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Suppose the current measurements in a strip of wire are assumed to follow
a normal distribution with a mean of 18 mm and a standard deviation of 3mm.
(a) What proportion of current measurements will exceed 21 mm?
(b) What proportion of current measurements are between 17 and 20 mm?
(c) What is the value of the current measurement of the top 15% of the strip wire.
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(a) notice that 21 mm is one standard deviation of 3 mm from the mean of 18 mm.
Use the empirical rule that says that 68% of the data falls within one standard deviation
from the mean.
In this case, it means that half of 68%, i.e. 68/2 = 34% falls between 18 mm and 21 mm.
Hence, 50% - 34% = 16% of data will exceed 21 mm.
It is the ANSWER to (a).
(b) To answer (b), you need to find the area under the normal curve
from z-score z1 = 17 mm to z-score z2 = 20 mm.
Use the standard function normcdf of the regular calculator TI-83/84
z1 z2 mean SD <<<---=== formatting pattern.
P = normcdf(17, 20, 18, 3)
You will get the value P = 0.3781 (rounded).
So, the answer to (b) is 0.3781 (rounded).
Alternatively, you may use an online calculator
https://onlinestatbook.com/2/calculators/normal_dist.html
which is free of charge and has extremely convenient interface.
(c) To answer (c), you need to find the z-score under the normal curve
such that the area of the normal curve on the right of z1 is 0.15.
Use the standard function invNorm of the regular calculator TI-83/84
area mean SD <<<---=== formatting pattern.
z = invNorm(0.85, 18, 3)
You will get the value z = 21.1093 (rounded).
So, the answer to (c) is 21.1093 mm (rounded).
Alternatively, you may use an online calculator
https://onlinestatbook.com/2/calculators/inverse_normal_dist.html
which is free of charge and has extremely convenient interface.
Thus the problem is solved in full and all questions are answered.
Notice that each question requires its own method.