Question

A sample of 10 claims in an insurance company had mean and variance of 5,478 and 1,723 , respectively. On reconciliation, it was found that one claim of 3,250 was wrongly written as \(4,250 .\) Calculate the mean and standard deviation of the sample with correct values.

Short answer

The correct mean of the sample is \(5,378\) and the corrected standard deviation is approximately \(42.1\).

Step-by-step solution

Calculate the Total Sum of the incorrect Sample

First, we will multiply the incorrect mean by the total number of observations in the sample to get the total sum of values. Using this formula: Sum = Mean * Total number of observations, which is, \( Sum = 5478 * 10 = 54780 . \)

Calculate the Correct Sum of Observations

Next, we subtract the incorrect value \((4,250)\) from the total sum, and then add the correct value \((3,250)\). The formula will be: New Sum = (Total Sum - Wrong Value) + Correct Value, hence, \(New Sum = (54780 - 4250 ) + 3250 = 54780 - 1000 = 53780 . \)

Calculate the Correct Mean

Now, we will divide the corrected total sum by the total number of observations to get the corrected mean. Using the formula: Correct Mean = New Sum / Total number of observations, which is, \( Correct Mean = 53780 / 10 = 5378 . \)

Calculate the Correct Variance

The formula for variance includes the squared difference of each observation from the mean. As only one observation was incorrect, we'll adjust the variance only for this observation. The new variance can be calculated as: Correct Variance = Initial Variance + (Wrong Value^2 - Correct Value^2)/ Total number of observations + (Initial Mean - Correct Mean)^2, which results in, \( Correct Variance = 1723 + (4250^2 - 3250^2)/10 + (5478 - 5378)^2 = 1772 . \)

Calculate the Correct Standard Deviation

Finally, the standard deviation is the square root of the variance. Hence, the correct standard deviation of the sample would be, \(Correct SD = \sqrt{Correct Variance} = \sqrt{1772} \approx 42.1 . \)

Key concepts

Mean Calculation

Calculating the mean is one of the fundamental steps in statistical analysis. The mean, often referred to as the average, represents the central value of a dataset. To find the mean, sum up all the numbers in a dataset and divide by the number of observations.

In this exercise, we first calculated the mean of the incorrect dataset. This was done by multiplying the incorrect mean by the total number of claims, resulting in a total sum of incorrect values. After correcting the erroneous data point from 4,250 to 3,250, a new sum was computed by adjusting the total sum. This new sum was divided by the number of observations, which gave us the corrected mean.

Remember, altering any single value in your dataset can shift the mean, highlighting the importance of accurate data entry and verification. An easy way to visualize the mean is by thinking of it as the balance point of your data, much like the balance point of a seesaw.

Variance Correction

Variance measures how much the numbers in a dataset differ from the average of the dataset. It provides insight into the spread or dispersion of the data. A higher variance means a wider spread of numbers. The formula for variance involves squaring the differences between each observation and the mean, then averaging those squared differences.

Correcting the variance in this exercise involved recalculating the squared differences for our corrected claim value. Since only one data point was wrong, the correction focused on adjusting the observed variance by considering both the change in the specific squared-term \((4250^2 - 3250^2)\) and the shift in the mean \((5478 - 5378)^2\). The difference was divided by the total number of claims, ensuring the updated variance accurately reflected the corrected dataset.

Understanding variance helps in appreciating the data's reliability. Even small changes in your dataset, like correcting a single wrong value, can impact the variance and, by extension, any decisions based on such statistical insights.

Standard Deviation

The standard deviation is an extension of variance that makes interpretation easier. It's simply the square root of the variance and has the same units as the original data, making it more intuitive to understand. Like variance, it informs you about the dispersion of the dataset - with a smaller standard deviation indicating that the data points are closer to the mean.

After obtaining the corrected variance in this exercise, we computed the corrected standard deviation by taking its square root. The standard deviation allows us to understand the extent to which the claims in the insurance data varied around the mean. It's a key metric for determining consistency within data.

In practical terms, a lower standard deviation means most of the data values are close to the mean, while a higher standard deviation indicates a more spread out dataset. Understanding the standard deviation is crucial for not just interpreting current data but also for predicting possible future occurrences based on your data's variation.