Question

If \(\mathrm{X}\) has a normal distribution with mean 9 and standard deviation 3 , find \(\mathrm{P}(5<\mathrm{X}<11)\).

Short answer

The probability of the variable X falling between 5 and 11 is approximately 0.6568 or 65.68%.

Step-by-step solution

Identify the mean and standard deviation

The problem states that X has a normal distribution with a mean (μ) of 9 and a standard deviation (σ) of 3.

Convert the values into z-scores

We need to find the probability P(5 < X < 11). To do this, we will convert the values 5 and 11 into z-scores using the formula: \(z = \frac{X - \mu}{\sigma}\) For X = 5, \(z_1 = \frac{5 - 9}{3} = \frac{-4}{3}\) For X = 11, \(z_2 = \frac{11 - 9}{3} = \frac{2}{3}\) Now, we need to find P(\(\frac{-4}{3}\) < Z < \(\frac{2}{3}\)), where Z is the standard normal variable.

Use the standard normal table to find the probabilities

Using the standard normal table, we look up the probabilities associated with the z-scores we calculated: P(Z < -\(\frac{4}{3}\)) ≈ 0.0918 P(Z < \(\frac{2}{3}\)) ≈ 0.7486

Calculate the probability within the range

Now, we need to find the probability of the standard normal variable Z falling between -\(\frac{4}{3}\) and \(\frac{2}{3}\). We do this by subtracting the smaller probability from the larger probability: P(5 < X < 11) = P(\(\frac{-4}{3}\) < Z < \(\frac{2}{3}\)) = P(Z < \(\frac{2}{3}\)) - P(Z < -\(\frac{4}{3}\)) P(5 < X < 11) ≈ 0.7486 - 0.0918 = 0.6568 So, the probability of the variable X falling between 5 and 11 is approximately 0.6568 or 65.68%.

Key concepts

Mean and Standard Deviation

The mean and standard deviation are fundamental concepts of statistics, especially when dealing with normal distributions. The mean, often symbolized by \( \mu \), represents the average or central value of a data set. It's like the balancing point of the data set. In our exercise, the mean is given as 9. This tells us that the average value around which the normal distribution is centered is 9.

The standard deviation, denoted by \( \sigma \), measures the amount of variation or dispersion in a set of values. A small standard deviation indicates that the values tend to be close to the mean, while a larger standard deviation indicates that the values are spread out over a wider range. For the given problem, the standard deviation is 3. This means that most of the data falls within 3 units of the mean.

Understanding these concepts helps us make inferences about the data set and is crucial in converting raw data to standardized values, or z-scores, which we'll explore next.

Z-Scores

Z-scores, also known as standard scores, are a way of describing a data point's distance from the mean in terms of standard deviations. When we measure how far away a score is from the average (mean), we can use z-scores to standardize this process. The formula to calculate a z-score is: \[ z = \frac{X - \mu}{\sigma} \]where \( X \) is the value, \( \mu \) is the mean, and \( \sigma \) is the standard deviation.

In the context of the normal distribution problem, to figure out the probability between two values (5 and 11), we first converted these values into z-scores. For \( X = 5 \), the z-score was calculated as:

\( z_1 = \frac{5 - 9}{3} = \frac{-4}{3} \)

This tells us that 5 is \( \frac{-4}{3} \) or about -1.33 standard deviations below the mean.

For \( X = 11 \), the z-score was:

\( z_2 = \frac{11 - 9}{3} = \frac{2}{3} \)

This indicates that 11 is \( \frac{2}{3} \) or approximately 0.67 standard deviations above the mean.

Z-scores allow us to translate different data metrics into a common language, enabling easy comparison and probability calculation using the standard normal table.

Standard Normal Table

The standard normal table, also known as the Z-table, is a mathematical table that allows us to find the probability that a standard normal variable is less than or equal to a given value. This table is essential when dealing with normal distributions because it translates z-scores into probability values.

In our original exercise, once we converted the given values (5 and 11) into z-scores, we used the standard normal table to look up the probabilities. For the z-score of \( \frac{-4}{3} \), the table provides the probability \( P(Z < \frac{-4}{3}) \approx 0.0918 \), indicating that there's about a 9.18% chance a value is below this point.

For \( \frac{2}{3} \), the standard normal table gives \( P(Z < \frac{2}{3}) \approx 0.7486 \). This probability value corresponds to about a 74.86% chance of being below this z-score.

• By subtracting these two probabilities, \( 0.7486 - 0.0918 \),

We determine that there's a 65.68% chance the data falls between the z-scores of \( \frac{-4}{3} \) and \( \frac{2}{3} \). The standard normal table is vital for transforming z-scores into definitive probability insights.