Mathematics
Assertion (A): For a grouped frequency distribution, we use Mean = A + × h to find the mean using step deviation method. Reason (R): Here t = . 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
Related Questions
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) 6032. 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) 67.834. 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) 1.8Directions :
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 :Weight of parcel (in grams) (x) Number of parcels Below 600 60 Below 500 58 Below 400 54 Below 300 35 Below 200 22 Below 100 10 35.How many parcels have weights in the range of 300 – 400 grams?
(a) 4
(b) 12
(c) 13
(d) 1936.How many parcels have weights in the range of 200 – 300 grams?
(a) 10
(b) 12
(c) 13
(d) 1937.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 – 60038.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 gAssertion (A): If xi’s are the mid-points of the class intervals of a grouped data, Σfi’s are the corresponding frequencies and x̄ is the mean, then Σfi(xi − x̄) = 1.
Reason (R): The sum of the deviations from the mean is 0.
A is true, R is false
A is false, R is true
Both A and R are true
Both A and R are false