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Chapter 4: Problem 7

If \(Y\) is \(b\left(100, \frac{1}{2}\right)\), approximate the value of \(P(Y=50)\).

Short Answer

Expert verified
The approximate value of \(P(Y=50)\) is 0.0797.

Step by step solution

01

Calculate the mean and standard deviation

Before proceeding with the approximation, it's necessary to calculate the mean \(\mu\) and standard deviation \(\sigma\) from the binomial formula. The mean \(\mu\) for a binomial distribution is given by the formula \(\mu = np\), and the standard deviation \(\sigma\) is given by \(\sigma = \sqrt{np(1-p)}\). For the given values, \(\mu = 100*0.5 = 50\), and the standard deviation \(\sigma = \sqrt{100*0.5*0.5} = 5\).
02

Approximate the binomial distribution using a normal distribution

The significant principle behind this step is the central limit theorem, which states that if sample size is large enough, the sampling distribution of the mean will be approximately normally distributed. So, \(Y\) can be approximated by \(N(\mu,\sigma^2)\), a normal distribution with a mean of 50 and variance of \(5^2\).
03

Calculate the required probability

The required probability \(P(Y=50)\) in the standard normal distribution would be \(P(49.5<Y<50.5)\) because of the Continuity Correction Factor, due to the fact we are approximating a discrete distribution using a continuous distribution. Converting to z-scores: \(P(\frac{49.5-\mu}{\sigma}<Z<\frac{50.5-\mu}{\sigma})\), substituting the known values we get \(P(-0.1<Z<0.1)\). Using Standard Normal tables, or as calculated by software or calculators, one finds that this corresponds to a probability of approximately 0.0797.

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Most popular questions from this chapter

Let \(Y\) be \(b\left(72, \frac{1}{3}\right)\). Approximate \(P(22 \leq Y \leq 28)\).

Let \(X_{1}, \ldots, X_{n}\) be a random sample from a uniform \((a, b)\) distribution. Let \(Y_{1}=\min X_{i}\) and let \(Y_{2}=\max X_{i} .\) Show that \(\left(Y_{1}, Y_{2}\right)^{\prime}\) converges in probability to the vector \((a, b)^{\prime}\).

Let \(X_{n}\) have a gamma distribution with parameter \(\alpha=n\) and \(\beta\), where \(\beta\) is not a function of \(n .\) Let \(Y_{n}=X_{n} / n .\) Find the limiting distribution of \(Y_{n}\).

Let \(\bar{X}_{n}\) denote the mean of a random sample of size \(n\) from a Poisson distribution with parameter \(\mu=1\) (a) Show that the mgf of \(Y_{n}=\sqrt{n}\left(\bar{X}_{n}-\mu\right) / \sigma=\sqrt{n}\left(\bar{X}_{n}-1\right)\) is given by \(\exp \left[-t \sqrt{n}+n\left(e^{t / \sqrt{n}}-1\right)\right]\) (b) Investigate the limiting distribution of \(Y_{n}\) as \(n \rightarrow \infty\). Hint: Replace, by its MacLaurin's series, the expression \(e^{t / \sqrt{n}}\), which is in the exponent of the mgf of \(Y_{n}\).

. Let \(X_{1}, X_{2}, \ldots, X_{n}\) be a random sample from a Poisson distribution with mean \(\mu .\) Thus \(Y=\sum_{i=1}^{n} X_{i}\) has a Poisson distribution with mean \(n \mu .\) Moreover, \(\bar{X}=Y / n\) is approximately \(N(\mu, \mu / n)\) for large \(n .\) Show that \(u(Y / n)=\sqrt{Y / n}\) is a function of \(Y / n\) whose variance is essentially free of \(\mu .\)
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