Problem 76 - Algebra 2

Question

A set of data with a mean of 39 and a standard deviation of 6.2 is normally distributed. Find each value, given its distance from the mean. \(-2\) standard deviations

Step-by-Step Solution

Answer

The value that is \(-2\) standard deviations from the mean is 26.6.

Step 1: Identify the known values

We are told that the mean (\(\mu\)) is 39 and the standard deviation (\(\sigma\)) is 6.2. We are asked to find the value that is \(-2\) standard deviations from the mean. So, \(Z = -2\).

Step 2: Substitute values into the formula

We input these known values into the formula \(X = \mu + Z\sigma\). Hence, the equation becomes: \(X = 39 + (-2)*6.2\).

Step 3: Perform the calculation

Multiply \(-2\) with \(6.2\) and then add \(39\) to the result. This will provide the value that is \(-2\) standard deviations from the mean.

Key Concepts

Standard Deviation

The concept of standard deviation is crucial in statistics, especially when dealing with a normal distribution. It measures the amount of variability or dispersion in a set of data. In simpler terms, it tells us how spread out the numbers in a data set are from the mean (average).

When data points are close to the mean, the standard deviation is small; when they are spread out, the standard deviation is larger. In our exercise, the standard deviation is 6.2, which gives us an idea of how much our data might fluctuate around the mean of 39.

• A lower standard deviation means data points are close to the mean.
• A higher standard deviation indicates data points are spread out over a wider range of values.

Understanding the standard deviation helps in predicting how values in the data are distributed. Knowing that something is "2 standard deviations" away gives a sense of how far from typical (the mean) a point lies.

Mean

The mean is one of the most common measures of central tendency, often simply referred to as the average. To find the mean, you sum all the data values and divide by the number of data points. It provides a central value around which data points tend to cluster. In the context of the normal distribution, the mean serves as the midpoint of the bell curve.

In our particular exercise, the mean is given as 39. This means that if we were to add up all the data values and divide by the number of data points, we'd get 39 as the result; this is where most of the values congregate. The mean can also be seen as the balance point of the distribution. It's important to note:

• The mean is sensitive to outliers; extremely high or low values can skew it.
• It is used in many statistical formulas, including those for calculating the z-score and standard deviation.

In a normal distribution, the mean is especially important because it's used to compute standard deviations and hence, to understand the entire distribution of the data.

Z-score

A z-score is a statistical measurement that describes a data point's relation to the mean of a group of data points. It is expressed in terms of standard deviations from the mean. A z-score tells you how many standard deviations an element is from the mean.

In our exercise, the z-score is egin{verbatim}−2egin{verbatim}, which implies that we are dealing with a value located two standard deviations below the mean. Understanding the z-score is crucial for identifying the position of a data point within a normal distribution.
Here are some key points about z-scores:

• A z-score of 0 indicates that the data point's score is identical to the mean score.
• A positive z-score indicates the data point is above the mean.
• A negative z-score means the data is below the mean.

Utilizing the z-score, we can convert data into a standard format, making it easier to compare data from different sets or scales. In practical terms, this exercise showed how the z-score is used to find a specific value within the distribution using the formula: \[ X = \mu + Z\sigma \].