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Chapter 4: Problem 1

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 \(P(49<\bar{X}<51)\).

Short Answer

Expert verified
The approximate value of P(49<\(\bar{X}\)<51) is \(1 - 2*0.1587 = 0.6826\).

Step by step solution

01

Identify population parameters

Start by noting that since we are dealing with a chi-square distribution with 50 degrees of freedom, the population mean (\(\mu\)) and variance (\(\sigma^{2}\)) are equal to the degrees of freedom and twice the degrees of freedom respectively, that is, \(\mu = 50\) and \(\sigma^{2} = 2*50 = 100\).
02

Calculate the mean and standard deviation of the sample mean

By the Central Limit Theorem, the mean of the sample means equals the population mean and the standard deviation of the sample mean (\(\sigma_{\bar{X}}\)) equals the standard deviation of the population divided by the square root of the sample size. Thus, the mean and variance of the sample mean are \(\mu_{\bar{X}} = \mu = 50\) and \(\sigma_{\bar{X}} = \sigma / \sqrt{n} = \sqrt{100/100} = 1\).
03

Standardize the sample mean and compute the probabilities

To compute the probabilities, we should convert the values to z-scores using the formula \( Z = (\bar{X} - \mu_{\bar{X}}) / \sigma_{\bar{X}} \). The value for \( P(49<\bar{X}<51) \) is equivalent to \( P(-11) \), which is generally known or can be looked up to be 0.1587. Thus, \( P(-1<Z<1) = 1 - 2*0.1587\).

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Most popular questions from this chapter

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