Directions : The marks obtained by 10 students in a class-test were as follows :

36, 64, 48, 52, 57, 73, 26, 39, 78, 67

31. The mean marks of the whole class is :

(a) 49.1
(b) 53.7
(c) 54
(d) 60

32. If the maximum marks in the test were 80, the mean percentage of marks obtained by the students is :

(a) 65%
(b) 67.5%
(c) 68%
(d) 72%

33. The mean marks of the top 5 scorers in the class is :

(a) 65.4
(b) 66.8
(c) 67.2
(d) 67.8

34. As per the Board’s instruction each student who obtained less than 50 marks was awarded 3 grace marks. The new mean of the marks thus obtained increases by :

(a) 0.9
(b) 1.2
(c) 1.5
(d) 1.8

Directions :
At a courier company, a daily report of parcels received for dispatch is prepared every evening, which classifies the parcels on the basis of their weights. The report of a certain day is as under :

35.How many parcels have weights in the range of 300 – 400 grams?
(a) 4
(b) 12
(c) 13
(d) 19

36.How many parcels have weights in the range of 200 – 300 grams?
(a) 10
(b) 12
(c) 13
(d) 19

37.In which of the following weight ranges, the number of parcels is the lowest?
(a) 100 – 200
(b) 200 – 300
(c) 400 – 500
(d) 500 – 600

38.The mean weight of parcels received on that particular day is :
(a) 226.78 g
(b) 251.67 g
(c) 284.28 g
(d) 302.16 g

Assertion (A): For a grouped frequency distribution, we use Mean = A + $\Big(\dfrac{\Sigma f t}{\Sigma f}\Big)$ × h to find the mean using step deviation method.

Reason (R): Here t = $\Big(\dfrac{x − A}{h}\Big)$.

1. A is true, R is false

2. A is false, R is true

3. Both A and R are true

4. Both A and R are false

Assertion (A): If x_i’s are the mid-points of the class intervals of a grouped data, Σf_i’s are the corresponding frequencies and x̄ is the mean, then Σf_i(x_i − x̄) = 1.

Reason (R): The sum of the deviations from the mean is 0.

1. A is true, R is false

2. A is false, R is true

3. Both A and R are true

4. Both A and R are false