Alisa says it is easier to compare the numbers in set A than the numbers in set B
2 answers:
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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.
Answer:
(a) <em> </em><em>n</em> : 20 50 100 500
P (-200 < <em>X</em> - <em>μ </em>< 200) : 0.2886 0.4444 0.5954 0.9376
(b) The correct option is (b).
Step-by-step explanation:
Let the random variable <em>X</em> represent the amount of deductions for taxpayers with adjusted gross incomes between $30,000 and $60,000 itemized deductions on their federal income tax return.
The mean amount of deductions is, <em>μ</em> = $16,642 and standard deviation is, <em>σ</em> = $2,400.
Assuming that the random variable <em>X </em>follows a normal distribution.
(a)
Compute the probability that a sample of taxpayers from this income group who have itemized deductions will show a sample mean within $200 of the population mean as follows:
- For a sample size of <em>n</em> = 20
- For a sample size of <em>n</em> = 50
- For a sample size of <em>n</em> = 100
- For a sample size of <em>n</em> = 500
<em> n</em> : 20 50 100 500
P (-200 < <em>X</em> - <em>μ </em>< 200) : 0.2886 0.4444 0.5954 0.9376
(b)
The law of large numbers, in probability concept, states that as we increase the sample size, the mean of the sample () approaches the whole population mean (
).
Consider the probabilities computed in part (a).
As the sample size increases from 20 to 500 the probability that the sample mean is within $200 of the population mean gets closer to 1.
So, a larger sample increases the probability that the sample mean will be within a specified distance of the population mean.
Thus, the correct option is (b).