The probability that A wakes up before the al | vedclass

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(C) Let $n$ be the number of days,so $n = 7$. Let $p$ be the probability that $A$ wakes up before the alarm,so $p = 0.4$. Let $q$ be the probability t

Vedclass · Accepted Answer

$2.8, 1.68$

Vedclass · Answer

(C) Let $n$ be the number of days,so $n = 7$. Let $p$ be the probability that $A$ wakes up before the alarm,so $p = 0.4$. Let $q$ be the probability that $A$ does not wake up before the alarm,so $q = 1 - p = 1 - 0.4 = 0.6$. This follows a binomial distribution $B(n, p)$. The mean of a binomial distribution is given by $E(X) = n \cdot p$. Mean $= 7 	imes 0.4 = 2.8$. The variance of a binomial distribution is given by $Var(X) = n \cdot p \cdot q$. Variance $= 7 	imes 0.4 	imes 0.6 = 2.8 	imes 0.6 = 1.68$. Thus,the mean and variance are $2.8$ and $1.68$ respectively.

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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