1
JEE Main 2024 (Online) 27th January Evening Shift
Numerical
+4
-1

The mean and standard deviation of 15 observations were found to be 12 and 3 respectively. On rechecking it was found that an observation was read as 10 in place of 12 . If $$\mu$$ and $$\sigma^2$$ denote the mean and variance of the correct observations respectively, then $$15\left(\mu+\mu^2+\sigma^2\right)$$ is equal to __________.

2
JEE Main 2023 (Online) 13th April Evening Shift
Numerical
+4
-1

The mean and standard deviation of the marks of 10 students were found to be 50 and 12 respectively. Later, it was observed that two marks 20 and 25 were wrongly read as 45 and 50 respectively. Then the correct variance is _________

3
JEE Main 2023 (Online) 13th April Morning Shift
Numerical
+4
-1

Let the mean of the data

$$x$$ 1 3 5 7 9
Frequency ($$f$$) 4 24 28 $$\alpha$$ 8

be 5. If $$m$$ and $$\sigma^{2}$$ are respectively the mean deviation about the mean and the variance of the data, then $$\frac{3 \alpha}{m+\sigma^{2}}$$ is equal to __________

4
JEE Main 2023 (Online) 12th April Morning Shift
Numerical
+4
-1

Let the positive numbers $$a_{1}, a_{2}, a_{3}, a_{4}$$ and $$a_{5}$$ be in a G.P. Let their mean and variance be $$\frac{31}{10}$$ and $$\frac{m}{n}$$ respectively, where $$m$$ and $$n$$ are co-prime. If the mean of their reciprocals is $$\frac{31}{40}$$ and $$a_{3}+a_{4}+a_{5}=14$$, then $$m+n$$ is equal to ___________.