If the mean and variance of the data 65, 68, 58, 44, 48, 45, 60,,, 60 where are 56 and 66.2 respectively, then ^2 + ^2 is equal to:

Given: The data are 65, 68, 58, 44, 48, 45, 60,,, 60 with mean x = 56 and variance ^2 = 66.2. Find: ^2 + ^2. Using the variance formula, ^2 = xi^2n - ( x)^2 with n = 10, ^2 = 66.2 and x = 56, 66.2 = xi^210 - 56^2 xi^210 = 66.2 + 3136 = 3202.2 xi^2 = 32022 Now compute the sum of squares of the known observations: 65^2 = 4225, 68^2 = 4624, 58^2 = 3364, 44^2 = 1936 48^2 = 2304, 45^2 = 2025, 60^2 = 3600 Since 60 occurs twice, 2 3600 = 7200 Therefore, 4225 + 4624 + 3364 + 1936 + 2304 + 2025 + 7200 = 25678 So, 25678 + ^2 + ^2 = 32022 Hence, ^2 + ^2 = 32022 - 25678 = 6344 Therefore, the correct option is A.

If the mean and variance of the data 65, 68, 58, 44, 48, 45, 60,,… — Solution | JEE Mathematics

MCQEasyJEE 2024Measures of Dispersion

If the mean and variance of the data $65, 68, 58, 44, 48, 45, 60, \alpha, \beta, 60$ where $\alpha > \beta$ are $56$ and $66.2$ respectively, then $\alpha^2 + \beta^2$ is equal to:

A
$6344$
B
$6450$
C
$6300$
D
$6200$

Answer

Correct answer:A

Step-by-step solution

Standard Method

Given: The data are $65, 68, 58, 44, 48, 45, 60, \alpha, \beta, 60$ with mean $\bar{x} = 56$ and variance $\sigma^2 = 66.2$ .

Find: $\alpha^2 + \beta^2$ .

Using the variance formula,

\sigma^2 = \frac{\sum x_i^2}{n} - (\bar{x})^2

with $n = 10$ , $\sigma^2 = 66.2$ and $\bar{x} = 56$ ,

66.2 = \frac{\sum x_i^2}{10} - 56^2

\frac{\sum x_i^2}{10} = 66.2 + 3136 = 3202.2

\sum x_i^2 = 32022

Detailed Computation

Now compute the sum of squares of the known observations:

65^2 = 4225, \quad 68^2 = 4624, \quad 58^2 = 3364, \quad 44^2 = 1936

48^2 = 2304, \quad 45^2 = 2025, \quad 60^2 = 3600

Since $60$ occurs twice,

2 \times 3600 = 7200

Therefore,

4225 + 4624 + 3364 + 1936 + 2304 + 2025 + 7200 = 25678

Direct Substitution

So,

25678 + \alpha^2 + \beta^2 = 32022

Hence,

\alpha^2 + \beta^2 = 32022 - 25678 = 6344

Therefore, the correct option is A.

Common mistakes

Using the wrong variance formula. The variance here is $\sigma^2 = \frac{\sum x_i^2}{n} - (\bar{x})^2$ , not just the average of the observations. Always convert the given mean and variance into an equation involving $\sum x_i^2$ .
Forgetting that $60$ appears twice. This makes the sum of known squares incorrect. Count both occurrences, so the contribution is $2 \times 60^2 = 7200$ .
Using $56^2$ incorrectly. Since $56^2 = 3136$ , any mistake here changes $\sum x_i^2$ and the final value. Evaluate the square carefully before substitution.

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