JEE Main 2023 — Mathematics Probability & Statistics

We have,

$$X={11,12,13,\ldots .,40,41}$$X=11,12,13,….,40,41

$$\Rightarrow n(X)=31$$⇒n(X)=31

$$Y={61,62,63,...,90,91}$$Y=61,62,63,...,90,91

$$\Rightarrow n(Y)=31$$⇒n(Y)=31

So,

$$\bar{x}=\frac{{\sum }_{i=11}^{41}i}{31}=\frac{1}{31}[\frac{41\times 42}{2}-\frac{10\times 11}{2}]=26$$xˉ=31∑i=1141​i​=311​[241×42​−210×11​]=26

$$\bar{y}=\frac{{\sum }_{i=61}^{91}i}{31}=\frac{1}{31}[\frac{91\times 92}{2}-\frac{60\times 61}{2}]=73$$yˉ​=31∑i=6191​i​=311​[291×92​−260×61​]=73

Combined mean,

$$\mu =\frac{31\times 26+31\times 76}{31+31}$$μ=31+3131×26+31×76​

$$\Rightarrow \mu =\frac{26+76}{2}=51$$⇒μ=226+76​=51

Also,

$${\sigma }^{2}=\frac{1}{62}\times (\sum _{i=1}^{31}{({x}_{i}-\mu )}^{2}+\sum _{i=1}^{31}{({y}_{i}-\mu )}^{2})$$σ2=621​×(i=1∑31​(xi​−μ)2+i=1∑31​(yi​−μ)2)

Since, ${x}_{i}\in X$xi​∈X are in A.P. with $31$31 elements common difference $1$1 , same is ${y}_{i}\in y$yi​∈y, when written in increasing order.

$$\therefore \sum _{i=1}^{31}{({x}_{i}-\mu )}^{2}=\sum _{i=1}^{31}{({y}_{i}-\mu )}^{2}={10}^{2}+{11}^{2}+\ldots .+{40}^{2}$$∴i=1∑31​(xi​−μ)2=i=1∑31​(yi​−μ)2=102+112+….+402

$$={10}^{2}+{11}^{2}+\ldots .+{40}^{2}$$=102+112+….+402

$$=\frac{40\times 41\times 81}{6}-\frac{9\times 10\times 19}{6}=21855$$=640×41×81​−69×10×19​=21855

Hence, ${\sigma }^{2}=\frac{21855}{62}+\frac{21855}{62}=705$σ2=6221855​+6221855​=705

$$\therefore |\bar{x}+\bar{y}-{\sigma }^{2}|=|26+76-705|=603$$∴∣xˉ+yˉ​−σ2∣=∣26+76−705∣=603