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The mean of two samples of sizes 200 and 300 were found to be 25 and 10 respectively. Their standard deviations were 3 and 4 respectively. The varience of combined sample size of 500 is
- a)64
- b)65.2
- c)67.2
- d)64.2
Correct answer is option 'C'. Can you explain this answer?
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The mean of two samples of sizes200 and 300were found to be 25and10 re...
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Given:
Sample 1:
- Sample size (n1) = 200
- Mean (μ1) = 25
- Standard deviation (σ1) = 3
Sample 2:
- Sample size (n2) = 300
- Mean (μ2) = 10
- Standard deviation (σ2) = 4
To find:
Variance of the combined sample size of 500.
Solution:
Step 1: Calculate the sum of squares of each sample:
- Sum of squares of Sample 1 (SS1) = (n1 - 1) * σ1^2
= (200 - 1) * 3^2
= 599 * 9
= 5391
- Sum of squares of Sample 2 (SS2) = (n2 - 1) * σ2^2
= (300 - 1) * 4^2
= 899 * 16
= 14384
Step 2: Calculate the combined sum of squares:
- Combined sum of squares (SST) = SS1 + SS2
= 5391 + 14384
= 19775
Step 3: Calculate the combined mean:
- Combined mean (μ) = (n1 * μ1 + n2 * μ2) / (n1 + n2)
= (200 * 25 + 300 * 10) / (200 + 300)
= (5000 + 3000) / 500
= 8000 / 500
= 16
Step 4: Calculate the combined variance:
- Combined variance (σ^2) = SST / (n1 + n2 - 1)
= 19775 / (200 + 300 - 1)
= 19775 / 499
≈ 39.64
Step 5: Calculate the combined standard deviation:
- Combined standard deviation (σ) = √(σ^2)
= √39.64
≈ 6.3
Step 6: Calculate the combined variance for the sample size of 500:
- Combined variance for sample size of 500 = σ^2 * (n1 + n2) / (n1 + n2 - 1)
= 39.64 * (200 + 300) / (200 + 300 - 1)
= 39.64 * 500 / 499
≈ 67.2
Therefore, the variance of the combined sample size of 500 is approximately 67.2. Hence, option C is the correct answer.
Sample 1:
- Sample size (n1) = 200
- Mean (μ1) = 25
- Standard deviation (σ1) = 3
Sample 2:
- Sample size (n2) = 300
- Mean (μ2) = 10
- Standard deviation (σ2) = 4
To find:
Variance of the combined sample size of 500.
Solution:
Step 1: Calculate the sum of squares of each sample:
- Sum of squares of Sample 1 (SS1) = (n1 - 1) * σ1^2
= (200 - 1) * 3^2
= 599 * 9
= 5391
- Sum of squares of Sample 2 (SS2) = (n2 - 1) * σ2^2
= (300 - 1) * 4^2
= 899 * 16
= 14384
Step 2: Calculate the combined sum of squares:
- Combined sum of squares (SST) = SS1 + SS2
= 5391 + 14384
= 19775
Step 3: Calculate the combined mean:
- Combined mean (μ) = (n1 * μ1 + n2 * μ2) / (n1 + n2)
= (200 * 25 + 300 * 10) / (200 + 300)
= (5000 + 3000) / 500
= 8000 / 500
= 16
Step 4: Calculate the combined variance:
- Combined variance (σ^2) = SST / (n1 + n2 - 1)
= 19775 / (200 + 300 - 1)
= 19775 / 499
≈ 39.64
Step 5: Calculate the combined standard deviation:
- Combined standard deviation (σ) = √(σ^2)
= √39.64
≈ 6.3
Step 6: Calculate the combined variance for the sample size of 500:
- Combined variance for sample size of 500 = σ^2 * (n1 + n2) / (n1 + n2 - 1)
= 39.64 * (200 + 300) / (200 + 300 - 1)
= 39.64 * 500 / 499
≈ 67.2
Therefore, the variance of the combined sample size of 500 is approximately 67.2. Hence, option C is the correct answer.
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The mean of two samples of sizes200 and 300were found to be 25and10 respectively. Their standard deviations were 3 and 4 respectively. The varience of combined sample size of 500 isa)64b)65.2c)67.2d)64.2Correct answer is option 'C'. Can you explain this answer?
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The mean of two samples of sizes200 and 300were found to be 25and10 respectively. Their standard deviations were 3 and 4 respectively. The varience of combined sample size of 500 isa)64b)65.2c)67.2d)64.2Correct answer is option 'C'. Can you explain this answer? for CA Foundation 2024 is part of CA Foundation preparation. The Question and answers have been prepared according to the CA Foundation exam syllabus. Information about The mean of two samples of sizes200 and 300were found to be 25and10 respectively. Their standard deviations were 3 and 4 respectively. The varience of combined sample size of 500 isa)64b)65.2c)67.2d)64.2Correct answer is option 'C'. Can you explain this answer? covers all topics & solutions for CA Foundation 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for The mean of two samples of sizes200 and 300were found to be 25and10 respectively. Their standard deviations were 3 and 4 respectively. The varience of combined sample size of 500 isa)64b)65.2c)67.2d)64.2Correct answer is option 'C'. Can you explain this answer?.
The mean of two samples of sizes200 and 300were found to be 25and10 respectively. Their standard deviations were 3 and 4 respectively. The varience of combined sample size of 500 isa)64b)65.2c)67.2d)64.2Correct answer is option 'C'. Can you explain this answer? for CA Foundation 2024 is part of CA Foundation preparation. The Question and answers have been prepared according to the CA Foundation exam syllabus. Information about The mean of two samples of sizes200 and 300were found to be 25and10 respectively. Their standard deviations were 3 and 4 respectively. The varience of combined sample size of 500 isa)64b)65.2c)67.2d)64.2Correct answer is option 'C'. Can you explain this answer? covers all topics & solutions for CA Foundation 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for The mean of two samples of sizes200 and 300were found to be 25and10 respectively. Their standard deviations were 3 and 4 respectively. The varience of combined sample size of 500 isa)64b)65.2c)67.2d)64.2Correct answer is option 'C'. Can you explain this answer?.
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