1
JEE Main 2024 (Online) 31st January Evening Shift
+4
-1

Let the mean and the variance of 6 observations $$a, b, 68,44,48,60$$ be $$55$$ and $$194$$, respectively. If $$a>b$$, then $$a+3 b$$ is

A
180
B
210
C
190
D
200
2
JEE Main 2024 (Online) 30th January Morning Shift
+4
-1

Let M denote the median of the following frequency distribution

Class 0 - 4 4 - 8 8 - 12 12 - 16 16 - 20
Frequency 3 9 10 8 6

Then 20M is equal to :

A
104
B
52
C
208
D
416
3
JEE Main 2024 (Online) 29th January Evening Shift
+4
-1

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

A
$$\frac{5}{4}$$
B
$$\frac{4}{5}$$
C
$$\frac{105}{4}$$
D
$$\frac{77}{12}$$
4
JEE Main 2024 (Online) 27th January Morning Shift
+4
-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 :
A
5
B
$\sqrt{115}$
C
10
D
$\sqrt{5}$
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