Problem 25 - 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. at least 80

Step-by-Step Solution

Answer

The probability of selecting a value at least 80 from the given normal distribution is approximately 0.9772.

Step 1: Calculate Z-Score

The first step involves converting the given value (80) into a z-score using the formula \(z = (X - μ) / σ\), where \(X\) is the value (80), \(μ\) is the mean (100), and \(σ\) is the standard deviation (10). Calculating, \(z = (80 - 100) / 10 = -2\).

Step 2: Find Probability From Z-Score

The next step is to find the probability that corresponds to the calculated z-score using the standard normal distribution table. However, since we are asked for the probability of the value being at least 80, we need the probability for \(z > -2\), which is simply \(1 - P(z < -2)\). Referencing the table for \(z=-2\), reference statistical tables give us the value as 0.0228.

Step 3: Calculate Final Probability

As noted in the previous step, our interest is in \(P(Z > -2)\), and since the total probability under the curve is 1, we get this as \(1 - P(Z < -2)\), which is \(1 - 0.0228 = 0.9772\).

Key Concepts

Understanding Z-Score

The z-score is a way to measure how many standard deviations a particular value is away from the mean of a distribution. It's especially useful in a normal distribution because it helps us understand where a specific value stands in relation to the entire distribution. The formula for calculating the z-score is:

• \( z = \frac{(X - \mu)}{\sigma} \)

Here, \(X\) refers to the value you are examining, \(\mu\) is the mean, and \(\sigma\) is the standard deviation.
A z-score can be either positive or negative. A positive z-score means the value is above the mean, while a negative z-score indicates it is below the mean. In our exercise, the value 80 resulted in a z-score of \( -2 \), suggesting it is 2 standard deviations below the mean of 100.

The Role of Standard Deviation

Standard deviation is a measure that helps us understand how spread out the values in a data set are around the mean. In simpler terms, it tells us how much variation or dispersion exists from the average (mean).

• A small standard deviation means that the values tend to be close to the mean.
• A larger standard deviation indicates the values are spread out over a wider range.

In the exercise, the standard deviation is specified as 10, which indicates that on average, the values deviate 10 units from the mean of 100.
Standard deviation is important because it allows us to calculate the z-score, which in turn helps determine the probability of a value occurring within a certain range of the normal distribution.

Exploring Probability

Probability measures the likelihood of a specific event occurring, and it ranges from 0 to 1. Whereas 0 indicates impossibility, 1 indicates certainty. In the context of normal distribution, probability allows us to understand how likely it is to randomly select a value within a certain range from our distribution.
For example, in the given exercise, once the z-score was calculated, determining the probability for \( z > -2 \) using statistical tables revealed the result as 0.9772.

• This means there is a 97.72% chance that a value will be 80 or higher.

Probability is powerful in statistics because it can guide decision-making processes by illustrating the likelihood of future events based on current data.

The Concept of Mean

The mean, commonly known as the average, is the sum of all the values in a data set divided by the number of values. It is a measure of central tendency, which helps summarize a set of data with a single number, indicating the central point where data values congregate.
In a normal distribution, the mean is also the midpoint of the symmetric bell curve. It represents the point around which all values are distributed equally.

• For example, in our exercise, the mean is 100.

Understanding the mean is critical as it acts as a reference point from which the standard deviation can illustrate the dispersion of values in the distribution.