The quality-assurance program for a certain adhesive formulation process involves measuring how well the adhesive sticks a piece of plastic to a glass surface. When the process is functioning correctly, the adhesive strength X is normally distributed with a mean of 200 N and a standard deviation of 10N. Each hour, you make one measurement of the adhesive strength. You are supposed to inform your supervisor if your measurement indicates that the process has strayed from its target distribution.

Question

a. Find P(X ? 160), under the assumption that the process is functioning correctly.
b. Based on your answer to part (a), if the process is functioning correctly, would a strength of 160 N be unusually small? Explain.
c. If you observed an adhesive strength of 160 N, would this be convincing evidence that the process was no longer functioning correctly? Explain.
d. Find P(X ? 203), under the assumption that the process is functioning correctly.
e. Based on your answer to part (d), if the process is functioning correctly, would a strength of 203 N be unusually large? Explain.
f. If you observed an adhesive strength of 203 N, would this be convincing evidence that the process was no longer functioning correctly? Explain.
g. Find P(X ? 195), under the assumption that the process is functioning correctly.
h. Based on your answer to part (g), if the process is functioning correctly, would a strength of 195 N be unusually small? Explain.
i. If you observed an adhesive strength of 195 N, would this be convincing evidence that the process was no longer functioning correctly? Explain.

Answer 1

(a) The z-score of 160 is (160 ? 200)/10 = ?4.00. The area to the left of z = ?4.00 is negligible.Therefore P(X?160)?0.
(b)Yes. If the procedure is functioning correctly, values of 160 N or less would almost never occur.
(c)Yes, because 160 N is unusually small if the process is functioning correctly.
(d)The z-score of 203 is (203 ? 200)/10 = 0.30. The area to the right of z = 0.30 is 1 ? 0.6179 = 0.3821.
Therefore P(X?203)=0.3821.
(e)No. If the procedure is functioning correctly, values of 203 N or more would occur about 38% of the time.
(f)No, because 203 N is not unusually large if the process is functioning correctly.
(g)The z-score of 195 is (195 ? 200)/10 = ?0.50. The area to the left of z = ?0.50 is 0.3085.
Therefore P(X?195)=0.3085. .
(h)No. If the procedure is functioning correctly, values of 195 N or less would occur about 31% of the time.
(i)No, because 195 N is not unusually small if the process is functioning correctly.