Using Standard Normal Table solutions

Using Standard Normal Table solutions - Stat 4714...

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Unformatted text preview: Stat 4714 In‐class Exercise Solutions– Using Standard Normal Table The line width for semiconductor manufacturing is assumed to be normally distributed with a mean of 0.5 micrometer and a standard deviation of 0.05 micrometer. (a) What is the probability that a line width is less than 0.49 micrometer? z= x−μ σ = .49 − .50 = −.2 .05 P(z < ‐0.2) = .4602 (b) What is the probability that a line width is greater than 0.60 micrometer? z= x−μ σ = .6 − .50 =2 .05 P(z > 2) = 1 ‐ P(z < 2) = 1 – .9772 = .0228 (c) If the specifications require that the line width is between 0.48 and 0.52 micrometer, what proportion meets specifications? z= .48 − .50 = −.4 σ .05 x − μ .52 − .50 z= = = .4 σ .05 = x−μ P(‐.4 < z < .4) = .6554‐ .3446 = .3108 (d) What is the line width associated with z.10 ? (Note: first you need to find z, and then plug z into the z‐score formula to find X). z.10 denotes a z‐value with an area to the right of it of .10. We look up 1‐.10 = .90 in the center of the table and find that the closest value is .8997 The corresponding z‐score is 1.28. To find X, we use the z‐score formula, solving for X: X = zσ+μ = 1.28*.05+.50 =.564 (e) What is the line width is associated with the 75% percentile (that is, the area to the left of z is .75)? (Note: first you need to find z, and then plug z into the z‐score formula to find X). We look up .75 in the center of the table. The two closest values are .7486 and .7517. The corresponding z‐scores are .67 and .68. (Either is fine.) To find X, we use the z‐score formula, solving for X: X = zσ+μ =.67*.05+.50 =.5335 or X= zσ+μ =.68*.05+.50 =.534 ...
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