1. A farmer divided a field into 1-foot by 1-foot sections and tested soil samples from 32 randomly selected sections in the fie
1 answer:
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Answer:
a) Upper Control Limit = 0.10
Lower Control Limit = 0.01
b) The tech assistance process is stable and in control.
Step-by-step explanation:
Given - After a number of complaints about its tech assistance, a
computer manufacturer examined samples of calls to determine
the frequency of wrong advice given to callers. Each sample
consisted of 100 calls.
SAMPLE 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16
Number of errors 5 3 5 7 4 6 8 4 5 9 3 4 5 6 6 7
To find - a. Determine 95 percent limits.
b. Is the tech assistance process stable (i.e., in control)
Proof -
z = 95% confidence interval
= 1.96
⇒z = 1.96
Proportion of defects,P = total defectives/total observations
= 0.0544
⇒P = 0.0544
Now,
Q = 1-P
= 0.9456
⇒Q = 0.9456
Now,
Average sample size, N = 100
Standard deviation =
= 0.0227
Now,
Upper Control Limit = P + z(Standard deviation)
= 0.0988
⇒Upper Control Limit = 0.0988 ≈ 0.10
And
Lower Control Limit = P - z(Standard deviation )
= 0.0099
⇒Lower Control Limit = 0.0099 ≈ 0.01
∴ we get
Upper Control Limit = 0.10
Lower Control Limit = 0.01
b.)
Now,
it is clear that the fraction defective values are wit in upper an lower control limits.
So, The tech assistance process is stable and in control.
Percent of red lights last between 2.5 and 3.5 minutes is 95.44% .
<u>Step-by-step explanation:</u>
Step 1: Sketch the curve.
The probability that 2.5<X<3.5 is equal to the blue area under the curve.
Step 2:
Since μ=3 and σ=0.25 we have:
P ( 2.5 < X < 3.5 ) =P ( 2.5−3 < X−μ < 3.5−3 )
⇒ P ( (2.5−3)/0.25 < (X−μ)/σ < (3.5−3)/0.25)
Since, Z = (x−μ)/σ , (2.5−3)/0.25 = −2 and (3.5−3)/0.25 = 2 we have:
P ( 2.5<X<3.5 )=P ( −2<Z<2 )
Step 3: Use the standard normal table to conclude that:
P ( −2<Z<2 )=0.9544
Percent of red lights last between 2.5 and 3.5 minutes is % .
