The following table shows the number of males (M) and females (F) (in thousands) in Towns X and Y during the five years from 2018 to 2022. Based on the data in the table, answer the questions. Year wise numbers of Males and Females in two Towns (in thousands) Years Town X Town Y Number of males Number of females Number of males Number of females 2018 50 49 53 50 2019 52 49 54 52 2020 55 52 55 54 2021 53 53 58 56 2022 55 52 62 55
Find the number of years in which the number of females in Town X and Y is less than their respective average numbers in these two towns.
Two, Two
The question asks us to identify the number of years in which the female population in Town X is less than the average female population of Town X over the five years, and similarly for Town Y. We need to first calculate the average number of females for each town from the provided data table.
Let's look at the given data table showing the female population in thousands for Town X and Town Y from 2018 to 2022.
| Years | Town X | Town Y | ||
|---|---|---|---|---|
| Number of males | Number of females | Number of males | Number of females | |
| 2018 | 50 | 49 | 53 | 50 |
| 2019 | 52 | 49 | 54 | 52 |
| 2020 | 55 | 52 | 55 | 54 |
| 2021 | 53 | 53 | 58 | 56 |
| 2022 | 55 | 52 | 62 | 55 |
To find the average number of females for each town, we sum up the number of females for all five years and divide by the number of years (which is 5).
The number of females in Town X (in thousands) over the years 2018, 2019, 2020, 2021, and 2022 are 49, 49, 52, 53, and 52 respectively.
Total number of females in Town X over 5 years = $49 + 49 + 52 + 53 + 52 = 255$ thousands.
Average female population in Town X = $\frac{\text{Total females}}{\text{Number of years}} = \frac{255}{5} = 51$ thousands.
The number of females in Town Y (in thousands) over the years 2018, 2019, 2020, 2021, and 2022 are 50, 52, 54, 56, and 55 respectively.
Total number of females in Town Y over 5 years = $50 + 52 + 54 + 56 + 55 = 267$ thousands.
Average female population in Town Y = $\frac{\text{Total females}}{\text{Number of years}} = \frac{267}{5} = 53.4$ thousands.
Now we compare the number of females in each year for Town X and Town Y with their respective average female populations.
The number of years in Town X where the female population is less than the average is 2 (2018 and 2019).
The number of years in Town Y where the female population is less than the average is 2 (2018 and 2019).
The number of years in which the female population in Town X is less than its average is 2. The number of years in which the female population in Town Y is less than its average is 2.
Therefore, the answer is Two for Town X and Two for Town Y.
| Town | Sum of Female Population (thousands) | Number of Years | Average Female Population (thousands) | Years Below Average | Count of Years Below Average |
|---|---|---|---|---|---|
| Town X | $49+49+52+53+52=255$ | 5 | $255/5=51$ | 2018, 2019 | 2 |
| Town Y | $50+52+54+56+55=267$ | 5 | $267/5=53.4$ | 2018, 2019 | 2 |
The average, or arithmetic mean, is a fundamental concept in statistics used to represent a typical value in a set of numbers. It is calculated by summing all the values in the set and dividing by the total number of values.
In this problem, calculating the average female population helps us find a central value for the female population in each town over the five years. Comparing the yearly population to this average allows us to see which years had a population size that was smaller than the typical level during this period.
Data interpretation questions like this often require calculating averages, percentages, ratios, or differences based on the data presented in tables or graphs. Understanding how to perform these basic calculations is crucial for analyzing data effectively.
The table shows District-wise data of a number of primary school teachers posted in schools of a city.
Study the table and answer the question:
District | Male teachers | Female teachers |
East | 1650 | 2375 |
North | 1075 | 2651 |
West | 1280 | 1520 |
South | 1170 | 1085 |
Central | 690 | 859 |
Table shows income (in Rs. ) received by 4 employees of a company during the month of December 2020 and all their income sources.
Source | Amit | Suresh | Nitin | Varun |
Salary | 35000 | 38500 | 29000 | 42000 |
Arrears | 6000 | 6300 | 5000 | 7500 |
Bonus | 1000 | 1100 | 1000 | 1240 |
Overtime | 1800 | 1950 | 1400 | 1500 |
Study the table and answer the question:
Income (Rs.) | No. of persons |
Less than 200 | 12 |
Less than 250 | 26 |
Less than 300 | 34 |
Less than 350 | 40 |
Less than 400 | 50 |
The following table shows the annual profit of a company (in Rs. lakh).
2014-2015 | 2015-2016 | 2016-0217 | 2017-2018 | 2018-2019 |
625 | 690 | 725 | 775 | 815 |
The period which has the maximum percentage increase in profit over the previous year is:
The table given below shows the number of persons participating in a survey from 6 different states.
| States | Persons |
| S1 | 100 |
| S2 | 200 |
| S3 | 400 |
| S4 | 500 |
| S5 | 600 |
| S6 | 800 |
What is the ratio of number of person participating in a survey from state S3 to the number of person participating in a survey from state S4?