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Problem 38

# 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$.

Expert verified
The value of $$n$$ should be approximately 33.
See the step by step solution

## 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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