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In certain manufacturing process, it was found that the mean length of the parts produced by the lathe was 20.05 mm, with the standard deviation of 0.02 mm. The lower and the upper specification limits for the length are 20.03 mm and 20.08 mm respectively.

a) What proportion of part s will be out of the specification limit.

b)Find the probability that a part selected at random would have a length less than 20.01?

c)The life time of a certain type of motor is normally distributed with the mean of 10 years and the standard deviation of 2 years. If the manufacturer is willing to replace only 3% percent of the motor that fail how long a guarantee should offer?

I was unable to get to the problem until now, but here at last is my answer.

Since the mean is 20.05 and the standard deviation is 0.02, how far away the limits are from the center can be found. Since 20.05 - 20.03 = 0.02, the lower limit is 1 standard deviation below the mean. Since 20.08 - 20.05 = 0.03, the upper limit is 1.5 standard deviations above the limit.

(a) To find the answer, find the probability that a normal variable is outside the range (-1, .5).

Take a normal distribution table, so I'll use

https://stat.duke.edu/~banks/111-FAQ.dir/StdNormCumDistTable.pdf

The value at -1 is 0.1587.

The value at 1.5 is one minus the value at -1.5, which is 1 - 0.0668 = 0.9332.

The chance the value is between -1 and 1.5 is 0.9332 - 0.1587 = 0.7745.

(b) Since 20.01 is 0.04 less than the average, which is 2 standard deviations away,

the probability of that is seen to be 0.0228.

(c) For this one, we have to find how many standard deviations give a chance of 0.03.

Reading the table, that is roughly at 1.88 standard deviations away from the normal.

Converting that back to the distribution we have means taking 10 - 2(1.88) = 10 - 3.76 = 6.24.

This is the same as roughly 6 years, 3 months. So if it goes out before then, he should replace it.

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