The table below shows the LOS for a sample of 11 discharged patients. 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-x̅) (x-x̅)2
1 1
2 3
3 5
4 3
5 2
6 29
7 3
8 4
9 2
10 1
11 2

Okay so I have the answer for A-D however I don't know the answers for E-F. Can anyone help me please. Thank you very much I appreciate it.

e) LOS = 29 for one of the patients. This would be considered an outlier.

f) Perhaps a better measure of central tendency would be the median.

A manager wishes to find out whether there is a relationship between the number of radio

advertisement aired per week and the amount of sales (in hundreds of Birr) of a product.
The data for the sample are given below.
Number of Advertisement (X) 2 5 8 8 10 12
Sales (Y) 2 4 7 6 9 10

i need answer

Sure! Let's break it down step by step.

a. Mean:
To calculate the mean, you need to sum up all the values in the dataset and divide it by the total number of values. In this case, you have 11 values. The sum of the LOS values is 60. So the mean would be 60/11 ≈ 5.5.

b. Range:
To find the range, you need to subtract the smallest value from the largest value. In this case, the smallest value is 1 and the largest value is 29. Therefore, the range is 29 - 1 = 28.

c. Variance:
To calculate the variance, you need to find the average of the squared differences between each value and the mean. Start by subtracting the mean from each value and then square the result. Then, calculate the mean of these squared differences. The variance is the average squared difference. In this case, the variance is 23.6 (rounded to one decimal place).

d. Standard deviation:
The standard deviation is the square root of the variance. So, taking the square root of the variance calculated in the previous step, the standard deviation is approximately 4.9 (rounded to one decimal place).

e. What value is affecting the mean and SD of this distribution?
From the given data, we can see that there is an outlier, which is the value 29. Since the mean is sensitive to extreme values, this outlier affects the mean. The standard deviation is also affected because it takes into account the distance of each value from the mean.

f. Does the mean adequately represent this distribution? If not, what would be a better measure of central tendency for this dataset?
The mean might not adequately represent this distribution because of the presence of the outlier (29), which significantly affects the mean. In situations like this where there are outliers that heavily influence the mean, a better measure of central tendency would be the median. The median is less affected by extreme values and provides a more representative measure for the central value of a skewed distribution like this one.