Question

TULIS. Question The piston rings are used to reject when a certain dimension is not within the specifications 2.0td. It is known that this measurement is normally distributed with mean 1.50 mm and standard deviation 0.20 mm. (4). Find the value d such that the specifications cover 90% of the measurements. (5). What is the probability that the measurement of a selected piston ring will be more than 3 mm? Question Laptops produced by a company last on an average of 5 years. The life span of each laptop follows

Answer 1

Here measurements ~ Normal (mean = 1.50, std.dev.= 0.20)

4) So let us find first there are which two value covers 90% of specifications.

middle 90% means the area below lower specifiction value is 0.05 and area below upper specification value is 0.95

so we can find this two values using excel command as follows , type this command as follows and press enter

lower specification value

=NORMINV(0.05,1.5,0.2)

So answer is

1.17102927

Upper Specification value

=NORMINV(0.95,1.5,0.2)

So answer is

1.82897073

So we have to find d such that specifications covers 90% measurements=1.1710 and 1.1890.

means $2.0\pm d$ covers the two values 1.1710 and 1.1890

so we hae to find d such that 2-d = 1.17 and 2+d = 1.19

as 2-d=1.17 so d = 0.83

and 2+d=1.19=-1.17 so d is = -1.17 is not possible

as d= 0.83 we got 2-d=2-0.83=1.17 and 2+d = 2+0.83=2.83 covers 90% of specifications

But between 1.17 and 2.83 the specification covers =0.95 means also covers 90% of specifications

note:

I think this question must ask find d such 95% specification covers

otherwise d is not possible which covers 90% specifications

since such d not exist 2-d and 2+d covers 90% of specifcations

5) Here we have to find

P( X>3) = 1 - P(X<3)

this probability can be find using below excel command , type as it is and press enter

=1-NORMDIST(3,1.5,0.2,1)

which gives answer

0

So probability of measurement of a selected piston ring will be more than 3 mm is 0