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?
Pts LOS

1 1
2 3
3 5
4 3
5 2
6 29
7 3
8 4
9 2
10 1
11 2
Total 55

Find the mean first = sum of scores/number of scores.

Range = highest score - lowest score.

Subtract each of the scores from the mean and square each difference. Find the sum of these squares. Divide that by the number of scores to get variance.

Standard deviation = square root of variance.

The mean acts like a fulcrum (balance point), therefore it is greatly modified by outlying scores (e.g., 29). If this is true, which of the remaining measures of central tendency would you use?

I'll let you do the calculations.

To calculate the mean, range, variance, and standard deviation, we will follow the following steps:

a. Mean:
To calculate the mean, we sum up all the values and divide by the total number of values. In this case, we sum up the Pts column.
Sum(Pts) = 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10 + 11 = 66
Total number of values = 11
Mean = Sum(Pts) / Total number of values = 66 / 11 = 6

b. Range:
To calculate the range, we find the difference between the highest and lowest values. In this case, the highest value is 29 (from the LOS column) and the lowest value is 1.
Range = Highest value - Lowest value = 29 - 1 = 28

c. Variance:
To calculate the variance, we need to calculate the difference between each value and the mean, square these differences, sum them up, and divide by the total number of values (minus 1). In this case, we will use the Pts column.
Mean = 6 (from the previous step)
Squared differences from the mean: (1-6)^2, (2-6)^2, (3-6)^2, (4-6)^2, (5-6)^2, (6-6)^2, (7-6)^2, (8-6)^2, (9-6)^2, (10-6)^2, (11-6)^2
Sum of squared differences = (1-6)^2 + (2-6)^2 + (3-6)^2 + (4-6)^2 + (5-6)^2 + (6-6)^2 + (7-6)^2 + (8-6)^2 + (9-6)^2 + (10-6)^2 + (11-6)^2 = 110
Total number of values = 11
Variance = Sum of squared differences / (Total number of values - 1)= 110 / 10 ≈ 11

d. Standard Deviation:
To calculate the standard deviation, we take the square root of the variance.
Standard Deviation = √Variance = √11 ≈ 3.3

e. The value affecting the mean and standard deviation of this distribution is the outlier value of 29 in the Pts column.

f. The mean does not adequately represent this distribution because it is heavily influenced by the outlier value of 29. A better measure of central tendency for this data set would be the median, as it is less affected by outliers.