OP Malhotra Class 11 Maths Solutions Chapter 28 Statistics Exercise 28 (A)

S Chand Class 11 ICSE Maths Solutions Chapter 28 Statistics Ex 28(a)

Question 1. One set of 100 observations has the mean 15 and another set of 150 observations has the mean 16. Find the mean of 250 observations by combining the two sets of given observations.
Answer: Here, for the first set, \( n_1 = 100 \) and \( \overline{x}_1 = 15 \). For the second set, \( n_2 = 150 \) and \( \overline{x}_2 = 16 \).
To find the combined mean of the two sets, we use the formula:
\( \overline{x}_{12} = \frac{n_1 \overline{x}_1 + n_2 \overline{x}_2}{n_1 + n_2} \)
Now, we put in the given values:
\( \overline{x}_{12} = \frac{100 \times 15 + 150 \times 16}{100 + 150} \)
\( \overline{x}_{12} = \frac{1500 + 2400}{250} \)
\( \overline{x}_{12} = \frac{3900}{250} \)
\( \overline{x}_{12} = 15.6 \)
So, the combined mean of the 250 observations is 15.6. This method helps in understanding the average when different groups are merged.
In simple words: We have two groups of numbers with their own average. To find the average of all numbers together, we multiply each group's count by its average, add them up, and then divide by the total number of items.

🎯 Exam Tip: Remember the formula for combined mean and make sure to correctly substitute the number of observations (\(n\)) and their respective means (\(\overline{x}\)).

Question 2. The mean age of 40 students is 6 years and the mean age of another group of 60 students is 20 years. Find out the mean age of the 100 students combined together.
Answer: We have two groups of students. For the first group, \( n_1 = 40 \) students with a mean age of \( \overline{x}_1 = 6 \) years. For the second group, \( n_2 = 60 \) students with a mean age of \( \overline{x}_2 = 20 \) years.
To find the combined mean age of all 100 students, we use the combined mean formula:
\( \overline{x}_{12} = \frac{n_1 \overline{x}_1 + n_2 \overline{x}_2}{n_1 + n_2} \)
Now, we substitute the values:
\( \overline{x}_{12} = \frac{40 \times 6 + 60 \times 20}{40 + 60} \)
\( \overline{x}_{12} = \frac{240 + 1200}{100} \)
\( \overline{x}_{12} = \frac{1440}{100} \)
\( \overline{x}_{12} = 14.4 \)
Therefore, the combined mean age of the 100 students is 14.4 years. This shows how the overall average shifts based on the size and average of each group.
In simple words: We calculated the average age of all 100 students by combining the total ages from both groups and dividing by the total number of students.

🎯 Exam Tip: Pay close attention to the number of items in each group as they significantly influence the combined mean. Double-check your arithmetic, especially when multiplying and adding.

Question 3. The mean of marks obtained in an examination by a group of 100 students is found to be 49.46. The mean of the marks obtained in the same examination by another group of 200 students was 52.32. Find the mean of the marks obtained by both the groups of students taken together.
Answer: For the first group of students, \( n_1 = 100 \) and the mean marks \( \overline{x}_1 = 49.46 \). For the second group, \( n_2 = 200 \) and the mean marks \( \overline{x}_2 = 52.32 \).
The total marks obtained by the first group is \( n_1 \overline{x}_1 \) and by the second group is \( n_2 \overline{x}_2 \).
To find the combined mean marks for both groups, we use the formula:
\( \overline{x}_{12} = \frac{n_1 \overline{x}_1 + n_2 \overline{x}_2}{n_1 + n_2} \)
Substituting the given values into the formula:
\( \overline{x}_{12} = \frac{100 \times 49.46 + 200 \times 52.32}{100 + 200} \)
\( \overline{x}_{12} = \frac{4946 + 10464}{300} \)
\( \overline{x}_{12} = \frac{15410}{300} \)
\( \overline{x}_{12} = 51.3666... \approx 51.367 \)
The combined mean marks for both groups of students is approximately 51.367. This shows the overall performance when all students are considered together.
In simple words: We calculated the average exam score for all students by adding up everyone's total scores from two different groups and then dividing by the total number of students.

🎯 Exam Tip: When dealing with decimals, carry calculations to sufficient decimal places to maintain accuracy, especially in intermediate steps, and round only the final answer if instructed.

