Answer:
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 )
