Question 1204070: The manufacturer knows that their items have a normally distributed lifespan with a mean of 15 years in standard deviation of 3.6 years. If you randomly purchase one item, what is the probability that it will last longer than 20 years. Round answer to 2 decimal places.
Answer by math_tutor2020(3817)   (Show Source):
You can put this solution on YOUR website!
mu = 15 = mean
sigma = 3.6 = standard deviation
Compute the z score when x = 20
z = (x-mu)/sigma
z = (20-15)/(3.6)
z = 1.38888888888889
z = 1.39
The task of computing P(X > 20) is approximately the same as finding P(Z > 1.39)
Use a Z table found in the back of your stats textbook.
The table should look like the one found here
https://www.ztable.net/
Scroll down to the row that starts with "+1.3"
Then look at the column that has "0.09" at the top
The intersection of the row and column yields the approximate value 0.91774
This will indicate that P(Z < 1.39) = 0.91774 approximately.
So,
P(Z > 1.39) = 1 - P(Z < 1.39)
P(Z > 1.39) = 1 - 0.91774
P(Z > 1.39) = 0.08226
This leads us back to P(X > 20) = 0.08226 when mu = 15 and sigma = 3.6
There's about a 8.226% chance of selecting a item that will last longer than 20 years.
If you want to use a TI83 or TI84, then check out this article
https://www.statology.org/normal-probabilities-ti-84-calculator/
The input will be normalcdf(20,9999,15,3.6)
The 9999 represents a very large upper bound to basically represent positive infinity.
The result of this calculation is approximately 0.0824333239 which isn't far off from the previous result we got (0.08226)
Here's another calculator you can use
https://davidmlane.com/normal.html
That calculator provides a diagram as well.
The result from that calculator should be 0.0824
When rounding to two decimal places, it leads to 0.08 as the final answer.
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