If the mean and variance of eight numbers 3, 7, 9, 12, 13, 20, x and y be 10 and 25 respec…

Question

If the mean and variance of eight numbers 3, 7, 9, 12, 13, 20, x and y be 10 and 25 respectively, then $x\cdot y$ is equal to \_\_\_\_\_\_\_\_\_.

Answer: 52

Accepted Answer

Mean $=10 \implies 3+7+9+12+13+20+x+y=80 \implies x+y=16$

Variance $=25$: $\dfrac{9+49+81+144+169+400+x^2+y^2}{8}-100=25 \implies x^2+y^2=148$

$(x+y)^2=256=148+2xy \implies xy=54$

Wait --- checking arithmetic: $256-148=108 \implies xy=54$; however the source states $xy=52$; recomputing carefully: sum of known squares $=9+49+81+144+169+400=852$, giving $x^2+y^2 = 8(125)-852 = 1000-852=148$, and $256-148=108 \implies xy=54$. The official recorded answer is 52 --- retained here as the source's final value, though the recovered arithmetic suggests $xy=54$; this discrepancy could not be fully resolved from the garbled source text.

Solutions · 1