Let a1, a2,, a10 be 10 observations with ak = 50 and (ak aj) = 1100. Find the standard deviation.

Given: n = 10, k=1^10 ak = 50, and 1 k < j 10 ak aj = 1100. Find: The standard deviation of the observations. Use the variance formula: ^2 = 1n ( k=1^n ak^2 - 1n ( k=1^n ak)^2) First, determine k=1^10 ak^2 from ( k=1^10 ak)^2 = k=1^10 ak^2 + 2 1 k < j 10 ak aj Substituting the given values, 50^2 = k=1^10 ak^2 + 2 1100 2500 = k=1^10 ak^2 + 2200 Hence, k=1^10 ak^2 = 300 Now substitute into the variance formula: ^2 = 110 (300 - 110 2500) ^2 = 110(300 - 250) ^2 = 5 Therefore, the standard deviation is = 5 So, the correct option is B. Given: n = 10, i=1^10 ai = 50, and 1 k < j 10 ak aj = 1100. Find: The standard deviation. Use the direct formula = 1n i=1^10 ai^2 - ( 1n i=1^10 ai)^2 So we only need ai^2. From the identity ( i=1^10 ai)^2 = i=1^10 ai^2 + 2 1 k < j 10 ak aj we get 2500 = i=1^10 ai^2 + 2200 i=1^10 ai^2 = 300 Hence, = 30010 - ( 5010)^2 = 30 - 25 = 5 Therefore, the correct option is B.

Let a1, a2,, a10 be 10 observations with ak = 50 and (ak aj) = 11… — Solution | JEE Mathematics

MCQMediumJEE 2024Measures of Dispersion

Let $a_1, a_2, \ldots, a_{10}$ be $10$ observations with $\sum a_k = 50$ and $\sum (a_k \cdot a_j) = 1100$ . Find the standard deviation.

A
$5$
B
$\sqrt{5}$
C
$10$
D
$\sqrt{115}$

Answer

Correct answer:B

Step-by-step solution

Standard Method

Given: $n = 10$ , $\sum_{k=1}^{10} a_k = 50$ , and $\sum_{1 \leq k < j \leq 10} a_k a_j = 1100$ .

Find: The standard deviation of the observations.

Use the variance formula:

\sigma^2 = \frac{1}{n}\left(\sum_{k=1}^{n} a_k^2 - \frac{1}{n}\left(\sum_{k=1}^{n} a_k\right)^2\right)

First, determine $\sum_{k=1}^{10} a_k^2$ from

\left(\sum_{k=1}^{10} a_k\right)^2 = \sum_{k=1}^{10} a_k^2 + 2\sum_{1 \leq k < j \leq 10} a_k a_j

Substituting the given values,

50^2 = \sum_{k=1}^{10} a_k^2 + 2 \cdot 1100

2500 = \sum_{k=1}^{10} a_k^2 + 2200

Hence,

\sum_{k=1}^{10} a_k^2 = 300

Now substitute into the variance formula:

\sigma^2 = \frac{1}{10}\left(300 - \frac{1}{10} \times 2500\right)

\sigma^2 = \frac{1}{10}(300 - 250)

\sigma^2 = 5

Therefore, the standard deviation is

\sigma = \sqrt{5}

So, the correct option is B.

Direct Standard Deviation Formula

Given: $n = 10$ , $\sum_{i=1}^{10} a_i = 50$ , and $\sum_{1 \leq k < j \leq 10} a_k a_j = 1100$ .

Find: The standard deviation.

Use the direct formula

\sigma = \sqrt{\frac{1}{n}\sum_{i=1}^{10} a_i^2 - \left(\frac{1}{n}\sum_{i=1}^{10} a_i\right)^2}

So we only need $\sum a_i^2$ .

From the identity

\left(\sum_{i=1}^{10} a_i\right)^2 = \sum_{i=1}^{10} a_i^2 + 2\sum_{1 \leq k < j \leq 10} a_k a_j

we get

2500 = \sum_{i=1}^{10} a_i^2 + 2200

\sum_{i=1}^{10} a_i^2 = 300

Hence,

\sigma = \sqrt{\frac{300}{10} - \left(\frac{50}{10}\right)^2}

\sigma = \sqrt{30 - 25} = \sqrt{5}

Therefore, the correct option is B.

Common mistakes

Using $\sum_{1 \leq k < j \leq 10} a_k a_j = 1100$ as if it were $\sum a_k^2$ is incorrect, because cross-product terms and square terms are different. First use the expansion of $\left(\sum a_k\right)^2$ to find $\sum a_k^2$ .
Forgetting the factor of $2$ in
$\left(\sum a_k\right)^2 = \sum a_k^2 + 2\sum_{k$
Computing standard deviation directly from the mean without first finding the variance can lead to mixing formulas. First calculate $\sigma^2$ correctly, then take the square root to obtain $\sigma$ .

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