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Normal distribution describes how values of a variable are spread around the mean. It forms a bell-shaped curve. For practical work, we convert normal variables into a standard normal variable, often denoted as \(Z\). This process is called standardization. The purpose of standardization is to compare different normal distributions or to use standard tables with pre-calculated probabilities.
The standard normal variable \(Z\) is defined using the formula:
\[ Z = \frac{X - \mu}{\sigma} \]
where:

• \(X\) is the original variable.
• \(\mu\) is the mean of the distribution.
• \(\sigma\) is the standard deviation of the distribution.

By applying this formula, we transform the original value into a standardized value. This value then corresponds to specific locations on the Z-table, making it easier to find related probabilities.

Calculating the Z-score is a critical step in many probability problems involving normal distribution. The Z-score tells us how many standard deviations an element is from the mean. To compute it, you can use the formula:
\[ Z = \frac{X - \mu}{\sigma} \]
Let's break it down with an example:
Given: \(X = 45\), \(\mu = 50\), and \(\sigma = 7\). Plug these values into the formula:
\[ Z = \frac{45 - 50}{7} \approx -0.7143 \]
Here, the Z-score of -0.7143 indicates that 45 is approximately 0.7143 standard deviations below the mean. This calculated Z-score can then be used to find probabilities using the Z-table.
The Z-score calculation helps standardize different variables, allowing us to use standard probability tables effectively.

Finding probability using the Z-table is the next step after calculating the Z-score. A Z-table, or standard normal table, provides the probability that a standard normal variable is less than or equal to a given Z-value. For our example, we calculated a Z-score of -0.7143.
Here's how to use the Z-table:

• First, locate the value -0.71 in the first column (Z-values up to the tenth place).
• Next, in the top row, locate 0.0043 (hundredth place of Z, the final digit of -0.7143).
• Find the intersection where the row and column meet. This gives the probability for \(Z \le -0.7143\).

The Z-table indicates that \(P(Z \le -0.7143)\) is approximately 0.2375. Thus, the probability that \(X \le 45\) in our original problem is 0.2375, which corresponds to the shaded area under the standard normal curve.