Times for a surgical procedure are normally distributed. Method A has a mean of 28 minutes and a standard deviation of 2 minutes for completion.
What is the probability that a surgical procedure conducted by method A takes at most 33 minutes to complete?
this is one of the best sites for your kind of question
http://en.wikipedia.org/wiki/List_of_countries_by_life_expectancy
enter
mean -- 28
SD ---2
click on below, enter 33
I get .99379
Is that site the one you meant to post?
no, sorry, must have missed the "copy" before the "paste"
try this one ....
http://davidmlane.com/hyperstat/z_table.html
To find the probability that a surgical procedure conducted by method A takes at most 33 minutes to complete, we need to calculate the z-score and then use a standard normal distribution table.
Step 1: Calculate the z-score.
The z-score formula is given by:
z = (x - μ) / σ
where x is the value we want to find the probability for (33 minutes), μ is the mean of the distribution (28 minutes), and σ is the standard deviation (2 minutes).
Substituting the values into the formula, we get:
z = (33 - 28) / 2
z = 5 / 2
z = 2.5
Step 2: Look up the z-score in the standard normal distribution table.
The standard normal distribution table provides the cumulative probability for different z-scores. We are interested in finding the cumulative probability for a z-score of 2.5.
Using the table, we can find that the cumulative probability (area under the curve) to the left of a z-score of 2.5 is approximately 0.9938.
Step 3: Calculate the probability.
Since we want to find the probability that the surgical procedure takes at most 33 minutes, which is equivalent to finding the cumulative probability to the left of 33 minutes on the distribution, we can use the cumulative probability from Step 2.
Therefore, the probability that a surgical procedure conducted by method A takes at most 33 minutes to complete is approximately 0.9938, or 99.38%.