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Difficulty: Very hardMeasures of Dispersion and Position (Range, IQR, Standard Deviation, Percentiles)

Dataset PP consists of nn numerical values with a mean of 4040 and a standard deviation of 66. Dataset QQ consists of nn numerical values with a mean of 6060 and a standard deviation of 66. Dataset RR is created by combining all nn values from Dataset PP and all nn values from Dataset QQ into a single dataset of 2n2n values. If σR\sigma_R represents the standard deviation of Dataset RR, which of the following is the exact value of σR\sigma_R?

  1. A
    66
  2. B
    626\sqrt{2}
  3. 2342\sqrt{34}Answer
  4. D
    1212
  5. E
    1616

Answer

The exact value of σR\sigma_R is 2342\sqrt{34}.
The total variance of a combined dataset is given by the law of total variance: σR2=Mean(σP2,σQ2)+Var(μP,μQ)\sigma_R^2 = \text{Mean}(\sigma_P^2, \sigma_Q^2) + \text{Var}(\mu_P, \mu_Q). Since σP=σQ=6\sigma_P = \sigma_Q = 6, the average within-group variance is 62=366^2 = 36. The combined mean is 5050, and both group means (4040 and 6060) lie 1010 units away from 5050, giving a between-group variance of 102=10010^2 = 100. Combining these yields σR2=36+100=136\sigma_R^2 = 36 + 100 = 136, so σR=136=234\sigma_R = \sqrt{136} = 2\sqrt{34}.

Step-by-Step Solution

1
Calculate the mean of the combined dataset RR.
μR=n(40)+n(60)2n=100n2n=50\mu_R = \frac{n(40) + n(60)}{2n} = \frac{100n}{2n} = 50
Since both datasets have equal size nn, the combined mean is the arithmetic average of the two group means.
2
Express the sum of squared deviations for Dataset PP around the combined mean μR=50\mu_R = 50.
\sum_{i=1}^n (p_i - 50)^2 = \sum_{i=1}^n ((p_i - 40) - 10)^2 = \sum_{i=1}^n (p_i - 40)^2 - 20\sum_{i=1}^n (p_i - 40) + 100n = 36n - 0 + 100n = 136n
The variance of PP gives ∑(pi−40)2=62n=36n\sum (p_i - 40)^2 = 6^2 n = 36n, and ∑(pi−40)=0\sum (p_i - 40) = 0 by the definition of the mean.
3
Express the sum of squared deviations for Dataset QQ around the combined mean μR=50\mu_R = 50.
\sum_{j=1}^n (q_j - 50)^2 = \sum_{j=1}^n ((q_j - 60) + 10)^2 = \sum_{j=1}^n (q_j - 60)^2 + 20\sum_{j=1}^n (q_j - 60) + 100n = 36n + 0 + 100n = 136n
The variance of QQ gives ∑(qj−60)2=62n=36n\sum (q_j - 60)^2 = 6^2 n = 36n, and ∑(qj−60)=0\sum (q_j - 60) = 0.
4
Compute the combined variance σR2\sigma_R^2 and take the square root to find σR\sigma_R.
\sigma_R^2 = \frac{136n + 136n}{2n} = \frac{272n}{2n} = 136 \implies \sigma_R = \sqrt{136} = 2\sqrt{34}
The total variance of a combined dataset equals the average within-group variance (3636) plus the between-group variance around the combined mean ((40−50)2=100(40-50)^2 = 100). Thus σR2=36+100=136\sigma_R^2 = 36 + 100 = 136.

Key Concept

Pooled standard deviation for combined datasets with differing means
Estimated Time:2m 0s
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Topics

Measures of Dispersion and Position (Range, IQR, Standard Deviation, Percentiles)

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