A video display tube for computer graphics terminals has a fine mesh screen behi

Question

A video display tube for computer graphics terminals has a fine mesh screen behind the viewing surface. The mesh is stretched and welded onto a metal frame during assembly. Too little tension at this stage will cause wrinkles, and too much tension will tear the mesh. Suppose that the tension reading of a randomly selected video display tube has a normal distribution with mean 275 and standard deviation 35 millivolts. Suppose further that the minimum acceptable tension is 230 millivolts and that the maximum acceptable tension is 320 millivolts. a) Determine and interpret the z-score for a tension of 320 millivolts b) Express the probability that a randomly selected video display tube has tension within the acceptable range in terms of , the cdf of a standard normal distribution c) Determine the probability asked about in part b) d) Re-answer b), using only positive values for inputs in the function e) Now suppose that the manufacturer wants to increase this probability of producing a mesh screen with an acceptable tension to 0.95 by keeping the mean the same and changing the standard deviation. Determine the value of the standard deviation needed to achieve this goal f) Continuing with part e), is the new value of the standard deviation larger or smaller than the original standard deviation? Explain why this makes sense intuitively.

Explanation / Answer

(a) Here = 275 mV

= 35 mV

Here Z - score for maximum acceptable tension is

Z = (320 - 275) / 0.35= 1.286

(B) Here if X is the random variable "tension reading of video display tube"

Pr(230 < X < 320) = NORM(230 < X < 320; 275 ; 35) = (Z2 ) - (Z1)

Z2  = (320 - 275)/ 35 = 1.286 and Z1 = (230 - 275)/ 35 = -1.286

so Pr(230 < X < 320) = (1.286) - (-1.286)

(c) Here (1.286) = 0.900 and (-1.286) = 0.100

so Pr(230 < X < 320) = 0.900 - 0.100 = 0.800

(d) Here (-1.286) = 1 - (1.286)

so Pr(230 < X < 320) = 2 * (1.286) -1 = 2 * 0.900 - 1 = 0.800

(e) Here we want that it shall have 95% acceptability. Let say new standard deviaton is 1

so Pr(230 < X < 320) = NORM (230 < X < 320) = 2 * (Z) -1 [ Taking concept of part (d) ]

here Z = (320 - 275)/ 1

so Pr(230 < X < 320) =0.95 as given in question to maintain specification

0.95 =  2 * (Z) -1

2 * (Z) = 1.95

(Z) = 0.975

so Z = -1 (0.975) = 1.96

so Z = (320 - 275)/ 1 = 1.96

1 = 45/1.96 = 22.96 mV

(f) Here the standard deviatonis smaller than the original standard devation as it covers more probbility than the erlier standard deviation which covers only 80% probability.

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