Question 4. The number of students in section X A and X B are 30 and 35 respectively. The mean in the mathematics test are as follows:

Find the mean score of XB.
Answer: We are given the following information:
For section X A: \( n_1 = 30 \) students, \( \overline{x}_1 = 70 \) (mean score).
For section X B: \( n_2 = 35 \) students, \( \overline{x}_2 = ? \) (mean score to be found).
For X A and X B combined: \( \overline{x}_{12} = 62 \) (combined mean score).
We know the formula for the combined mean:
\( \overline{x}_{12} = \frac{n_1 \overline{x}_1 + n_2 \overline{x}_2}{n_1 + n_2} \)
Now, we substitute the known values into the formula:
\( 62 = \frac{30 \times 70 + 35 \times \overline{x}_2}{30 + 35} \)
First, calculate the sum in the denominator:
\( 62 = \frac{2100 + 35 \overline{x}_2}{65} \)
Multiply both sides by 65 to clear the denominator:
\( 62 \times 65 = 2100 + 35 \overline{x}_2 \)
\( 4030 = 2100 + 35 \overline{x}_2 \)
Subtract 2100 from both sides:
\( 4030 - 2100 = 35 \overline{x}_2 \)
\( 1930 = 35 \overline{x}_2 \)
Divide by 35 to find \( \overline{x}_2 \):
\( \overline{x}_2 = \frac{1930}{35} \)
\( \overline{x}_2 \approx 55.1428... \)
Therefore, the mean score of section X B is approximately 55.14. This calculation helps determine an unknown group average when the combined average is known.
In simple words: We used the total average and the average of one group to figure out the average of the second group. We did this by working backward from the combined mean formula.

🎯 Exam Tip: When a combined mean problem involves finding a missing mean, set up the combined mean formula and solve for the unknown variable using algebraic manipulation.

Question 5. Two samples of sizes 50 and 100 are given. The mean of these samples respectively are 56 and 50. Find the mean of size 150 by combining.
Answer: We have two samples. For the first sample, \( n_1 = 50 \) and its mean is \( \overline{x}_1 = 56 \). For the second sample, \( n_2 = 100 \) and its mean is \( \overline{x}_2 = 50 \).
We need to find the combined mean of these two samples, which together form a sample of size \( 50 + 100 = 150 \).
Using the combined mean formula:
\( \overline{x}_{12} = \frac{n_1 \overline{x}_1 + n_2 \overline{x}_2}{n_1 + n_2} \)
Substitute the given values into the formula:
\( \overline{x}_{12} = \frac{50 \times 56 + 100 \times 50}{50 + 100} \)
\( \overline{x}_{12} = \frac{2800 + 5000}{150} \)
\( \overline{x}_{12} = \frac{7800}{150} \)
\( \overline{x}_{12} = 52 \)
So, the combined mean of the two samples is 52. This process is useful for averaging data from different sources.
In simple words: We took two groups of numbers, each with its own average. To find the average of all numbers put together, we added up the total values from both groups and divided by the total number of items.

🎯 Exam Tip: Ensure that the values for \( n_1 \), \( \overline{x}_1 \), \( n_2 \), and \( \overline{x}_2 \) are correctly identified from the problem statement before applying the formula.

Question 6. The mean and standard deviation of the marks obtained by the groups of students, consisting of 50 each, are given below. Calculate the mean and standard deviation of the marks obtained by all the 100 students.

Answer: We have two groups of students, each with 50 students. For Group 1: \( n_1 = 50 \), \( \overline{x}_1 = 60 \), \( \sigma_1 = 8 \).
For Group 2: \( n_2 = 50 \), \( \overline{x}_2 = 55 \), \( \sigma_2 = 7 \).
Total number of students \( N = n_1 + n_2 = 50 + 50 = 100 \).

**1. Calculate the Combined Mean (\( \overline{x}_{12} \)):**
We use the formula: \( \overline{x}_{12} = \frac{n_1 \overline{x}_1 + n_2 \overline{x}_2}{n_1 + n_2} \)
Substitute the values:
\( \overline{x}_{12} = \frac{50 \times 60 + 50 \times 55}{50 + 50} \)
\( \overline{x}_{12} = \frac{3000 + 2750}{100} \)
\( \overline{x}_{12} = \frac{5750}{100} \)
\( \overline{x}_{12} = 57.5 \)

**2. Calculate the deviations (\( d_1 \) and \( d_2 \)):**
\( d_1 = \overline{x}_{12} - \overline{x}_1 = 57.5 - 60 = -2.5 \)
\( d_2 = \overline{x}_{12} - \overline{x}_2 = 57.5 - 55 = 2.5 \)

