The given table represents the monthly income of 100 families of a locality. Monthly income range (in Rs.) Number of families Income more than Rs. 10,000 100 Income more than Rs. 13,000 85 Income more than Rs. 16,000 69 Income more than Rs. 19,000 50 Income more than Rs. 22,000 33 Income more than Rs. 25,000 15 The number of families having income range (in Rs.) 19000 - 22000 is:
17
The question provides a table showing the cumulative frequency distribution of monthly incomes for 100 families in a locality. The data is presented in a "more than" format, indicating the number of families whose income exceeds a certain threshold.
Here is the data given in the problem:
| Monthly income range (in Rs.) | Number of families |
|---|---|
| Income more than Rs. 10,000 | 100 |
| Income more than Rs. 13,000 | 85 |
| Income more than Rs. 16,000 | 69 |
| Income more than Rs. 19,000 | 50 |
| Income more than Rs. 22,000 | 33 |
| Income more than Rs. 25,000 | 15 |
This table tells us, for example, that 100 families have an income more than Rs. 10,000, and 85 families have an income more than Rs. 13,000, and so on.
We are asked to find the number of families whose monthly income is in the range of Rs. 19,000 - 22,000. To find the number of families in a specific range from a "more than" cumulative frequency table, we can use the following logic:
In this case, the range is Rs. 19,000 - 22,000.
From the table:
The number of families with income in the range Rs. 19,000 - 22,000 is the number of families with income more than Rs. 19,000 minus the number of families with income more than Rs. 22,000.
Calculation:
\text{Number of families in range } 19000 - 22000 = (\text{Families with income } > 19000) - (\text{Families with income } > 22000)
\text{Number of families in range } 19000 - 22000 = 50 - 33
\text{Number of families in range } 19000 - 22000 = 17
Therefore, the number of families having an income range of Rs. 19,000 - 22,000 is 17.
Based on the cumulative frequency data provided in the table, we determined that 17 families fall within the monthly income bracket of Rs. 19,000 to Rs. 22,000.
| Concept | Explanation | Application in this problem |
|---|---|---|
| Cumulative Frequency | The total frequency of a variable below a certain value or above a certain value. | The table gives "more than" cumulative frequencies. |
| "More Than" Cumulative Frequency | The total number of observations whose value is greater than the given value. | Each entry shows how many families earn more than the stated amount. |
| Finding Class Frequency from Cumulative Frequency | For a range (a - b), the frequency is (Cumulative frequency > a) - (Cumulative frequency > b) for a "more than" table. | Used to find families in the Rs. 19000 - 22000 range. |
Data can be presented in various frequency distribution formats to make it easier to understand and analyze. Two common types are:
Converting between these forms is a common task in statistics. To get the simple frequency for a class from a "more than" table, you subtract the cumulative frequency of the upper boundary from the cumulative frequency of the lower boundary.
Find the mode for the given distribution (rounded off to two decimal places).
| Class Interval | 5-10 | 10-15 | 15-20 | 20-25 | 25-30 | 30-35 |
| Frequency | 8 | 7 | 6 | 9 | 11 | 10 |
The arithmetic mean of the following data is _________.
23, 17, 20, 19, 21
The median of the following data will be _________.
32, 25, 33, 27, 35, 29 and 30
For a sample data, mean = 60 and median = 48. For this distribution, the mode is:
The mode of the following data is __________.
13, 15, 31, 12, 27, 13, 27, 30, 27, 28 and 16
The median of a set of 11 distinct observations is 73.2. If each of the largest five observations of the set is increased by 3, then the median of the new set:
What is the mode of the given data?
5, 7, 9, 7, 3, 7, 5, 7, 8, 6, 7
In tossing a coin, let the probability of turning up a head be p . The hypotheses are H 0∶ p = 0.4 vs H 1 ∶ p = 0.6. H 0is rejected if there are five or more heads in six tosses. Then the power of the test is:
For ANOVA two-way classification, to test two types of cloth in fashion trends, we have the following table.
Source of Variations | SS | Df | MSS | F-Ratio |
Variety A | 280 | 2 | 140 | 42.04 |
Variety B | α | 3 | 34.03 | |
Error | 20 | β | 3.33 | |
Total | 640 | 11 |
The arithematic average of Edgeworth - Marshal index number
What is the mode of the given data?
3, 0, 1, 0, 2, 1, 2, 0, 1, 2, 1, 1, 1, 3, 2What is the mode of the given data?
21, 22, 23, 23, 24, 21, 22, 23, 21, 23, 24, 23, 21, 23A bowler has taken 0, 3, 2, 1, 5, 3, 4, 5, 5, 2, 2, 0, 0, 1 and 2 wickets in 15 consecutive matches. What is the mode of the given data?
The data given below shows the number of people who have saved a certain amount of money.
Saving (In Rs.) | Number of people |
5 | 1 |
15 | 3 |
20 | 4 |
25 | 2 |
30 | 1 |
35 | 1 |
40 | 2 |
What is the median of the given data?
If the ratio of mean and median of a certain data is 4 : 5, then find the ratio of its mean and mode.