JEE Main 2022 · 29 Jun · Shift 1 · Mathematics Q20

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MathematicsSingle correct+4 / -1Question 20 of 90

Let the mean and the variance of 5 observations $x_1,x_2,x_3,x_4,x_5$ be $\dfrac{24}5$ and $\dfrac{194}{25}$ respectively. If the mean and variance of the first 4 observation are $\dfrac72$ and $a$ respectively, then $(4a+x_5)$ is equal to :

Answer and worked solution

Official answer: B

Sums. $\sum x=24$ , first four 14, so $x_5=10$ . $\sum x^2=5\left(\frac{194}{25}+\frac{576}{25}\right)=154$ ; first four 54.

Variance of four. $\frac{54}4-\frac{49}4=\frac54=a$ .

$4a+x_5=15$ .

(B)

Topics:Variance and standard deviation Measures of central tendency

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JEE Main 2022 · 29 Jun · Shift 1, Mathematics question 20