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The probability that A wakes up before the alarm rings is 0.4 . Then, the mean and variance of the number of times A wakes up before the alarm rings, in the next 7 days respectively are:
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The correct answer is:
$2.8,1.68$
We are given that $x=7$
$\begin{aligned} & \mathrm{p}=0.7, \mathrm{q}=1-0.4=0.6 \\ & \text { Now mean }=x \cdot \mathrm{p}=7 \times 0.4=2.8 \\ & \text { Variance }=x p q=2.8 \times 0.6=1.68\end{aligned}$
$\begin{aligned} & \mathrm{p}=0.7, \mathrm{q}=1-0.4=0.6 \\ & \text { Now mean }=x \cdot \mathrm{p}=7 \times 0.4=2.8 \\ & \text { Variance }=x p q=2.8 \times 0.6=1.68\end{aligned}$
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