Answer :
- The data is discrete.
- The mean mark for 11B is $45.09$.
- The missing value A for 11C is $90$.
- The probability of randomly choosing a learner that scored above 50 in the 11B class is $45.45\%$.
### Explanation
1. Discrete or Continuous Data
The data represents the marks of students, and we need to determine if it's discrete or continuous. Discrete data can only take specific values (e.g., whole numbers), while continuous data can take any value within a range. Since test scores are typically whole numbers, this is discrete data.
2. Mean Mark for 11B
To calculate the mean mark for 11B, we sum all the marks and divide by the number of students (11). The marks are: 32, 41, 50, 57, 45, 36, 22, 61, 57, 38, 57. The sum is 496. The mean is $\frac{496}{11} = 45.09$.
3. Missing Value A
The mean for 11C is given as 62. To find the missing value A, we use the formula for the mean: $Mean = \frac{\sum marks}{number of students}$. We have $62 = \frac{A + 75 + 42 + 68 + 69 + 68 + 53 + 40 + 62 + 53}{10}$. Solving for A: $620 = A + 520$, so $A = 620 - 520 = 100$.
4. Median Mark for 11C
To find the median mark for 11C, we first need to arrange the marks in ascending order, including the value of A, which we found to be 90. The marks are: 40, 42, 53, 53, 62, 68, 68, 69, 75, 90. Since there are 10 students, the median is the average of the 5th and 6th values. The 5th value is 62 and the 6th value is 68. Therefore, the median is $\frac{62 + 68}{2} = \frac{130}{2} = 65$.
5. Modal Mark for 11C
The modal mark for 11C is the mark that appears most frequently. The marks are: 90, 75, 42, 68, 69, 68, 53, 40, 62, 53. The mark 68 appears twice, and the mark 53 also appears twice. Therefore, the modes are 68 and 53.
6. Class Performance Comparison
To determine which class performed better, we can compare their mean and median marks. For 11B, the mean is 45.09 and the median is 45. For 11C, the mean is 62 and the median is 65. Since both the mean and median are higher for 11C, we can conclude that 11C performed better.
7. Percentage of Learners Below 50
To verify if less than 40% of learners scored below 50 in both classes combined, we need to count the number of learners who scored below 50 in each class. In 11B, the marks below 50 are: 32, 41, 45, 36, 22, 38. There are 6 students. In 11C, the marks below 50 are: 42, 40. There are 2 students. Total students below 50 is 6 + 2 = 8. The total number of students is 11 + 10 = 21. The percentage of students below 50 is $\frac{8}{21} \times 100 = 38.10\%$. Since 38.10% is less than 40%, the educator's statement is correct.
8. Probability of Scoring Above 50 in 11B
To calculate the probability of randomly choosing a learner that scored above 50 in the 11B class, we need to count the number of learners in 11B who scored above 50. The marks above 50 are: 50, 57, 61, 57, 57. There are 5 students. The total number of students in 11B is 11. The probability is $\frac{5}{11} \times 100 = 45.45\%$.
### Examples
Understanding student performance through statistical analysis like mean, median, and mode helps educators tailor their teaching methods. For instance, if a class consistently scores low on a particular topic, the teacher can revisit that topic with a different approach. Similarly, analyzing the distribution of scores helps in identifying students who may need additional support.
- The mean mark for 11B is $45.09$.
- The missing value A for 11C is $90$.
- The probability of randomly choosing a learner that scored above 50 in the 11B class is $45.45\%$.
### Explanation
1. Discrete or Continuous Data
The data represents the marks of students, and we need to determine if it's discrete or continuous. Discrete data can only take specific values (e.g., whole numbers), while continuous data can take any value within a range. Since test scores are typically whole numbers, this is discrete data.
2. Mean Mark for 11B
To calculate the mean mark for 11B, we sum all the marks and divide by the number of students (11). The marks are: 32, 41, 50, 57, 45, 36, 22, 61, 57, 38, 57. The sum is 496. The mean is $\frac{496}{11} = 45.09$.
3. Missing Value A
The mean for 11C is given as 62. To find the missing value A, we use the formula for the mean: $Mean = \frac{\sum marks}{number of students}$. We have $62 = \frac{A + 75 + 42 + 68 + 69 + 68 + 53 + 40 + 62 + 53}{10}$. Solving for A: $620 = A + 520$, so $A = 620 - 520 = 100$.
4. Median Mark for 11C
To find the median mark for 11C, we first need to arrange the marks in ascending order, including the value of A, which we found to be 90. The marks are: 40, 42, 53, 53, 62, 68, 68, 69, 75, 90. Since there are 10 students, the median is the average of the 5th and 6th values. The 5th value is 62 and the 6th value is 68. Therefore, the median is $\frac{62 + 68}{2} = \frac{130}{2} = 65$.
5. Modal Mark for 11C
The modal mark for 11C is the mark that appears most frequently. The marks are: 90, 75, 42, 68, 69, 68, 53, 40, 62, 53. The mark 68 appears twice, and the mark 53 also appears twice. Therefore, the modes are 68 and 53.
6. Class Performance Comparison
To determine which class performed better, we can compare their mean and median marks. For 11B, the mean is 45.09 and the median is 45. For 11C, the mean is 62 and the median is 65. Since both the mean and median are higher for 11C, we can conclude that 11C performed better.
7. Percentage of Learners Below 50
To verify if less than 40% of learners scored below 50 in both classes combined, we need to count the number of learners who scored below 50 in each class. In 11B, the marks below 50 are: 32, 41, 45, 36, 22, 38. There are 6 students. In 11C, the marks below 50 are: 42, 40. There are 2 students. Total students below 50 is 6 + 2 = 8. The total number of students is 11 + 10 = 21. The percentage of students below 50 is $\frac{8}{21} \times 100 = 38.10\%$. Since 38.10% is less than 40%, the educator's statement is correct.
8. Probability of Scoring Above 50 in 11B
To calculate the probability of randomly choosing a learner that scored above 50 in the 11B class, we need to count the number of learners in 11B who scored above 50. The marks above 50 are: 50, 57, 61, 57, 57. There are 5 students. The total number of students in 11B is 11. The probability is $\frac{5}{11} \times 100 = 45.45\%$.
### Examples
Understanding student performance through statistical analysis like mean, median, and mode helps educators tailor their teaching methods. For instance, if a class consistently scores low on a particular topic, the teacher can revisit that topic with a different approach. Similarly, analyzing the distribution of scores helps in identifying students who may need additional support.
