JEE Main 2021 — Mathematics Probability & Statistics

Given, mean$=\frac{309}{22}$=22309​

$$\Rightarrow \frac{3a+9b+180+189+135}{a+b+26}=\frac{309}{22}$$⇒a+b+263a+9b+180+189+135​=22309​

$$\Rightarrow \frac{504+3a+9b}{a+b+26}=\frac{309}{22}$$⇒a+b+26504+3a+9b​=22309​

$$\Rightarrow 66a+198b+11088=309a+309b+8034$$⇒66a+198b+11088=309a+309b+8034

$$\Rightarrow 243a+111b=3054$$⇒243a+111b=3054

$$\Rightarrow 81a+37b=1018...(1)$$⇒81a+37b=1018...(1)

Now, median$=14,$=14, which lies in the interval $12-18,$12−18, thus $12-18$12−18 is the median class and we have $=l+\frac{(\frac{N}{2}-c)\times h}{f},$=l+f(2N​−c)×h​,

Where, $l$l is the lower limit of the median class i.e. $l=12,c$l=12,c is the cumulative frequency of the class above the median class i.e. $c=a+b,f$c=a+b,f is the frequency of the median class i.e. $f=12$f=12 and $h$h is the length of the interval i.e. $h=6.$h=6.

Thus, we have $12+(\frac{(\frac{a+b+26}{2})-(a+b)}{12})\times 6=14$12+(12(2a+b+26​)−(a+b)​)×6=14

$$\Rightarrow \frac{(\frac{a+b+26-2a-2b}{2})}{2}=14-12$$⇒2(2a+b+26−2a−2b​)​=14−12

$$\Rightarrow \frac{26-a-b}{4}=2$$⇒426−a−b​=2

$$\Rightarrow 26-a-b=8$$⇒26−a−b=8

$$\Rightarrow a+b=18...(2)$$⇒a+b=18...(2)

On solving the equations $(1)$(1) and $(2),$(2), we get $a=8,b=10.$a=8,b=10.

$$\therefore {(a-b)}^{2}={(8-10)}^{2}=4.$$∴(a−b)2=(8−10)2=4.