Using the data in the table, calculate the mean, range, variance, and standard deviation, and then answer questions e and f. Round the variance and standard deviation to one decimal place.
a. Mean
b. Range
c. Variance
d. Standard deviation
e. What value is affecting the mean and SD of this distribution?
f. Does the mean adequately represent this distribution? If not, what would be a better measure of central tendency for this data set?
Patient Length of Stay LOS-Mean(5) (LOS –mean)2
(X-X)
1 1
2 3
3 5
4 3
5 2
6 29
7 3
8 4
9 2
10 1
11 2
To calculate the mean, range, variance, and standard deviation, we need to follow these steps:
a. Mean:
- Add up all the values in the "Length of Stay" column: 1 + 3 + 5 + 3 + 2 + 29 + 3 + 4 + 2 + 1 + 2 = 55
- Divide the sum by the total number of values: 55 / 11 = 5 (rounded to one decimal place)
- The mean of the "Length of Stay" is 5.
b. Range:
- Find the minimum value in the "Length of Stay" column: 1
- Find the maximum value in the "Length of Stay" column: 29
- Calculate the range by subtracting the minimum value from the maximum value: 29 - 1 = 28
- The range of the "Length of Stay" is 28.
c. Variance:
- Subtract the mean from each value in the "Length of Stay" column and square the result.
- Sum up all these squared differences: (1-5)^2 + (3-5)^2 + (5-5)^2 + (3-5)^2 + (2-5)^2 + (29-5)^2 + (3-5)^2 + (4-5)^2 + (2-5)^2 + (1-5)^2 + (2-5)^2 = 322
- Divide the sum by the total number of values: 322 / 11 = 29.3 (rounded to one decimal place)
- The variance of the "Length of Stay" is 29.3.
d. Standard deviation:
- Take the square root of the variance calculated in the previous step: sqrt(29.3) = 5.4 (rounded to one decimal place)
- The standard deviation of the "Length of Stay" is 5.4.
e. The value of 29 in patient 6 is significantly larger compared to the other values. This value affects both the mean and standard deviation of the distribution.
f. No, the mean does not adequately represent this distribution as it is greatly influenced by the outlier value of 29. A better measure of central tendency for this data set would be the median, which is less affected by outliers.
To calculate the mean, range, variance, and standard deviation of the given data, follow these steps:
a. Mean:
To find the mean (average) of the Length of Stay (LOS) data, sum up all the values and divide it by the total number of values.
Sum of LOS = 1 + 3 + 5 + 3 + 2 + 29 + 3 + 4 + 2 + 1 + 2 = 55
Mean = Sum of LOS / Number of Values = 55 / 11 = 5
b. Range:
To find the range, subtract the minimum value from the maximum value.
Minimum LOS = 1
Maximum LOS = 29
Range = Maximum LOS - Minimum LOS = 29 - 1 = 28
c. Variance:
To calculate the variance, first find the squared deviation from the mean for each value and then find the average of those squared deviations.
Squared Deviation from the Mean = (LOS - Mean)^2
(1-5)^2 = 16
(3-5)^2 = 4
(5-5)^2 = 0
(3-5)^2 = 4
(2-5)^2 = 9
(29-5)^2 = 576
(3-5)^2 = 4
(4-5)^2 = 1
(2-5)^2 = 9
(1-5)^2 = 16
(2-5)^2 = 9
Sum of Squared Deviations = 16 + 4 + 0 + 4 + 9 + 576 + 4 + 1 + 9 + 16 + 9 = 658
Variance = Sum of Squared Deviations / Number of Values = 658 / 11 ≈ 59.8 (rounded to one decimal place)
d. Standard Deviation:
To calculate the standard deviation, take the square root of the variance.
Standard Deviation = √(Variance) = √(59.8) ≈ 7.7 (rounded to one decimal place)
e. What value is affecting the mean and SD of this distribution?
The value 29 has a significant impact on both the mean and standard deviation. Being an outlier, it drags the values upwards, causing the mean to be higher than a typical representation of the data. Additionally, the large difference between this outlier and the other values increases the standard deviation.
f. Does the mean adequately represent this distribution? If not, what would be a better measure of central tendency for this data set?
No, the mean does not accurately represent this distribution since it is heavily influenced by the outlier value of 29. A better measure of central tendency for this data set would be the median, which represents the middle value of an ordered list. In this case, the median would be less affected by the outlier and provide a better reflection of the central tendency of the data.