JEE Main 2024 (Online) 29th January Evening Shift | Statistics Question 12 | Mathematics

If the mean and variance of five observations are $$\frac{24}{5}$$ and $$\frac{194}{25}$$ respectively and the mean of the first four observations is $$\frac{7}{2}$$, then the variance of the first four observations in equal to

$$\frac{5}{4}$$

$$\frac{4}{5}$$

$$\frac{105}{4}$$

$$\frac{77}{12}$$

JEE Main 2024 (Online) 27th January Morning Shift

MCQ (Single Correct Answer)

-1

Let $\mathrm{a}_1, \mathrm{a}_2, \ldots \mathrm{a}_{10}$ be 10 observations such that $\sum\limits_{\mathrm{k}=1}^{10} \mathrm{a}_{\mathrm{k}}=50$ and $\sum\limits_{\forall \mathrm{k} < \mathrm{j}} \mathrm{a}_{\mathrm{k}} \cdot \mathrm{a}_{\mathrm{j}}=1100$. Then the standard deviation of $\mathrm{a}_1, \mathrm{a}_2, \ldots, \mathrm{a}_{10}$ is equal to :

$\sqrt{115}$

$\sqrt{5}$

JEE Main 2023 (Online) 15th April Morning Shift

MCQ (Single Correct Answer)

-1

The mean and standard deviation of 10 observations are 20 and 8 respectively. Later on, it was observed that one observation was recorded as 50 instead of 40. Then the correct variance is :

JEE Main 2023 (Online) 11th April Evening Shift

MCQ (Single Correct Answer)

-1

Let the mean of 6 observations $$1,2,4,5, \mathrm{x}$$ and $$\mathrm{y}$$ be 5 and their variance be 10 . Then their mean deviation about the mean is equal to :

$$\frac{10}{3}$$

$$\frac{8}{3}$$

$$\frac{7}{3}$$

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