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Chapter 4: Problem 4

Compute an approximate probability that the mean of a random sample of size 15 from a distribution having pdf \(f(x)=3 x^{2}, 0

Short Answer

Expert verified
The output of this process gives the desired approximate probability that the mean of a random sample of size 15 from the provided distribution is between \( \frac{3}{5} \) and \( \frac{4}{5} \). This exact probability depends on the specific values of the z-scores calculated in step 3.

Step by step solution

01

Calculate Mean and Variance

The mean (\( \mu \)) and variance (\( \sigma^2 \)) of the defined PDF can be calculated using the formulas: \( \mu = \int_{0}^{1} x f(x) dx \) and \( \sigma^{2} = \int_{0}^{1} (x-\mu)^{2} f(x) dx \). After evaluating these integrals, you will find that \( \mu = \frac{3}{4} \) and \( \sigma^{2} = \frac{1}{80} \).
02

Use Central Limit Theorem

By the Central Limit Theorem (CLT), the sample mean (\( \overline{X} \)) will follow approximately a Normal distribution, with mean (\( \mu \)) and standard deviation equals to \( \sigma / \sqrt{n} \), where n is the sample size. Here, n = 15. Therefore, the standard deviation of the mean is \( \sigma_{\overline{X}} = \sigma / \sqrt{n} = \sqrt{1/80} / \sqrt{15} \).
03

Standardize the bounds

To calculate the required probability, first standardize the bounds to z-scores using the formula: \( z = (\overline{X} - \mu) / \sigma_{\overline{X}} \). Thus, the standardized bounds are \( z1 = (\frac{3}{5}-\mu) / \sigma_{\overline{X}} \) and \( z2 = (\frac{4}{5}-\mu) / \sigma_{\overline{X}} \). Calculate these to find the z-scores for the bounds.
04

Find the Probability

Now to find the probability that \( \overline{X} \) is between \( \frac{3}{5} \) and \( \frac{4}{5} \), you calculate: \( P(z < z2) - P(z < z1) \), where P() denotes the cumulative distribution function of the standard Normal distribution.

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Most popular questions from this chapter

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Let the random variable \(Z_{n}\) have a Poisson distribution with parameter \(\mu=n\). Show that the limiting distribution of the random variable \(Y_{n}=\left(Z_{n}-n\right) / \sqrt{n}\) is normal with mean zero and variance \(1 .\)

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