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Chapter 2: Problem 69

If \(X\) is normally distributed with mean 1 and variance 4 , use the tables to find \(P\\{2

Short Answer

Expert verified
The probability that \(X\) falls between 2 and 3 is approximately 0.1498.

Step by step solution

01

Identify the parameters

In this exercise, we are given the mean (\(\mu\)) as 1 and the variance (\(\sigma^2\)) as 4. So, we have: - Mean: \(\mu = 1\), - Variance: \(\sigma^2 = 4\), - Standard deviation: \(\sigma = \sqrt{4} = 2\). We want to find the probability that a value of \(X\) lies between 2 and 3: \(P(2 < X < 3)\).
02

Calculate Z-scores

We need to convert the given values of \(X\) (2 and 3) into Z-scores using the formula: \(Z = \frac{X - \mu}{\sigma}\). 1. For X = 2: \(Z_1 = \frac{2 - 1}{2} = \frac{1}{2} = 0.5\) 2. For X = 3: \(Z_2 = \frac{3 - 1}{2} = \frac{2}{2} = 1\) So now we want to find the probability \(P(0.5 < Z < 1)\).
03

Look up Z-scores in the Z-table

Now we look up the Z-scores in the standard normal distribution table and find the corresponding probabilities: 1. For \(Z_1 = 0.5\), the table shows a probability of 0.6915. 2. For \(Z_2 = 1\), the table shows a probability of 0.8413.
04

Calculate the final probability

In the final step, we subtract the smaller probability from the larger one to find the probability of \(Z\) lying between 0.5 and 1: \(P(0.5 < Z < 1) = P(Z < 1) - P(Z < 0.5)\) \(P(0.5 < Z < 1) = 0.8413 - 0.6915\) \(P(0.5 < Z < 1) = 0.1498\) So the probability that \(X\) falls between 2 and 3 is approximately 0.1498.

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Most popular questions from this chapter

Let \(X_{1}, X_{2}, X_{3}\), and \(X_{4}\) be independent continuous random variables with a common distribution function \(F\) and let $$ p=P\left\\{X_{1}X_{3}

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Let \(\phi\left(t_{1}, \ldots, t_{n}\right)\) denote the joint moment generating function of \(X_{1}, \ldots, X_{n}\). (a) Explain how the moment generating function of \(X_{i}, \phi_{X_{i}}\left(t_{i}\right)\), can be obtained from \(\phi\left(t_{1}, \ldots, t_{n}\right)\). (b) Show that \(X_{1}, \ldots, X_{n}\) are independent if and only if $$ \phi\left(t_{1}, \ldots, t_{n}\right)=\phi_{x_{1}}\left(t_{1}\right) \cdots \phi_{X_{n}}\left(t_{n}\right) $$

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