Understanding the Normal Distribution Problem

The question asks us to find the probability P(|X| \le 3) for a normal random variable X with mean zero and variance 9. We are given that a = P(X \ge 3).

First, let's identify the key parameters of the normal distribution:

• Mean (\mu): 0
• Variance (\sigma^2): 9
• Standard Deviation (\sigma): \sqrt{9} = 3

The normal distribution with mean 0 is symmetric around zero. This symmetry is crucial for solving this problem.

Converting to Standard Normal Variable (Z)

To work with probabilities for a general normal variable, we usually convert it to the standard normal variable Z using the formula:

Z = \frac{X - \mu}{\sigma}

In this case, \mu = 0 and \sigma = 3, so Z = \frac{X - 0}{3} = \frac{X}{3}.

Analyzing the given probability 'a'

We are given a = P(X \ge 3). Let's convert the value X=3 to a Z-score:

Z = \frac{3}{3} = 1

So, a = P(Z \ge 1).

Analyzing the probability to find P(|X| \le 3)

We need to find P(|X| \le 3). The inequality |X| \le 3 means -3 \le X \le 3. Let's convert these X values to Z-scores:

• For X = -3, Z = \frac{-3}{3} = -1.
• For X = 3, Z = \frac{3}{3} = 1.

So, P(|X| \le 3) is equivalent to P(-1 \le Z \le 1).

Using Symmetry of the Standard Normal Distribution

The standard normal distribution (Z) is symmetric around its mean, which is 0. This means the area under the curve from 0 to any value z is the same as the area from -z to 0. Also, the area in the right tail P(Z \ge z) is equal to the area in the left tail P(Z \le -z).

We know a = P(Z \ge 1). Due to symmetry, P(Z \le -1) = P(Z \ge 1) = a.

The total area under the standard normal curve is 1. The area P(-1 \le Z \le 1) can be found by taking the total area and subtracting the areas in the two tails:

P(-1 \le Z \le 1) = 1 - P(Z < -1) - P(Z > 1)

For continuous distributions, P(Z < -1) = P(Z \le -1) and P(Z > 1) = P(Z \ge 1).

So, P(-1 \le Z \le 1) = 1 - P(Z \le -1) - P(Z \ge 1).

Substituting the value of a:

P(-1 \le Z \le 1) = 1 - a - a = 1 - 2a.

Therefore, P(|X| \le 3) = 1 - 2a.

Conclusion on the Probability Calculation

Based on the properties of the normal distribution centered at zero and the given information a = P(X \ge 3), we found that P(|X| \le 3) equals 1 - 2a.

Revision Table: Normal Distribution Concepts

Additional Information: Applications of Normal Distribution

The normal distribution is one of the most important probability distributions in statistics. It is used to model many natural phenomena, such as:

• Heights and weights of people.
• Measurement errors in experiments.
• Test scores.
• Financial market movements.

Understanding how to calculate probabilities using Z-scores and utilizing the symmetry property is fundamental for working with normal distributions.

The problem highlights how symmetry simplifies probability calculations for normal distributions centered at zero. P(|X| \le c) for a normal distribution with mean 0 and standard deviation \sigma can be written as P(-c/\sigma \le Z \le c/\sigma), which, by symmetry, is 1 - 2 \times P(Z \ge c/\sigma).