Question

For a particular disease, the probability that a patient passes away within the x^th year is given by the following table:

xx	P(x)P(x)
0	0.013
1	0.043
2	0.309
3	0.261
4	0.325
5	0.041
6	0.008

This means that the probability that someone passes away within the first year is 0.013. The probability that they pass away after one year would be 0.043.

What is the mean number of years someone survives?  (round to two decimal places)

What is the standard deviation for the number of years someone survives?  (round to two decimal places)

The Range Rule of Thumb for survival would be:  years to  years. (Round each to two decimal places.)

In your own words, interpret what the Range Rule of Thumb tells us about this disease. (If you don't remember what the Range Rule of Thumb is, review the heading "Unusual Values" in the Learning Activities.)

Answer 1

Solution:

The formulas for mean and standard deviations are given as below:

Mean = ∑ x*P(x)

Standard deviation = sqrt[∑(x - mean)^2*P(x)]

The calculation table is given as below:

x	P(x)	x*P(x)	(x - mean)^2*P(x)
0	0.013	0	0.116766117
1	0.043	0.043	0.171484387
2	0.309	0.618	0.307148781
3	0.261	0.783	2.349E-06
4	0.325	1.3	0.326952925
5	0.041	0.205	0.164492369
6	0.008	0.048	0.072144072
Total	1	2.997	1.158991

What is the mean number of years someone survives?  (round to two decimal places)

Solution:

Mean = ∑ x*P(x)

Mean = 2.997

Mean = 3.00

What is the standard deviation for the number of years someone survives?  (round to two decimal places)

Standard deviation = sqrt[∑(x - mean)^2*P(x)]

Standard deviation = sqrt[1.158991]

Standard deviation = 1.076564443

Standard deviation = 1.08

The Range Rule of Thumb for survival would be:  

Range = 4*Standard deviation

Range = 4*1.08

Range = 4.32

The Range Rule of Thumb for survival would be: 0.00 years to 4.32 years

In your own words, interpret what the Range Rule of Thumb tells us about this disease.

The range rule of thumb explains that the patient will be surviving for a particular disease for 0.00 years to 4.32 years.