PlanetMaster
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1. I like to take the step-by-step approach here. We need to find probability that there are exactly 2 of 8 days \(\text{(}^8C_{2}\text{)}\) on which martin is playing computer games when his friend phones. So there are 2 days when he calls in his gaming time and 6 days when he might call outside this window i.e \(\text{(}0.25^{2}\times0.75^{6}\text{)}\).
So \(p=^8C_{2}\times0.25^{2}\times0.75^{6}\)
\(p=0.3115\)
2. Recall that normal distribution can be used as an approximation to the binomial distribution if X ~ B(n, p) and if n is large and/or p is close to ½, then X is approximately N(np, npq) (where q = 1 - p).
Since \(12\times0.25=3\) is quite small, using normal approximation isn't a great idea here.
3. Recall that \(\text{Z-score}=\frac{X-\mu}{\sigma}\).
Since we need P(X ≥ 13), we can use continuity correction P(X ≥ n) = P(X > n – 0.5) so
\(\text{Z-score}>\frac{12.5-10}{\sqrt{7.5}}=0.91241\)
\(P(X<12.5) = 0.81922\)
\(P(X>12.5) = 1 - P(x<12.5) = 0.18078\)
Hope this helps!
