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 10 N. 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.
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.

A) P ( X ≤ 160 ) = 0

B) Unusually small

C) process was no longer functioning correctly

D) P ( X ≥ 203 ) = 0.3821

E) Not unusually large

F) No-Evidence

G) P ( X ≤ 195 ) = 0.3085

H) Not unusually small

I) No evidence

Step-by-step explanation:

Given:-

- The random variable (X) denotes the adhesive strength is normally distributed with mean u = 200 N and standard deviation s.d = 10 N.

                            X ~ N ( 200 , 10^2 )

Solution:-

A) Find P(X ≤ 160), under the assumption that the process is functioning correctly.

- Determine the Z-score value:

                      Z-score = ( X - u ) / s.d

                                    = ( 160 - 200 ) / 10

                                    = -4

- Use the standardized z-table to determine the probability:

                      P ( X ≤ 160 ) = P ( Z ≤ -4 )

                                           = 0

- Assuming the process is functioning properly then the adhesive strength of X = 160 N would be considered unusually small since the probability of occurrence is approximately 0.

- If we were to observe an adhesive strength process that gives us the value of 160 N can imply that the process is not functioning properly as its outside the 3 standard deviations from the mean value. ( Conclusive Evidence )

D) Find P(X ≥ 203), under the assumption that the process is functioning correctly.

- Determine the Z-score value:

                      Z-score = ( X - u ) / s.d

                                    = ( 203 - 200 ) / 10

                                    = 0.3

- Use the standardized z-table to determine the probability:

                      P ( X ≥ 203 ) = P ( Z ≥ 0.3 )

                                           = 0.3821

- Assuming the process is functioning properly then the adhesive strength of X = 203 N would be not be considered unusually small since the probability of occurrence is in the heart of the bell curve.

- If we were to observe an adhesive strength process that gives us the value of 203 N can imply that the process is functioning properly as its within 1 standard deviation from the mean value. ( No evidence )

G) Find P(X ≤ 195), under the assumption that the process is functioning correctly.

- Determine the Z-score value:

                      Z-score = ( X - u ) / s.d

                                    = ( 195 - 200 ) / 10

                                    = -0.5

- Use the standardized z-table to determine the probability:

                      P ( X ≤ 195 ) = P ( Z ≤ -0.5 )

                                           = 0.3085

- Assuming the process is functioning properly then the adhesive strength of X = 195 N would be not be considered unusually small since the probability of occurrence is in the heart of the bell curve.

- If we were to observe an adhesive strength process that gives us the value of 195 N can imply that the process is functioning properly as its within 1 standard deviation from the mean value. ( No evidence )

the reporter's right

step-by-step explanation:

1/4 = .25 / 25%