JEE Main 2026 — Mathematics Probability & Statistics

Given $\displaystyle \sum_{i=1}^{20}(x_i + 5)^2 = 2500$i=1∑20​(xi​+5)2=2500 and $\displaystyle \sum_{i=1}^{20}(x_i - 5)^2 = 100$i=1∑20​(xi​−5)2=100.

Expanding the first equation:

$$\sum_{i=1}^{20} (x_i^2 + 10x_i + 25) = 2500$$i=1∑20​(xi2​+10xi​+25)=2500

$$\sum_{i=1}^{20} x_i^2 + 10 \sum_{i=1}^{20} x_i + 500 = 2500$$i=1∑20​xi2​+10i=1∑20​xi​+500=2500

$\displaystyle \sum_{i=1}^{20} x_i^2 + 10 \sum_{i=1}^{20} x_i = 2000$i=1∑20​xi2​+10i=1∑20​xi​=2000 --- (1)

Expanding the second equation:

$$\sum_{i=1}^{20} (x_i^2 - 10x_i + 25) = 100$$i=1∑20​(xi2​−10xi​+25)=100

$$\sum_{i=1}^{20} x_i^2 - 10 \sum_{i=1}^{20} x_i + 500 = 100$$i=1∑20​xi2​−10i=1∑20​xi​+500=100

$\displaystyle \sum_{i=1}^{20} x_i^2 - 10 \sum_{i=1}^{20} x_i = -400$i=1∑20​xi2​−10i=1∑20​xi​=−400 --- (2)

Subtracting (2) from (1):

$$20 \sum_{i=1}^{20} x_i = 2400 \Rightarrow \sum_{i=1}^{20} x_i = 120$$20i=1∑20​xi​=2400⇒i=1∑20​xi​=120

Adding (1) and (2):

$$2 \sum_{i=1}^{20} x_i^2 = 1600 \Rightarrow \sum_{i=1}^{20} x_i^2 = 800$$2i=1∑20​xi2​=1600⇒i=1∑20​xi2​=800

Mean ($\mu$μ) is given by:

$$\mu = \dfrac{\sum_{i=1}^{20} x_i}{20} = \dfrac{120}{20} = 6$$μ=20∑i=120​xi​​=20120​=6

Variance ($\sigma^2$σ2) is given by:

$$\sigma^2 = \dfrac{\sum_{i=1}^{20} x_i^2}{20} - \mu^2 = \dfrac{800}{20} - 6^2 = 40 - 36 = 4$$σ2=20∑i=120​xi2​​−μ2=20800​−62=40−36=4

Standard deviation ($\sigma$σ) is $\sqrt{4} = 2$4​=2.

The ratio of mean to standard deviation is:

$$\dfrac{\mu}{\sigma} = \dfrac{6}{2} = \dfrac{3}{1}$$σμ​=26​=13​

Answer: $3:1$3:1