Advertisements
Advertisements
Question
Calculate the mean of the following data:
| Class | 4 – 7 | 8 – 11 | 12 – 15 | 16 – 19 |
| Frequency | 5 | 4 | 9 | 10 |
Solution
Since, given data is not continuous, so we subtract 0.5 from the lower limit and add 0.5 in the upper limit of each class.
Now, we first find the class mark xi of each class and then proceed as follows.
| Class | Class marks `(bb(x_i))` |
Frequency `(bb(f_i))` |
`bb(f_ix_i)` |
| 3.5 – 7.5 | 5.5 | 5 | 27.5 |
| 7.5 – 11.5 | 9.5 | 4 | 38 |
| 11.5 – 15.5 | 13.5 | 9 | 121.5 |
| 15.5 – 19.5 | 17.5 | 10 | 175 |
| `sumf_i = 28` | `sumf_ix_i = 362` |
Therefore, mean `(barx) = (sumf_ix_i)/(sumf_i)`
= `362/28`
= 12.93
Hence, mean of the given data is 12.93.
APPEARS IN
RELATED QUESTIONS
The following table gives the literacy rate (in percentage) of 35 cities. Find the mean literacy rate.
| Literacy rate (in %) | 45 − 55 | 55 − 65 | 65 − 75 | 75 − 85 | 85 − 95 |
| Number of cities | 3 | 10 | 11 | 8 | 3 |
Frequency distribution of daily commission received by 100 salesmen is given below :
|
Daily Commission (in Rs.) |
No. of Salesmen |
| 100-120 | 20 |
| 120-140 | 45 |
| 140-160 | 22 |
| 160-180 | 09 |
| 180-200 | 04 |
Find mean daily commission received by salesmen, by the assumed mean method.
Find the mean of each of the following frequency distributions
| Class interval | 0 - 8 | 8 - 16 | 16 - 24 | 24 - 32 | 32 - 40 |
| Frequency | 5 | 9 | 10 | 8 | 8 |
Find the mean of the following data using step-deviation method:
| Class | 500 – 520 | 520 – 540 | 540 – 560 | 560 – 580 | 580 – 600 | 600 – 620 |
| Frequency | 14 | 9 | 5 | 4 | 3 | 5 |
Find the mean of the following distribution:
| x | 10 | 30 | 50 | 70 | 89 |
| f | 7 | 8 | 10 | 15 | 10 |
Find the mean wage of a worker from the following data:
| Wages (In ₹) | 1400 | 1450 | 1500 | 1550 | 1600 | 1650 | 1700 |
| Number of workers | 15 | 20 | 18 | 27 | 15 | 3 | 2 |
Mean of 100 items is 49. It was discovered that three items which should have been 60, 70, 80 were wrongly read as 40, 20, 50 respectively. The correct mean is ______.
If xi’s are the midpoints of the class intervals of grouped data, fi’s are the corresponding frequencies and `barx` is the mean, then `sum(f_ix_i - barx)` is equal to ______.
In calculating the mean of grouped data, grouped in classes of equal width, we may use the formula `barx = a + (sumf_i d_i)/(sumf_i)` where a is the assumed mean. a must be one of the mid-points of the classes. Is the last statement correct? Justify your answer.
The mean of 6 distinct observations is 6.5 and their variance is 10.25. If 4 out of 6 observations are 2, 4, 5 and 7, then the remaining two observations are ______.
