A massage therapist provides sessions that last exactly 27 minutes, but as a reward to his customers, every fifth session is 42 minutes. If every customer comes for a very large number of sessions, what is the standard deviation of the massage time? Round your answer to three decimal places.
Please transfer the question to the suitable forum, moderators.barca14 wrote:A massage therapist provides sessions that last exactly 27 minutes, but as a reward to his customers, every fifth session is 42 minutes. If every customer comes for a very large number of sessions, what is the standard deviation of the massage time? Round your answer to three decimal places.
The massage time (in minute) to a customer runs as below for sessions:
27, 27, 27, 27, 42, 27, 27, 27, 27, 42, 27, 27, 27, 27, 42, ...
If we only had enough time to take 5 massage time, then in statistical terms this means we have a sample size of 5 and in this case we use the standard deviation equation for a sample of a population:
s = √[∑ (x - M)^2/(N - 1)]
Where in, s is the standard deviation, x is each value in sample, M is the mean of the values, and N is the number of values (the sample size)
The rest of this work will be done in the case where we have a sample size of 5 massage time, therefore we will be using the standard deviation equation for a sample of a population.
Here are the 5 massage time
27, 27, 27, 27, 42
Now, let's calculate the standard deviation:
1. Calculate the mean:
(27 + 27 + 27 + 27 + 42)/5 = 30
2. Calculate x - M for each value in the sample:
27 - 30 = -3
27 - 30 = -3
27 - 30 = -3
27 - 30 = -3
42 - 30 = 12
3. Calculate ∑ (x - M)^2 = (-3)^2 + (-3)^2 + (-3)^2 + (-3)^2 + (12)^2 = 36 + 144 = 180.
4. Calculate the standard deviation,: s = √[∑ (x - M)^2/(N - 1)] = √[180/4)] = √45 = [spoiler]6.71[/spoiler] approx
