Advertisements
Advertisements
Question
Find the mean of the distribution:
| Class | 1 – 3 | 3 – 5 | 5 – 7 | 7 – 10 |
| Frequency | 9 | 22 | 27 | 17 |
Solution
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)` |
| 1 – 3 | 2 | 9 | 18 |
| 3 – 5 | 4 | 22 | 88 |
| 5 – 7 | 6 | 27 | 162 |
| 7 – 10 | 8.5 | 17 | 144.5 |
| `sumf_i = 75` | `sumf_ix_i = 412.5` |
Therefore, mean `(barx) = (sumf_ix_i)/(sumf_i)`
= `412.5/75`
= 5.5
Hence, mean of the given distribution is 5.5.
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 |
If the mean of the following data is 20.6. Find the value of p.
| x | 10 | 15 | P | 25 | 35 |
| f | 3 | 10 | 25 | 7 | 5 |
If the mean of the following data is 15, find p.
| x | 5 | 10 | 15 | 20 | 25 |
| f | 6 | P | 6 | 10 | 5 |
The arithmetic mean of the following data is 14. Find the value of k
| x1 | 5 | 10 | 15 | 20 | 25 |
| f1 | 7 | k | 8 | 4 | 5 |
For the following distribution, calculate mean using all suitable methods:
| Size of item | 1 - 4 | 4 - 9 | 9 - 16 | 16 - 27 |
| Frequency | 6 | 12 | 26 | 20 |
The mean of the following frequency distribution is 62.8 and the sum of all the frequencies is 50. Compute the missing frequency f1 and f2.
| Class | 0 - 20 | 20 - 40 | 40 - 60 | 60 - 80 | 80 - 100 | 100 - 120 |
| Frequency | 5 | f1 | 10 | f2 | 7 | 8 |
If the mean of the following distribution is 27, find the value of p.
| Class | 0 - 10 | 10 - 20 | 20 - 30 | 30 - 40 | 40 - 50 |
| Frequency | 8 | p | 12 | 13 | 10 |
If the mean of 5 observation x, x + 2, x + 4, x + 6and x + 8 , find the value of x.
Find the mean of the following data, using assumed-mean method:
| Class | 0 – 20 | 20 – 40 | 40 – 60 | 60 – 80 | 80 – 100 | 100 - 120 |
| Frequency | 20 | 35 | 52 | 44 | 38 | 31 |
The following table shows the age distribution of patients of malaria in a village during a particular month:
| Age (in years) | 5 – 14 | 15 – 24 | 25 – 34 | 35 – 44 | 45 – 54 | 55 - 64 |
| No. of cases | 6 | 11 | 21 | 23 | 14 | 5 |
Find the average age of the patients.
The loans sanctioned by a bank for construction of farm ponds are shown in the following table. Find the mean of the loans.
|
Loan
(Thousand Rupees)
|
40 - 50 | 50 - 60 | 60 - 70 | 70 - 80 | 80 - 90 |
| No. of farm ponds | 13 | 20 | 24 | 36 | 7 |
There are three dealers A, B and C in Maharashtra. Suppose, the trade of each of them in september 2018 was as shown in the following table.
The rate of GST on each transaction was 5%.
Read the table and answer the questions below it.
| Dealer | GST collected on the sale |
GST paid at the time of purchase |
ITC | Tax paid to the Govt. |
Taxbalance with the Govt. |
| A | Rs.5000 | Rs. 6000 | Rs. 5000 | Rs. 0 | Rs. 1000 |
| B | Rs 5000 | Rs. 4000 | Rs. 4000 | Rs. 1000 | Rs. 0 |
| C | Rs.5000 | Rs. 5000 | Rs. 5000 | Rs. 0 | Rs. 0 |
(i) How much amount did the dealer A get by sale ?
(ii) For how much amount did the dealer B buy the articles ?
(iii) How much is the balance of CGST and SGST left with the government that was paid by A ?
The mean of n observation is `overlineX` .f the first item is increased by 1, second by 2 and so on, then the new mean is
If the mean of 6, 7, x, 8, y, 14 is 9, then ______.
Mean of a certain number of observation is `overlineX`. If each observation is divided by m(m ≠ 0) and increased by n, then the mean of new observation is
In the formula
The measurements (in mm) of the diameters of the head of the screws are given below :
| Diameter (in mm) | no. of screws |
| 33 - 35 | 9 |
| 36 - 38 | 21 |
| 39 - 41 | 30 |
| 42 - 44 | 22 |
| 45 - 47 | 18 |
Calculate the mean diameter of the head of a screw by the ' Assumed Mean Method'.
Find the mean of the following distribution:
| x | 10 | 30 | 50 | 70 | 89 |
| f | 7 | 8 | 10 | 15 | 10 |
Mean of n numbers x1, x2, … xn is m. If xn is replaced by x, then new mean is ______.
If mean = (3median - mode) . k, then the value of k is ______.
(For an arranged data) The median is the ______.
Is it true to say that the mean, mode and median of grouped data will always be different? Justify your answer
Find the mean for the following distribution:
| Class Interval | 20 - 40 | 40 - 60 | 60 - 80 | 80 - 100 |
| Frequency | 4 | 7 | 6 | 3 |
The distribution given below shows the runs scored by batsmen in one-day cricket matches. Find the mean number of runs.
| Runs scored |
0 – 40 | 40 – 80 | 80 – 120 | 120 – 160 | 160 – 200 |
| Number of batsmen |
12 | 20 | 35 | 30 | 23 |
The mean of the following frequency distribution is 25. Find the value of f.
| Class | 0 – 10 | 10 – 20 | 20 – 30 | 30 – 40 | 40 – 50 |
| Frequency | 5 | 18 | 15 | f | 6 |
Find the mean of the following frequency distribution:
| Class | 1 – 5 | 5 – 9 | 9 – 13 | 13 – 17 |
| Frequency | 4 | 8 | 7 | 6 |
Find the mean of the following frequency distribution:
| Class: | 10 – 15 | 15 – 20 | 20 – 25 | 25 – 30 | 30 – 35 |
| Frequency: | 4 | 10 | 5 | 6 | 5 |
The lengths of 40 leaves of a plant are measured correct to the nearest millimeter, and the data obtained is represented in the following table:
| Length (in mm) | Number of leaves |
| 118−126 | 3 |
| 127–135 | 5 |
| 136−144 | 9 |
| 145–153 | 12 |
| 154–162 | 5 |
| 163–171 | 4 |
| 172–180 | 2 |
Find the mean length of the leaves.
