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Suppose that components have weights that are normally distributed with u = 100 and o = 10. An experimenter measures the weights of a random sample of 25 components in order to estimate u. What is the probability that the experimenter's estimate of u will be ess than 95? 1. 0 0,9772 0.9772 2. O 0.0062 0.0062 3. O 0.0228 0.0228 4. O 0.9938 0.9938

Suppose that components have weights that are normally distributed with u = 100 and o = 10. An experimenter measures the weights of a random sample of 25 components in order to estimate u. What is the probability that the experimenter's estimate of u will be ess than 95? 1. 0 0,9772 0.9772 2. O 0.0062 0.0062 3. O 0.0228 0.0228 4. O 0.9938 0.9938

Algebra & Trigonometry with Analytic Geometry
Algebra & Trigonometry with Analytic Geometry
13th Edition
ISBN:9781133382119
Author:Swokowski
Publisher:Swokowski
Chapter10: Sequences, Series, And Probability
Section10.8: Probability
Problem 32E
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Suppose that components have weights that are normally distributed with u = 100 and o = 10. An experimenter measures the weights of a random sample of 25 components in order to estimate u. What is the probability that the experimenter's estimate of u will be ess than 95? 1. O 0.9772 0.9772 2. O 0.0062 0.0062 3. O 0.0228 0.0228 4. O 0.9938 0.9938
Transcribed Image Text:Suppose that components have weights that are normally distributed with u = 100 and o = 10. An experimenter measures the weights of a random sample of 25 components in order to estimate u. What is the probability that the experimenter's estimate of u will be ess than 95? 1. O 0.9772 0.9772 2. O 0.0062 0.0062 3. O 0.0228 0.0228 4. O 0.9938 0.9938
To test the similarity of variance of the two populations I and II, a sample of 10 were taken from population I and the variance was 24.7, while a sample of 13 were taken from population II and the variance was 37.2 For the 0.05 level of significance and f(1-a)(12;9) 1/(fa(9,12) then the interval estimator for the ratio of variance o/o? is 1. O ketiganya salah None of the above (Sın)^2/(s)^2. f{a/2}[9;12) < (on)?/(01)*2 < (S1)/(sı)2. f(1-a/2|9:12) (sı)^2/(sı)^2. f(a/2)(9;12) < (o1)^2/(01)^2 < (Sı)^2(s)^2. f(1-a/2)(9;12) 2. O (Sı)^2/(s)^2+ f(1-a/2){12;9) < (O1)^2/(01)^2 < (Si)^2/(si)^2. (Sı)^2/(s)^2. f(1-a/2)(12,9) < (o1)^2(01)^2 < (Sı)^2/(si)^2. f(a/2)(12;9) f(a/2)(12;9) * 3. O (S)^2/(sı)^2. f(a/2)(12,9) < (O1)^2/(01)^2 < (S1)^2/(si)^2. f¢1-a/2)(12;9) (Sı)"2/(si)^2. f(a/2)(12;9) < (o1)^2/(01)^2 < (S1)^?/(si)^2» f(1-a/2){12;9) 4. O
Transcribed Image Text:To test the similarity of variance of the two populations I and II, a sample of 10 were taken from population I and the variance was 24.7, while a sample of 13 were taken from population II and the variance was 37.2 For the 0.05 level of significance and f(1-a)(12;9) 1/(fa(9,12) then the interval estimator for the ratio of variance o/o? is 1. O ketiganya salah None of the above (Sın)^2/(s)^2. f{a/2}[9;12) < (on)?/(01)*2 < (S1)/(sı)2. f(1-a/2|9:12) (sı)^2/(sı)^2. f(a/2)(9;12) < (o1)^2/(01)^2 < (Sı)^2(s)^2. f(1-a/2)(9;12) 2. O (Sı)^2/(s)^2+ f(1-a/2){12;9) < (O1)^2/(01)^2 < (Si)^2/(si)^2. (Sı)^2/(s)^2. f(1-a/2)(12,9) < (o1)^2(01)^2 < (Sı)^2/(si)^2. f(a/2)(12;9) f(a/2)(12;9) * 3. O (S)^2/(sı)^2. f(a/2)(12,9) < (O1)^2/(01)^2 < (S1)^2/(si)^2. f¢1-a/2)(12;9) (Sı)"2/(si)^2. f(a/2)(12;9) < (o1)^2/(01)^2 < (S1)^?/(si)^2» f(1-a/2){12;9) 4. O
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