Week9 Practice Problems Weibull Xbar Distr ZCI Answers PDF

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Week9 Practice Problems Weibull Xbar Distr ZCI Answers...

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STA 220, Winter 15 Week 9 – Practice Problems Answers

1. a. X = shelf life of selected battery X ~ Weibull(shape = 2, scale = 10) P(X < 15) = .8946 b. P(12 < X < 15) = P(X ≤ 15) – P(X ≤ 12) = .8946 - .7631 = .1315 c. P(X > x*) = .50 => x* = 8.3255 months 2. In general, n ≥ 30 is considered large enough for the CLT. 3. False – The CLT tells us that in this situation X (not the population) will be approximately normal. 4. a. Since n ≥ 30, by the CLT the distribution of X is (approximately) normal. b. The mean will be µ = 35 c. The standard error will be (σ/sqrt(n)) = 2.5/sqrt(50) = 0.3536 5. Note that the population is normal, thus the sample mean is normal. a. X = mean fill of the 10 selected bottles

The random variable must be X , not X.

X ~ N(20.2, (.07/sqrt(10)) = .0221) => X ~ N(20.2, .0221) P( X < 20.15) = .0118

20.15

20.2

b. X = mean fill of the 15 selected bottles (Note that the RV must be X , not X )

X ~ N(20.2, (.07/sqrt(15)) = .0181) => X ~ N(20.2, .0181) P(20.21 < X < 20.22) = .1557

20.2 20.21 20.22 6. Note that the sample size is greater than 30 (n=40), and thus the sample mean is approximately normal.

X = mean width of the 40 selected rings (Note that the RV must be X , not X ) X ~ N(2.5, (.2/sqrt(40)) = .03162) => X ~ N(2.5, .03126) P( X < 2.47 or X > 2.53) = 2P( X < 2.47) = 2(.1686) = .3372

by symmetry

2.47

2.5

2.53

7. a. Note: Since n > 30, we can assume X is approximately normal Note: It is assumed that the standard deviation is known, thus the z α/2 confidence interval is used. 138 ± 1.960*1.6/sqrt(47) => 138 ± .457 => (137.543, 138.457) b. m.o.e. = .457

8. a. Note: Since the population is stated as normal, we can assume X is normal Note: The standard deviation is assumed to be known as 3.7, so zα/2 confidence interval is used. 99.5 ± 1.645*3.7/sqrt(25) => 99.5 ± 1.2173 => (98.2827, 100.7173) b. No - Based on the confidence interval, the mean size could be as low as 98.2827 microns (which is less than 98.3) or as large as 100.72 microns (which is larger than 100.3).

9. a. Recall that the interval is: (point estimate – m.o.e. , point estimate + m.o.e.). Thus the point estimate is the midpoint of the interval: point estimate = (48 + 54)/2 = 51 b. m.o.e. = 54 – 51 = 3...