Problem 9-22 (modified). A quality control inspector accepts
shipments of 500 precision .5" steel rods if the mean diameter of a
sample of 81 falls between .4995" and .5005". Previous evaluations
have established that the standard deviation for individual rod
diameters is .003".
What is the probability the inspector will accept an
out-of-tolerance shipment having mu=.5003? (Note: we aren't told
the tolerance, but for simplicity assume that it is .0002 so that
mu=.5003 is out-of-tolerance. The acceptance standard of between
.4995" and .5005" relates to the sample mean, not to the population
mean mu.)
Probability=
What is the probability the inspector will reject a near-perfect
shipment having mu=.4999?
Probability=
1)
for normal distribution z score =(X-μ)/σx | |
here mean= μ= | 0.5003 |
std deviation =σ= | 0.003 |
sample size =n= | 81 |
std error=σx̅=σ/√n= | 0.0003 |
probability the inspector will accept an out-of-tolerance shipment:
probability = | P(0.4995<X<0.5005) | = | P(-2.4<Z<0.6)= | 0.7257-0.0082= | 0.7175 |
2)
probability the inspector will reject a near-perfect shipment having mu=.4999:
probability =1-P(-0.4995<X<0.5005)=1-P(-1.2<Z<1.8)=1-(0.9641-0.1151)=0.1510 |
Get Answers For Free
Most questions answered within 1 hours.