Question 1199334: A company that manufactures batteries guarantees them a life of 24 months.
(i) If the average life has been found in tests to be 33 months and a S. D. of 4 months, how many will have to be replaced under guarantee if a normal distribution is assumed for battery lifetimes ?
Found 2 solutions by Theo, ikleyn:
Answer by Theo(13342)   (Show Source):
You can put this solution on YOUR website! mean is 33 months.
standard deviation is 4 months.
guarantee is for 24 months.
z = (x - m) / s = (24 - 33) / 4 = -2.25
z is the z-score
x is the test score to be compared to the mean
m is the mean
s is the standard deviation
the guarantee is for 24 months.
if it last longer than 24 months, then the battery doesn't have to be replaced under guarantee.
if it last less than 24 months, then it will have to be replaced under guarantee.
the area to the left of a z-score of -2.25 is equal to .0122244334.
that's equal to 1.22% rounded to 2 decimal points.
that's the percent of batteries sold that would have to be replaced under guarantee.
Answer by ikleyn(52864)  (Show Source):
You can put this solution on YOUR website! .
A company that manufactures batteries guarantees them a life of 24 months.
If the average life has been found in tests to be 33 months and a S. D. of 4 months,
how many will have to be replaced under guarantee if a normal distribution
is assumed for battery lifetimes ?
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We are given that the lifetime is normally distributed with the mean of 33 months
and SD of 4 months.
They want you determine the probability P(L < 24 months), where L is the continuous
random variable of lifetime in months.
Every normal distribution curve is a bell shaped curve.
In this problem, they want you determine the area under this specific normal curve
on the left from the raw mark of 24 months. This area is the sought probability.
Go to web-site https://onlinestatbook.com/2/calculators/normal_dist.html
and find there free of charge online calculator, specially developed for this purpose.
Input the mean value 33 and the standard deviation value 4;
input 24 in the window "Below"; then click "Recalculate".
You will get the ANSWER 0.0122 in the window "Probability".
The auxiliary plot will show you the area of the interest.
Now, when you know "what to do", I will tell you that you can get the same result
in other way, using your calculator TI-83 or TI-84 (or whatever).
For it, use the calculator' standard function normalcdf with parameters
z1 z2 mean SD
P = normalcdf(-9999, 24, 33, 4) = 0.0122.
You will get the same value of the probability 0.0122 from your calculator.
Solved.
At this point, the solution is completed.
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