Roughly speaking, the standard deviation indicates how far, on average, the observations are from the mean. Do you think that for the data set below the standard deviation will give a good indication of the typical deviation from the mean?
2, 3, 4, 4, 5, 5, 6, 6, 100
What drawback of the standard deviation is illustrated by this example?
I don't see any drawbacks
The standard deviation includes Outliers like "100", and may give a false picture of the data.
To determine if the standard deviation gives a good indication of the typical deviation from the mean in the given dataset, let's calculate the standard deviation and analyze the results.
1. Calculate the mean:
Add up all the values in the dataset and divide the sum by the total number of observations.
Mean = (2 + 3 + 4 + 4 + 5 + 5 + 6 + 6 + 100) / 9 = 141/9 = 15.67 (approximately)
2. Calculate the deviation from the mean for each observation:
Subtract the mean from each observation.
Deviation from Mean = (2-15.67, 3-15.67, 4-15.67, 4-15.67, 5-15.67, 5-15.67, 6-15.67, 6-15.67, 100-15.67)
= (-13.67, -12.67, -11.67, -11.67, -10.67, -10.67, -9.67, -9.67, 84.33)
3. Calculate the squared deviation for each observation:
Square each deviation calculated in step 2.
Squared Deviation = (-13.67^2, -12.67^2, -11.67^2, -11.67^2, -10.67^2, -10.67^2, -9.67^2, -9.67^2, 84.33^2)
= (187.16, 160.48, 136.12, 136.12, 113.84, 113.84, 93.73, 93.73, 7106.73)
4. Calculate the variance:
Add up all the squared deviations and divide by the total number of observations.
Variance = (187.16 + 160.48 + 136.12 + 136.12 + 113.84 + 113.84 + 93.73 + 93.73 + 7106.73) / 9
= 8071.35 / 9 ≈ 896.82
5. Calculate the standard deviation:
Take the square root of the variance calculated in step 4.
Standard Deviation = √(896.82) ≈ 29.94
Now, let's analyze the result. The standard deviation of approximately 29.94 is quite high, reflecting the large deviation of the value 100 from the mean of 15.67. However, it is important to note that this large deviation is primarily due to the outlier value of 100. The other observations have relatively small deviations compared to the outlier.
Therefore, in this example, the standard deviation does not provide a good indication of the typical deviation from the mean because it is greatly influenced by the presence of the outlier value. The drawback of the standard deviation is that it is sensitive to extreme values, leading to distorted results when outliers are present in the dataset.