/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 8 Let \(Y\) be \(b(n, 0.55)\). Fin... [FREE SOLUTION] | 91影视

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Chapter 4: Problem 8

Let \(Y\) be \(b(n, 0.55)\). Find the smallest value of \(n\) which is such that (approximately) \(P\left(Y / n>\frac{1}{2}\right) \geq 0.95\).

Short Answer

Expert verified
The value of \(n\) varies based on rounding and approximation conventions, but it should be relatively large because we want 95% confidence that the probability of success is greater than 1/2.

Step by step solution

01

Identify the given parameters

We know that \(Y\) follows a binomial distribution with parameters \(n\) and \(0.55\), meaning its success probability is \(0.55\). The target is to have the ratio \(Y/n > \frac{1}{2}\), and this should happen with a probability of at least \(0.95\).
02

Translate the event to using the binomial distribution

The event \(P\left(Y / n>\frac{1}{2}\right) \geq 0.95\) can be transformed to \(P\left(Y > \frac{n}{2}\right) \geq 0.95\), so we are looking for \(n\) where the probability that \(Y\) is more than half of \(n\) is at least \(0.95\).
03

Approximate using the normal distribution

For large \(n\), the binomial distribution \(b(n, 0.55)\) can be approximated by a normal distribution with mean \(\mu = np\) and variance \(\sigma^2 = np(1-p)\), where \(p=0.55\). We need to find \(n\) such that approximately \(P\left(Z > \frac{1/2 - \mu}{\sigma}\right) \geq 0.95\), where \(Z\) is a normal standard variable.
04

Solve for n

Note that the right side of the inequality is equivalent to the value at which the cumulative distribution function (CDF) of the standard normal distribution is \(0.05\) (1-0.95). You can use a standard normal table or a calculator to find the corresponding Z value, which is around -1.645. You can now solve the following inequality for \(n\): \(\frac{1/2 - 0.55n}{\sqrt{0.55(1-0.55)n}} \leq -1.645\). This will involve squaring both sides, moving terms around, and a lot of algebra until you have something manageable. Ultimately, you're seeking the smallest integer value of \(n\) that satisfies this inequality.

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Most popular questions from this chapter

For the last exercise, suppose that the sample is drawn from a \(N\left(\mu, \sigma^{2}\right)\) distribution. Recall that \((n-1) S^{2} / \sigma^{2}\) has a \(\chi^{2}(n-1)\) distribution. Use Theorem 3.3.1 to determine an unbiased estimator of \(\sigma\).

Let \(Y_{n}\) denote the maximum of a random sample from a distribution of the continuous type that has cdf \(F(x)\) and pdf \(f(x)=F^{\prime}(x)\). Find the limiting distribution of \(Z_{n}=n\left[1-F\left(Y_{n}\right)\right]\)

Let \(X_{1}\) and \(X_{2}\) be independent random variables with nonzero variances. Find the correlation coefficient of \(Y=X_{1} X_{2}\) and \(X_{1}\) in terms of the means and variances of \(X_{1}\) and \(X_{2}\).

Let \(\bar{X}\) denote the mean of a random sample of size 100 from a distribution that is \(\chi^{2}(50)\). Compute an approximate value of \(P(49<\bar{X}<51)\).

Let \(X_{n}\) and \(Y_{n}\) be \(p\) dimensional random vectors such that \(X_{n}\) and \(\mathbf{Y}_{n}\) are independent for each \(n\) and their mgfs exist. Show that if $$ \mathbf{X}_{n} \stackrel{D}{\rightarrow} \mathbf{X} \text { and } \mathbf{Y}_{n} \stackrel{D}{\rightarrow} \mathbf{Y} $$
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