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Answered Same Day Dec 21, 2021

Solution

David answered on Dec 21 2021
121 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
) 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
) 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.
) 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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