = 531/9 = 59

_{1}+ x_{2}+…+ x_{n})/nHere n = 8

∴ x̄ = (11 + 14 + 23 + 26 + 10 + 12 + 18 + 6)/8 = 120/8 = 15

**Question 2: Marks obtained in mathematics) by 9 students are given below:****60, 67, 52, 76, 50, 51, 74, 45 and 56****(a) find the arithmetic mean****(b) if marks of each student be increased by 4; what will be the new value of arithmetic mean.**

**Solution 2:**(a) Here n = 9x̄ = (x

_{1}+ x_{2}+……+ x_{n})/n∴ x̄ = (60 + 67 + 52 + 76 + 50 + 51 + 74 + 45 + 56)/9 = 531/9 = 59

(b) If marks of each student be increased by 4 then new arithmetic mean will be = 59 + 4 = 63

Here n = 10

x̄ = (x

∴ x̄ = (3 + 4 + 5 + 6 + 7 + 8 + 9 + 10 + 11 + 12)/10 = 75/10 = 7.5

x̄ = (7 + 11 + 6 + 5 + 6)/5 = 35/5 = 7

(b) If we subtract 2 from each number, then the mean will be 7-2 = 5

⇒ 9324 + 148f = 8645 + 245f

⇒ 245f – 148f = 9324 – 8645

⇒ f = 679/97

⇒ f = 7

Let the assumed mean A = 72.5

**Question 3: Find the mean of the natural numbers from 3 to 12.**

**Solution 3:**Numbers between 3 to 12 are 3, 4, 5, 6, 7, 8, 9, 10, 11 and 12.Here n = 10

x̄ = (x

_{1}+ x_{2}+…….x_{n})/n∴ x̄ = (3 + 4 + 5 + 6 + 7 + 8 + 9 + 10 + 11 + 12)/10 = 75/10 = 7.5

**Question 4:****(a) Find the mean of 7, 11, 6, 5, and 6****(b) If each number given in (a) is diminished by 2, find the new value of the mean.**

**Solution 4:**(a) The mean of 7, 11, 6, 5 and 6x̄ = (7 + 11 + 6 + 5 + 6)/5 = 35/5 = 7

(b) If we subtract 2 from each number, then the mean will be 7-2 = 5

**Question 5: The mean of the following frequency distribution is 21 1/7. Find the value of ‘f’.**

**Solution 5:**x̄ = 21 1/7 = 148/7
x̄ = Î£f

⇒ 148 = (1235 + 35f)/(63 + f)_{i}x_{i}/Î£f_{i}= (1235 + 35f)/(63 + f)⇒ 9324 + 148f = 8645 + 245f

⇒ 245f – 148f = 9324 – 8645

⇒ f = 679/97

⇒ f = 7

**Question 6: Using step-deviation method, calculate the mean marks of the following distribution.****Solution 6:**Let the assumed mean A = 72.5

Mean = A + Î£f

x̄ = Î£f

(i) Median = (120 + 1)/2 = 60.5

Through mark 60.5, draw a parallel line to x-axis which meets the curve at A. From A draw a perpendicular to x-axis meeting it at B.

The value of point B is the median = 43

_{i}d_{i}/Î£f_{i}= 72.5 + (-280/80) = 69

**Question 7: Using the information given in the adjoining histogram, calculate the mean.****Solution 7:**_{x}/Î£f = 2850/75 = 38

**Question 8: By drawing an ogive, estimate the median for the following frequency distribution:**

**Solution 8:**
Number of terms = 55

∴ Median = (55 + 1

^{th})/2 term = 28^{th}term
Through mark of 28 on the y-axis, draw a line parallel to x-axis which meets the curve at A, From A, draw a perpendicular to x-axis, which meets x-axis at B.

The value of B is the median which is 18.4 kg

**Question 9: The marks obtained by 120 students in a mathematics test is given below:****Draw an ogive for the given distribution on a graph sheet. Use a suitable scale for your ogive. Use your ogive to estimate:****(i) the median****(ii) the number of students who obtained more than 75% in test.****(iii) the number of students who did not pass in the test if the pass percentage was 40.****(iv) the lower quartile**

**Solution 9:**^{th}termThrough mark 60.5, draw a parallel line to x-axis which meets the curve at A. From A draw a perpendicular to x-axis meeting it at B.

The value of point B is the median = 43

(ii) Number of students who obtained up to 75% marks in the test = 120

Number of students who obtained more than 75% mark in the test = 120 – 110 = 10

Number of students who obtained more than 75% mark in the test = 120 – 110 = 10

(iii) Number of students who obtained less than 40% marks in the test = 52(from the graph;x = 40, y = 52)

(iv) Lower quartile = Q

_{1}= 120 × ¼ = 30^{th }term = 30

**Question 10: Find the mean for the following frequency distribution:****Solution 10:**
Mean = Î£f

_{x}/Î£f = 7150/50 = 143