Problem 20 - Algebra 2

Question

A normal distribution has a mean of 100 and a standard deviation of \(10 .\) Find the probability that a value selected at random is in the given interval. from 80 to 100

Step-by-Step Solution

Answer

The probability that a randomly selected value from a normally distributed dataset lies between 80 and 100 is 47.72%

Step 1: Calculate the Z-scores

The Z-score needs to be calculated for values of 80 and 100. The formula to find the Z-score is \( z = (x - \mu) / \sigma \), where x is the value from the dataset, \(\mu\) is the mean and \(\sigma\) is the standard deviation. For x=80, the Z-score will be \( z_{80} = (80 - 100) / 10 = -2 \). For x=100, the Z-score will be \( z_{100} = (100 - 100) / 10 = 0 \).

Step 2: Consult the Standard Normal (Z) Table

The next step is to use a Standard Normal (Z) Table to find the probability that a z-score is below a certain value, giving the area to the left under the curve. The entry for z=-2 in the Z-table is 0.0228 which means that there is 0.0228 or 2.28% probability that a value is below 80. Similarly, for z=0, the value from the table is 0.5000, suggesting that there is a 50% probability that a value is below 100.

Step 3: Calculate the Required Probability

The final step is to calculate the probability that a value lies between 80 and 100. This involves subtracting the probability of the value being below 80 from the probability of the value being below 100. Therefore, the calculation will be: Probability = 0.5000 - 0.0228 = 0.4772 or 47.72%.

Key Concepts

Z-score

A Z-score is a way of quantifying how far a particular data point is from the mean of a distribution. It tells us the number of standard deviations a value is from the mean.
To calculate a Z-score, use the formula:

• \( z = \frac{x - \mu}{\sigma} \)

Here, \( x \) is the data point you are interested in, \( \mu \) is the mean of the data set, and \( \sigma \) is the standard deviation.
In the context of our problem, we found two Z-scores:

• The Z-score for 80 was -2, meaning 80 is 2 standard deviations below the mean.
• The Z-score for 100 was 0, indicating it is exactly the mean.

These Z-scores help us determine how likely a random value from the distribution will be between these points.

Standard Deviation

Standard deviation is a measure of the amount of variation or dispersion in a set of values.
It indicates how spread out the values in a dataset are around the mean (average) value.
A small standard deviation means that the values tend to be close to the mean, while a large standard deviation indicates that the values are spread out over a wider range.

• For a perfectly normal distribution, approximately 68% of the data falls within one standard deviation from the mean.
• About 95% is within two standard deviations, and 99.7% is within three standard deviations.

In our example, the standard deviation was \( 10 \), showing that for the numbers 80 and 100:

• 80 is two standard deviations below the mean.
• 100 is exactly at the mean.

Understanding standard deviation helps in assessing the probability of finding values within certain ranges.

Z-table

The Z-table is a chart that shows the total probability (area under the curve) to the left of a given Z-score in a standard normal distribution.
It's an essential tool for finding probabilities associated with normal distributions.

• The Z-table tells you how much of the distributions lie below a given Z-score.

For instance, in the step-by-step example:

• A Z-score of -2 corresponds to an approximate probability of 0.0228, suggesting that 2.28% of the data is below this point.
• A Z-score of 0 gives a probability of 0.5000, meaning that 50% of the data is below the mean.

By using the Z-table, one can find the probability of random values occurring within specific ranges, like from 80 to 100 in our scenario.

Probability

Probability, in the context of normal distributions, refers to the likelihood of a particular value or range of values occurring.
We use Z-scores and the Z-table to determine these probabilities.

• The task is to find how likely values fall between two points on a distribution curve.

From the previous calculations:

• The probability that a value is below 80 is 0.0228.
• The probability that a value is below 100 is 0.5000.

To find the probability of a value being between 80 and 100, we subtract 0.0228 from 0.5000, resulting in 0.4772, or 47.72%.
This means there is a 47.72% chance a random value from the distribution will fall within this range.

• This technique offers powerful insights in fields like statistics and data analysis.