Explore Westonci.ca, the top Q&A platform where your questions are answered by professionals and enthusiasts alike. Get quick and reliable answers to your questions from a dedicated community of professionals on our platform. Get immediate and reliable solutions to your questions from a community of experienced professionals on our platform.

15. तल दिइएको तालिकामा 30 जना खेलाडीहरूको उमेर (वर्षमा) लाई उल्लेख गरिएको छ।
In the table given below, the ages (in years) of the 30 players are mentioned.

\begin{tabular}{|l|c|c|c|c|c|}
\hline उमेर वर्षमा (Age in years) & [tex]$0-10$[/tex] & [tex]$10-20$[/tex] & [tex]$20-30$[/tex] & [tex]$30-40$[/tex] & [tex]$40-50$[/tex] \\
\hline खेलाडी सङ्या (Number of players) & 6 & 4 & 5 & 4 & 11 \\
\hline
\end{tabular}

a) दिइएको तथ्याङ्कको रीत पर्ने श्रेणी कति हुन्छ? लेख्नुहोस्।
What is the modal class of the given data? Write it.

b) दिइएको तथ्याङ्कको मध्यिका श्रेणी पत्ता लगाउनुहोस्।
Find the median class of the given data.

c) दिइएको तथ्याङ्कको पहिलो चतुर्थांश गणना गर्नुहोस्।
Calculate the first quartile of the given data.

d) 20 वर्ष मुनिका खेलाडीहरूको औसत उमेर कति हुन्छ? गणना गर्नुहोस्।
What is the average age of players under 20 years? Calculate it.

Sagot :

Sure, let's analyze the given data step by step to find the answers to the questions.

Given Data:

| उमेर वर्षमा (Age in years) | 0-10 | 10-20 | 20-30 | 30-40 | 40-50 |
|----------------------------|------|-------|-------|-------|-------|
| खेलाडी सङ्या (Number of players) | 6 | 4 | 5 | 4 | 11 |

a) Modal Class of the Given Data:

The modal class is the class interval with the highest frequency. In our data:

- Number of players in 0-10 years: 6
- Number of players in 10-20 years: 4
- Number of players in 20-30 years: 5
- Number of players in 30-40 years: 4
- Number of players in 40-50 years: 11

The highest frequency is 11, which corresponds to the class interval 40-50 years. Therefore, the modal class is:

(40, 50)

b) Median Class of the Given Data:

To find the median class, we need to determine the class interval that contains the median.

- The total number of players: 6 + 4 + 5 + 4 + 11 = 30
- The median position (n/2) = 30/2 = 15

We need to find the class interval where the cumulative frequency reaches or exceeds 15:

1. Cumulative frequency for 0-10 years: 6
2. Cumulative frequency for 10-20 years: 6 + 4 = 10
3. Cumulative frequency for 20-30 years: 10 + 5 = 15
4. Cumulative frequency for 30-40 years: 15 + 4 = 19
5. Cumulative frequency for 40-50 years: 19 + 11 = 30

The cumulative frequency first reaches 15 in the class interval 20-30 years. Therefore, the median class is:

(20, 30)

c) First Quartile (Q1) of the Given Data:

The first quartile (Q1) position is at (n/4) = 30/4 = 7.5

We need to find the class interval where the cumulative frequency reaches or exceeds 7.5:

1. Cumulative frequency for 0-10 years: 6
2. Cumulative frequency for 10-20 years: 6 + 4 = 10
3. Cumulative frequency for 20-30 years: 10 + 5 = 15
4. Cumulative frequency for 30-40 years: 15 + 4 = 19
5. Cumulative frequency for 40-50 years: 19 + 11 = 30

The cumulative frequency first reaches or exceeds 7.5 in the class interval 10-20 years. Therefore, the first quartile (Q1) lies in:

(10, 20)

d) Average Age of Players Under 20 Years:

To find the average age of players under 20 years, we need to consider the players in the class intervals 0-10 and 10-20 years:

- Number of players in 0-10 years: 6
- Number of players in 10-20 years: 4

Total number of players under 20 years = 6 + 4 = 10

We calculate the midpoint (average age within each class interval) and then find the weighted average:

- Midpoint of 0-10 years = (0 + 10) / 2 = 5 years
- Midpoint of 10-20 years = (10 + 20) / 2 = 15 years

Weighted average age:
[tex]\( \text{Average age} = \frac{(6 \times 5) + (4 \times 15)}{6 + 4} \)[/tex]

= [tex]\( \frac{30 + 60}{10} \)[/tex]

= [tex]\( \frac{90}{10} \)[/tex]

= 9 years

Therefore, the average age of players under 20 years is:

9 years