Question 1204070
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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
<a href = "https://www.ztable.net/">https://www.ztable.net/</a>
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
<a href = "https://www.statology.org/normal-probabilities-ti-84-calculator/">https://www.statology.org/normal-probabilities-ti-84-calculator/</a>
The input will be <font color=red>normalcdf(20,9999,15,3.6)</font>
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
<a href = "https://davidmlane.com/normal.html">https://davidmlane.com/normal.html</a>
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 <font color=red size=4>0.08</font> as the final answer.
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