Question 1200295: A manufacturer knows that their items have a normally distributed length, with a mean of 13.5 inches, and standard deviation of 3.1 inches.
If one item is chosen at random, what is the probability that it is less than 7.9 inches long?
Found 2 solutions by math_tutor2020, ikleyn:
Answer by math_tutor2020(3817)   (Show Source):
You can put this solution on YOUR website!
Answer: 0.03515 (approximate)
Work Shown:
z = (x-mu)/sigma
z = (7.9-13.5)/3.1
z = -1.80645161290322 approximately
z = -1.81
Then use a table such as this one to find that
P(Z < -1.81) = 0.03515
In the link I posted, scroll down to the section labeled "How to Read The Z Table" for an example.
Answer by ikleyn(52824)  (Show Source):
You can put this solution on YOUR website! .
A manufacturer knows that their items have a normally distributed length,
with a mean of 13.5 inches, and standard deviation of 3.1 inches.
If one item is chosen at random, what is the probability that it is less than 7.9 inches long?
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Below is slightly different method of computing, without making manual calculations on the way.
This probability is the area under the normal curve
on the left of the assigned raw mark 7.9.
Use function normalcdf of your calculator.
normalcdf is the " Normal Cumulative distribution function ".
The format of this function is normalcdf(low mark, high mark, mean, SD).
So, P = normalcdf(-9999, 7.9, 13.5, 3.1) = 0.0354. ANSWER
Alternatively, you may use free of charge online calculator at this web-site
https://onlinestatbook.com/2/calculators/normal_dist.html
It shows the shaded are of interest, so you clearly see what you are doing
and can understand what is going. It prevents you of making errors.
If you are a novice in such calculations, I recommend you to start
with the online calculator and then to switch to the graphing calculator.
Solved.
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If you need instructions on using your graphing calculator for normalcdf function,
find all necessary instructions from this video-lesson
Normal Distribution: Calculating Probabilities {TI 84 Plus CE}
https://www.youtube.com/watch?v=IyEKEL9nm28
or from this textual description
https://cosmosweb.champlain.edu/people/stevens/WebTech/TIFiles/Chap6-TI-83.pdf
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