Given the following set of numbers: x=7.1, 1.8, 2.1, 1.5, 2.1.
Find the median, mean and standard deviation. State the formulas (or procedure) for these values, plug the numbers into the formula and evaluate the results. Compare the mean and median. Which is more resistant to outliers?
Your notes or text book should have given you the steps and formulas to answer this question.
if not, here is a page that shows the steps for a problem much like yours
http://hubpages.com/hub/stddev
To find the median, mean, and standard deviation of a set of numbers, let's follow these steps.
Step 1: Arrange the numbers in ascending order.
The given set of numbers: x = 7.1, 1.8, 2.1, 1.5, 2.1.
Arranged in ascending order: 1.5, 1.8, 2.1, 2.1, 7.1.
Step 2: Find the Median.
The median is the middle value in a sorted set of numbers. If there is an odd number of values, the median is the middle one. If there is an even number, it is the average of the two middle values.
In this case, there are five numbers, so the median is the third value, which is 2.1.
Step 3: Find the Mean.
The mean (or average) of a set of numbers is calculated by summing all the values and dividing by the total count of values.
The sum of the numbers in this case is 1.5 + 1.8 + 2.1 + 2.1 + 7.1 = 14.6.
Since there are five numbers, the mean is 14.6 / 5 = 2.92.
Step 4: Find the Standard Deviation.
The standard deviation measures the dispersion of data points from the mean. It is calculated using the following formula:
Standard Deviation = √[(Σ(xi - x̄)^2) / N]
Where:
- Σ represents the sum of
- (xi - x̄) is the difference between each value (xi) and the mean (x̄)
- N is the total count of values
Using the formula and plugging in the given values:
Standard Deviation = √[( (1.5 - 2.92)^2 + (1.8 - 2.92)^2 + (2.1 - 2.92)^2 + (2.1 - 2.92)^2 + (7.1 - 2.92)^2) / 5]
Standard Deviation = √[(2.9104 + 1.3396 + 0.6724 + 0.6724 + 16.4944) / 5]
Standard Deviation = √[22.09 / 5]
Standard Deviation ≈ √4.418 ≈ 2.10
Finally, let's compare the mean and median:
The mean value is 2.92, while the median value is 2.1. In this case, the mean is higher than the median.
Regarding resistance to outliers, the median is more resistant. Outliers are extreme values that deviate significantly from the other values. Since the median only considers the middle value(s), it is less affected by extreme values and thus more resistant to outliers. On the other hand, the mean includes all values, so outliers can significantly affect its value.