Question

Let \(X\) be \(N(5,10)\). Find \(P\left[0.04<(X-5)^{2}<38.4\right]\).

Short answer

The probability \(P[-\sqrt{38.4} < X - 5 < \sqrt{38.4}]\) corresponds to the difference between the cumulative standard normal probabilities \(\Phi(\sqrt{38.4}/\sqrt{10}) - \Phi(-\sqrt{38.4}/\sqrt{10})\).

Step-by-step solution

Interpret the given inequality

We have the inequality \(0.04 < (X - 5)^{2} < 38.4\). This can be rewritten as \(\sqrt{0.04} < X - 5 < \sqrt{38.4}\), or in other terms \( -\sqrt{38.4} < X - 5 < -\sqrt{0.04}\) and \(\sqrt{0.04} < X - 5 < \sqrt{38.4}\), essentially yielding the double inequality \(-\sqrt{38.4} < X - 5 < \sqrt{38.4}\) as the condition we are interested in. This is equivalent to asking for the random variable to be within certain number of standard deviation units from the mean.

Standardize the inequalities

For a normal random variable, we usually standardize the variable by subtracting the mean and dividing by the stdandard deviation to get the Z-score. In this case, \(Z = \frac{(X - 5)}{\sqrt{10}}\). Hence our inequality becomes \(-\frac{\sqrt{38.4}}{\sqrt{10}} < Z < \sqrt{38.4}/\sqrt{10}\).

Use the Z-table

The Z-table gives the cumulative probability that a standard normal random variable equals a value less than Z. Therefore, we need to find \(\Phi(\sqrt{38.4}/\sqrt{10}) - \Phi(-\sqrt{38.4}/\sqrt{10})\), where \(\Phi\) represents the cumulative standard normal distribution.

Calculation

Finally, if you look up \(\sqrt{38.4}/\sqrt{10}\) and \(-\sqrt{38.4}/\sqrt{10}\) in the Z-table, we get the respective probabilities which can be subtracted to get the result.

Key concepts

Standard Normal Variable

When delving into the realm of probability and statistics, a standard normal variable is a cornerstone concept. It is a random variable that has a normal distribution with a mean of 0 and a standard deviation of 1. In simpler terms, it's a special case of the more general normal distributions, which can have any mean or standard deviation.
Standard normal variables are used as a comparison tool to assess other normal distributions, as they can be converted to the standard normal form through a process called standardization. This involves transforming the data so that its mean and variation match those of the standard normal variable, making it easier to analyze and interpret.

• Conversion to a standard normal variable enables comparison across different normal distributions.
• For a variable X with mean \( \mu \) and standard deviation \( \sigma \) the transformation \( Z = \frac{X - \mu}{\sigma} \) gives us the corresponding standard normal variable, Z.
• The standard normal distribution provides a reference for determining how unusual or extreme an observation is.

This process is essential when dealing with exercises like the one given, where transforming to a standard normal variable allows us to work within a standardized framework for probability calculation and to make use of widely available Z-tables.

Z-score

The Z-score is a statistical measurement that describes a value's relationship to the mean of a group of values, measured in terms of standard deviations from the mean. In essence, it is a standard score that indicates how many standard deviations an element is from the mean.

In the exercise provided, calculating the Z-score allows us to express how far and in what direction a value deviates from the mean, after accounting for the spread of the distribution. The Z-score is calculated using the formula:
\[ Z = \frac{(X - \mu)}{\sigma} \]
Where \( X \) is the value from the original normal distribution, \( \mu \) is the mean, and \( \sigma \) is the standard deviation. This score tells us where the value lies in relation to the standard normal distribution:

• A Z-score of 0 corresponds to a value that is exactly at the mean.
• A Z-score greater than 0 indicates the value is above the mean.
• A Z-score less than 0 signifies the value is below the mean.

Once we have the Z-scores, we can compare them to a standard normal distribution to understand the probability of observing such a value within that distribution, which leads us directly to the notion of cumulative standard normal distribution.

Cumulative Standard Normal Distribution

When we talk about the cumulative standard normal distribution, we refer to the probability that a standard normal variable \( Z \) will have a value less than or equal to a given value \( z \). It is essentially the cumulative area under the standard normal curve up to point \( z \).

In the context of our exercise, after finding the standardized Z-scores, we use the cumulative standard normal distribution, often represented by \( \Phi(z) \) in formulas, to find the probabilities associated with these scores. This requires a Z-table, also known as a standard normal probability table, which lists the cumulative probabilities for different Z-score values.
\[ P(a < Z < b) = \Phi(b) - \Phi(a) \]
For the given exercise, after converting the variable \( X \) to Z, we look up the cumulative probabilities for both \( \Phi(\sqrt{38.4}/\sqrt{10}) \) and \( \Phi(-\sqrt{38.4}/\sqrt{10}) \) in the Z-table and subtract them to find \( P(0.04<(X-5)^{2}<38.4) \) which is the required probability. This cumulative distribution is crucial in areas such as hypothesis testing and confidence interval estimation, representing the block upon which a large part of inferential statistics is built.