It didn't come out right, the numbers 1,3,5,3,2, 29,3, 4, 2, 1, 2 are supposed to be under the column of length of stay. Columns Los-mean (5) and LOS-Mean)2 are intentionally left empty. And the x-x) and (x-x)2 signs are under the LOS mean columns.
I need help, someone please help me with this.
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
I honestly don't understand how to do this. If someone can help me out and guide me in the right direction I would appreciate it. I need to get this right on a test.
2 answers
a.
mean = (1+3+5+3+2+29+3+4+2+1+2)/11 =55/11= 5
(1-5) =
x (x-x̅) (x-x̅)^2
1 -4 16
3 -2 4
5 0 0
3 -2 4
2 -3 9
29 24 576
3 -2 4
4 -1 1
2 -3 9
1 -4 16
2 -3 9
b.
Range= highest value - lowest value
Range =29-1 = 28
c. variance = (x-x̅)^2 = 648
648/(11-) = 64.8
d. standard deviation = Sqrt(variance) = sqrt(64.8) =8.04984
mean = (1+3+5+3+2+29+3+4+2+1+2)/11 =55/11= 5
(1-5) =
x (x-x̅) (x-x̅)^2
1 -4 16
3 -2 4
5 0 0
3 -2 4
2 -3 9
29 24 576
3 -2 4
4 -1 1
2 -3 9
1 -4 16
2 -3 9
b.
Range= highest value - lowest value
Range =29-1 = 28
c. variance = (x-x̅)^2 = 648
648/(11-) = 64.8
d. standard deviation = Sqrt(variance) = sqrt(64.8) =8.04984
3 answers