**3. Calculate the Combined Standard Deviation (\( \sigma_{12} \)):**
We use the formula:
\( \sigma_{12} = \sqrt{\frac{n_1 (\sigma_1^2 + d_1^2) + n_2 (\sigma_2^2 + d_2^2)}{n_1 + n_2}} \)
Substitute the calculated values:
\( \sigma_{12} = \sqrt{\frac{50 (8^2 + (-2.5)^2) + 50 (7^2 + (2.5)^2)}{50 + 50}} \)
\( \sigma_{12} = \sqrt{\frac{50 (64 + 6.25) + 50 (49 + 6.25)}{100}} \)
\( \sigma_{12} = \sqrt{\frac{50 (70.25) + 50 (55.25)}{100}} \)
\( \sigma_{12} = \sqrt{\frac{3512.5 + 2762.5}{100}} \)
\( \sigma_{12} = \sqrt{\frac{6275}{100}} \)
\( \sigma_{12} = \sqrt{62.75} \)
\( \sigma_{12} \approx 7.92 \)
The combined mean for all 100 students is 57.5, and their combined standard deviation is approximately 7.92. This gives a full statistical picture of the merged group.
In simple words: First, we found the average score of all 100 students together. Then, we calculated how spread out their scores were from this new average. This helps us understand both the typical score and the variation in scores for the whole group.

🎯 Exam Tip: Be careful with the signs of \( d_1 \) and \( d_2 \), as squaring them makes them positive. Ensure all terms are correctly substituted into the combined standard deviation formula, as a small error can significantly change the result.

Question 7. The mean and standard deviation of distribution of 100 and 150 items are 50, 5 and 40, 6 respectively. Find the mean and standard deviation of all the 250 items taken together.
Answer: We have two distributions. For Distribution 1: \( n_1 = 100 \), \( \overline{x}_1 = 50 \), \( \sigma_1 = 5 \).
For Distribution 2: \( n_2 = 150 \), \( \overline{x}_2 = 40 \), \( \sigma_2 = 6 \).
Total number of items \( N = n_1 + n_2 = 100 + 150 = 250 \).

**1. Calculate the Combined Mean (\( \overline{x}_{12} \)):**
We use the formula: \( \overline{x}_{12} = \frac{n_1 \overline{x}_1 + n_2 \overline{x}_2}{n_1 + n_2} \)
Substitute the values:
\( \overline{x}_{12} = \frac{100 \times 50 + 150 \times 40}{100 + 150} \)
\( \overline{x}_{12} = \frac{5000 + 6000}{250} \)
\( \overline{x}_{12} = \frac{11000}{250} \)
\( \overline{x}_{12} = 44 \)

**2. Calculate the deviations (\( d_1 \) and \( d_2 \)):**
\( d_1 = \overline{x}_{12} - \overline{x}_1 = 44 - 50 = -6 \)
\( d_2 = \overline{x}_{12} - \overline{x}_2 = 44 - 40 = 4 \)

**3. Calculate the Combined Standard Deviation (\( \sigma_{12} \)):**
We use the formula:
\( \sigma_{12} = \sqrt{\frac{n_1 (\sigma_1^2 + d_1^2) + n_2 (\sigma_2^2 + d_2^2)}{n_1 + n_2}} \)
Substitute the calculated values:
\( \sigma_{12} = \sqrt{\frac{100 (5^2 + (-6)^2) + 150 (6^2 + 4^2)}{100 + 150}} \)
\( \sigma_{12} = \sqrt{\frac{100 (25 + 36) + 150 (36 + 16)}{250}} \)
\( \sigma_{12} = \sqrt{\frac{100 \times 61 + 150 \times 52}{250}} \)
\( \sigma_{12} = \sqrt{\frac{6100 + 7800}{250}} \)
\( \sigma_{12} = \sqrt{\frac{13900}{250}} \)
\( \sigma_{12} = \sqrt{55.6} \)
\( \sigma_{12} \approx 7.46 \)
The combined mean for all 250 items is 44, and their combined standard deviation is approximately 7.46. Understanding these combined values is crucial for data analysis across different sets.
In simple words: First, we found the overall average for all 250 items from both distributions. Then, we calculated how much the individual items varied from this new overall average, giving us the combined spread of the data.

🎯 Exam Tip: Carefully compute \( d_1 \) and \( d_2 \) including their signs, as incorrect signs will lead to errors when calculating \( d^2 \) in the combined standard deviation formula. Ensure you correctly apply the squares to both the original standard deviations and the deviations from the combined mean.