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## Here's the Solution to this Question

We can assume that, there are five lamps in my flat. And I have the probabilities of the number of lamps which are switched on in different room of my flat, then we need calculate the mean, variance and standard deviation of the data obtained.

Let n is the number of switched on lamps.

Probability distribution:

$P(n=0)=0.1$

$P(n=1)=0.2$

$P(n=2)=0.25$

$P(n=3)=0.15$

$P(n=4)=0.2$

$P(n=5)=0.1$

So,

The mean (the average number of switched on lamps):

$E(X)=(1\times0.2)+(2\times0.25)+(3\times0.15)+(4\times0.2)+(5\times0.1)=2.45$

Variance (how far the number of switched lamps is spread out from their average value):

$Var(X)=[E(X^2)-(E(X))^2]$

$E(X^2)=(0.2)+(4\times0.25)+(9\times0.15)+(16\times0.2)+(25\times0.1)=8.25$

So,

$Var(X)=8.25-2.45^2=2.25$

and Standard deviation (sd):

$\sigma=\sqrt{Var(X)}=\sqrt{2.25}=1.5$

Hence, we can conclude that, there are three numbers of switched on lamps (1,2,3) within one standard deviation (range is from 0.95 to 3.95).