I need help with this problem A plant manager orders a process shutdown and sett
ID: 3248261 • Letter: I
Question
I need help with this problem
A plant manager orders a process shutdown and setting readjustment wheneve the pH of the final product falls above 7.20 or below 6.80. The sample pH is normally distributed with an unknown mean mu and a standard deviation of sigma = 0.10. Determine the following probabilities: (a) Of readjusting when the process is operating as intended with mu = 7.0. (b) Of readjusting when the process is slightly off target with the mean pH mu = 7.05. (c) Of failing to readjust when the process is too alkaline and the mean pH is mu = 7.25. (d) Of failing to readjust when the process is too acidic and the mean pH is mu = 6.75.Explanation / Answer
1. Solution:
(a) P(X<6.8 or X>7.2)
=P(X<6.8)+P(X>7.2)
=P((X-)/ <(6.8-7)/0.10) + P((X-)/ >(7.2-7)/0.10)
=P(Z < -2)+P(Z > 2)
= 0.0228+0.0228 (check normal table)
= 0.0456
(b) P(X<6.8 or X>7.2)
=P(X<6.8)+P(X>7.2)
=P((X-)/ <(6.8-7.05)/0.10) + P((X-)/ >(7.2-7.05)/0.10)
= P(Z < -2.5)+P(Z > 1.5)
= 0.00621 + 0.00621
= 0.01242
(c) P(6.8 X 7.2)
=P(X 7.2)-P(X< 6.8)
=P((X-)/ <=(7.2-7.25)/0.10) -P((X-)/ < (6.8-7.25)/0.10)
=P(Z<= -0.5)-P(Z<-4.5)
=0.3085-0.000003 (check normal table)
= 0.3085
(d) P(6.8 X 7.2)
=P(X<= 7.1)-P(X< 6.9)
=P((X-)/ (7.2-6.75)/0.10) -P((X-)/ < (6.8-6.75)/0.10)
=P(Z 4.5)-P(Z < 0.5)
= 0.999997-0.691462 (check normal table)
= 0.308535