Question

If \(X\) is \(N\left(\mu, \sigma^{2}\right)\), find \(b\) so that \(P[-b<(X-\mu) / \sigma

Short answer

The value of \(b\) that satisfies the equation \(P[-b<(X-\mu) / \sigma<b]=0.90\) is approximately \(1.645\) by using Table II and the 'qnorm' command in R gives \(b = 1.6448536269514722\).

Step-by-step solution

Understand the Normal Distribution and Standardize the random variable

First, recall that a normal random variable \(X\) can be standardized by subtracting the mean from the variable and dividing by the standard deviation, also known as creating a Z-score. This brings the random variable to a standard normal distribution with mean 0 and standard deviation of 1. This implies that \( \frac{(X-\mu)}{\sigma} \) is a standard normal random variable.

Apply the Properties of the Standard Normal Distribution

Next, use the properties of a standard normal distribution to solve for \(b\). Since the distribution is symmetric around \(0\), you know that the probability that a standard normal variable is less than \(b\) is \(0.90 / 2 = 0.45\) for the range \([0,b]\) and similarly for \([-b,0]\). The sum of these two gives \(0.90\) as required.

Calculate the Z-score using Table II or R qnorm

Finally, to calculate \(b\), find the z-score that corresponds to a cumulative probability just over \(0.45\) in the z-score table or by using the 'qnorm' command in R. This z-score is the required value of \(b\). The 'qnorm' command in R gives more accurate results as it calculates the exact value rather than an approximation from the table.