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Let \(X \sim \mathrm{B}(n, p)\). Use the CLT to find \(n\) such that: $P[X>n / 2] \leq 1-\alpha\(. Calculate the value of \)n\( when \)\alpha=0.90\( and \)p=0.45$.

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The value of \(n\) should be approximately 33.
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Step 1 - Understand the problem

We're given a variable \(X\) with a binomial distribution \(B(n, p)\). We're to find \(n\) so that \(P[X > n/2] ≤ 1-\alpha\). The central limit theorem (CLT) can be applied as it states that the sum of a large number of independent and identically distributed random variables will approximate a normal distribution, regardless of the shape of the original distribution.

Step 2 - Apply the Central Limit Theorem (CLT)

With \(p=0.45\) and \((1-p)=0.55\), the mean \(\mu = np\) and the standard deviation \(\sigma = \sqrt{np(1-p)}\). We have \(x = n/2\). By the CLT, when \(n\) is large, \(X\) approximates a normal distribution which means \(Z = (x-\mu)/\sigma\) where \(Z\) follows a standard normal distribution.

Step 3 - Setup the inequality

We set up the provided inequality according to the parameters of the problem. Given \(P[X > n/2] ≤ 1-\alpha\), we substitute into the CLT expression to get \(P[Z > (n/2 - np) / \sqrt{np(1-p)}] ≤ 1-\alpha\). Normalize \(Z\) to reference standard normal distribution tables.

Step 4 - Solve the inequality

Use standard normal distribution tables where \(Z\) score for \(\alpha=0.90\) (one-tailed test) is approximately 1.28. Solve for \(n\) by equating the left side of the inequality to 1.28, giving us the equality \((n/2 - np) / \sqrt{np(1-p)} \approx 1.28\). As a result, we can solve for \(n\) with the given \(p\) and \(\alpha\) values, ending up with \(n \approx 33\)

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