A student gets at least 8 hours of sleep 45% of the nights; the sleeping schedule is independent from night to night. Let Xi indicate whether the student gets at least 8 hours of sleep during the next 4 night respectively for i = 1, 2, 3, and 4. Let X = X1 + X2 + X3 + X4. Find the variance of X.
I'm struggling to understand what kind of distribution this problem is follow. I think it would be binomial. In that case wouldn't the variance of having x successes in 4 nights (trials) be modelled by the binomial. Hence the variance of X is (npq)^0.5 ? However, I feel like the question wants me to do some RV algebra. But a bit stuck.
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$\begingroup$There are four requirements for a binomial distribution:
- Fixed number of trials
- Only two possible outcomes per trial
- Same success probability for every trial
- Trials are independent of each other.
All four conditions are satisfied. $n = 4, p = 0.45$
$\endgroup$ 1 $\begingroup$Each Xi is bernoulli and the entire X is binomial. A binomial random variable is the sum of independent, identically distributed bernoulli random variables. That is, $X_i\sim \mathscr{Bern}(.45)$ for i=1,2,3,4 and $X\sim \mathscr{Bino}(4,.45)$.
Now you can find the variance of the sum of the Xi's or the X directly. We have $Var(X)=np(1-p)=4(.45)(.55)$. Also, the Xi's are independent, so they are uncorrelated (independence is stronger than correlation), thus $Var(X_1+X_2+X_3+X_4)=p(1-p)+...+p(1-p)=.45(.55)+...+.45(.55)=4(.45)(.55)$.
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