Problem 10 - Algebra 2 and Trigonometry

Question

In \(10-13,\) the mean and standard deviation of a normal distribution are given. Find each probability to the nearest hundredth. mean \(=80,\) standard deviation \(=10, P(50 \leq x \leq 95)\)

Step-by-Step Solution

Answer

The probability \( P(50 \leq x \leq 95) \) is approximately 0.93.

Step 1: Calculate Z-scores

The formula for a Z-score is \( Z = \frac{X - \mu}{\sigma} \), where \( X \) is the value, \( \mu \) is the mean, and \( \sigma \) is the standard deviation. Substitute for \( X = 50 \) and \( X = 95 \) to find the Z-scores.\[ Z_1 = \frac{50 - 80}{10} = -3 \] \[ Z_2 = \frac{95 - 80}{10} = 1.5 \]

Step 2: Use Z-table for Probability

Using the standard normal distribution table (Z-table), find the probabilities corresponding to the Z-scores. \( P(Z < -3) \) corresponds to a probability of 0.0013, and \( P(Z < 1.5) \) corresponds to a probability of 0.9332.

Step 3: Calculate the Probability Between Two Z-scores

The probability \( P(50 \leq X \leq 95) \) is given by \( P(Z < 1.5) - P(Z < -3) \). Substituting the probabilities from the Z-table: \( 0.9332 - 0.0013 = 0.9319 \).

Step 4: Round the Probability

Round the probability obtained in the previous step to the nearest hundredth. This gives us \( P(50 \leq X \leq 95) \approx 0.93 \).

Key Concepts

Mean and Standard Deviation

In the context of a normal distribution, the mean and standard deviation are critical components. The mean, denoted as \( \mu \), indicates the central point of the data distribution, essentially serving as the 'average'. It reflects where most data points are likely to be situated along the number line.
The standard deviation, represented by \( \sigma \), measures the spread of the data around the mean. A small standard deviation implies that data points are closely clustered around the mean, while a larger standard deviation signifies a broader spread. Understanding these two parameters helps in analyzing the distribution shape. With a mean of 80 and a standard deviation of 10, the exercise you're working on shows how the data values are distributed around 80 and how dispersed they are.

Z-scores

Z-scores are a way to convert individual data points into a standard normal distribution, which always has a mean of 0 and a standard deviation of 1. This conversion helps in determining how far and in what direction, a data point lies from the mean. The formula for computing the Z-score is:

• \( Z = \frac{X - \mu}{\sigma} \)

where \( X \) is the data point, \( \mu \) is the mean, and \( \sigma \) is the standard deviation.
In your exercise, by applying this formula to \( X = 50 \) and \( X = 95 \), the Z-scores are calculated as -3 and 1.5, respectively. These Z-scores tell us that the value of 50 lies three standard deviations below the mean and 95 lies one and a half standard deviations above the mean.

Probability Calculation

Once you have the Z-scores for the boundary values of your range, calculating the probability that a value falls within this range is straightforward. The probability calculation for defined limits \(P(a \leq X \leq b)\) can be found using the probabilities of the corresponding Z-scores.
For this exercise, you've determined the Z-scores of -3 and 1.5. The probability calculation is given by finding the difference between the probabilities for these two Z-scores. In our case, this results in:

• \( P(Z < 1.5) - P(Z < -3) = 0.9332 - 0.0013 = 0.9319 \)

This indicates about 93.19% of values lie within this range before rounding.

Z-table

A Z-table helps read off probabilities corresponding to Z-scores under the standard normal distribution. These tables typically show the probabilities that a standard normal variable will take a value less than a given Z-score.
When dealing with a Z-table, you simply look up your calculated Z-scores to find the associated probabilities. In the current problem, for a Z-score of -3, the table gives a probability of 0.0013, and for a Z-score of 1.5, the probability is 0.9332.
Z-tables are essential tools that simplify the transformation of Z-scores into probabilities, without the need for intricate calculations. They are extensively used as a reference in statistics for probability estimation, providing quick and easy access to the information you need.