Mathematics
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 :
| 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) 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
Measures of Central Tendency
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Answer
Table :
| Class interval (x) | Number of parcels (f) | Class mark (x) | fx |
|---|---|---|---|
| 0-100 | 10 | 50 | 500 |
| 100-200 | 12 | 150 | 1800 |
| 200-300 | 13 | 250 | 3250 |
| 300 - 400 | 19 | 350 | 6650 |
| 400 - 500 | 4 | 450 | 1800 |
| 500 - 600 | 2 | 550 | 1100 |
| Total | ∑ f = 60 | ∑ fx = 15100 |
35. From table,
The number of parcels for the 300 – 400 range = 19 (54 - 35).
Hence, option (d) is the correct option.
36.From table,
The number of parcels for the 200 – 300 range = 13 (35 - 22).
Hence, option (c) is the correct option.
37.The lowest frequency is 2, which occurs in the 500 – 600 range.
Hence, option (d) is the correct option.
38. By formula,
Hence, option (b) is the correct option.
Answered By
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Related Questions
Which of the following cannot be determined graphically?
Mean
Median
Mode
none of these
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) 1.8Assertion (A): For a grouped frequency distribution, we use Mean = A + × h to find the mean using step deviation method.
Reason (R): Here t = .
A is true, R is false
A is false, R is true
Both A and R are true
Both A and R are false
Assertion (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