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Can you solve the following problems:3-204,3-208,4-36,4-38,4-58,4-54,4-60,4-80,


Michael Oval Michael Oval Michael Oval Michael Oval Michael Oval Michael Sticky Note Unmarked set by Michael Michael Oval Michael Oval Michael Oval Michael Oval photo1 photo2 photo3 photo4 photo5 photo6
Answered Same DayDec 21, 2021

Answer To: Michael Oval Michael Oval Michael Oval Michael Oval Michael Oval Michael Sticky Note Unmarked set by...

David answered on Dec 21 2021
119 Votes
Solution 3-204:
P (2.1 < X-bar < 2.5) = 1/3 + 1/3 = 2/3 or 0.67
Solution 3-208:
We have µ = 3.5 and σ =1
a) The respective
Z-score with X-bar = 3.25 is
Z = (X-bar - µ)/ (σ/√n)
= (3.25 – 3.5)/ (1/√25)
= -1.25
Using Z-tables, the probability is
P [Z > -1.25] = 1 – 0.1057 = 0.8943
b) The respective Z-score with X-bar = 3.25 is
Z = (X-bar - µ)/ (σ/√n)
= (3.25 – 3.5)/ (1/√100)
= -2.5
Using Z-tables, the probability is
P [Z > -2.5] = 1 – 0.0062 = 0.9938
c) As the sample size increases, the probability also increases.
Solution 4-36:
a) Sample mean, X-bar = (38.02 + 61.98)/2 = 50
b) 95% confidence interval is 38.02 to 61.98 because it is wider than 90% confidence
interval of 39.95 to 60.05.
Solution 4-38:
a) Null Hypothesis (Ho): µ ≥ 100
Alternative Hypothesis (Ha): µ < 100
Test Statistics
Z = (X-bar - µ)/ (σ/√n)
= (100.6 – 100)/ (2/√9)
= 0.9
P [Z < 0.9] = 0.8159
Since p-value is greater than 0.05, we fail to reject Ho.
There is sufficient evidence to conclude that the fiber be judged acceptable.
b) The respective Z-score with X-bar = 102 is
Z = (X-bar - µ)/ (σ/√n)
= (102 – 100)/ (2/√9)
= 3
Using Z-tables, the probability is
P [Z ≥ 3] = 1 – 0.9987 = 0.0013
c) 95% confidence interval is given...
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