(A) Find the Mean and Standard Deviation of the Sampling

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The Answer is (A) E( (A) E( ) = = 450, and SE( ) = = 5 (B) P( < 445) = P(Z < -1) = 0.5000 - 0.3413 = 0.1587 (C) P(445< <455) = P(-1.0 < Z < 1.0) = 2(0.3413) = 0.6826 (D) P( > 460) = P(Z > 2.0) = 0.5000 - 0.4772 = 0.0228 (E) The sample size is large; n = 100 is greater than 30, so the distribution of the average weekly earnings for employees is at least approximately normal.

(A) E( ) = = 450, and SE( ) = = 5 (B) P( < 445) = P(Z < -1) = 0.5000 - 0.3413 = 0.1587 (C) P(445< <455) = P(-1.0 < Z < 1.0) = 2(0.3413) = 0.6826 (D) P( > 460) = P(Z > 2.0) = 0.5000 - 0.4772 = 0.0228 (E) The sample size is large; n = 100 is greater than 30, so the distribution of the average weekly earnings for employees is at least approximately normal.

) =

= 450, and SE( (A) E( ) = = 450, and SE( ) = = 5 (B) P( < 445) = P(Z < -1) = 0.5000 - 0.3413 = 0.1587 (C) P(445< <455) = P(-1.0 < Z < 1.0) = 2(0.3413) = 0.6826 (D) P( > 460) = P(Z > 2.0) = 0.5000 - 0.4772 = 0.0228 (E) The sample size is large; n = 100 is greater than 30, so the distribution of the average weekly earnings for employees is at least approximately normal.

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= 5
(B) P( (A) E( ) = = 450, and SE( ) = = 5 (B) P( < 445) = P(Z < -1) = 0.5000 - 0.3413 = 0.1587 (C) P(445< <455) = P(-1.0 < Z < 1.0) = 2(0.3413) = 0.6826 (D) P( > 460) = P(Z > 2.0) = 0.5000 - 0.4772 = 0.0228 (E) The sample size is large; n = 100 is greater than 30, so the distribution of the average weekly earnings for employees is at least approximately normal.

< 445) = P(Z < -1) = 0.5000 - 0.3413 = 0.1587
(C) P(445< (A) E( ) = = 450, and SE( ) = = 5 (B) P( < 445) = P(Z < -1) = 0.5000 - 0.3413 = 0.1587 (C) P(445< <455) = P(-1.0 < Z < 1.0) = 2(0.3413) = 0.6826 (D) P( > 460) = P(Z > 2.0) = 0.5000 - 0.4772 = 0.0228 (E) The sample size is large; n = 100 is greater than 30, so the distribution of the average weekly earnings for employees is at least approximately normal.

<455) = P(-1.0 < Z < 1.0) = 2(0.3413) = 0.6826
(D) P( (A) E( ) = = 450, and SE( ) = = 5 (B) P( < 445) = P(Z < -1) = 0.5000 - 0.3413 = 0.1587 (C) P(445< <455) = P(-1.0 < Z < 1.0) = 2(0.3413) = 0.6826 (D) P( > 460) = P(Z > 2.0) = 0.5000 - 0.4772 = 0.0228 (E) The sample size is large; n = 100 is greater than 30, so the distribution of the average weekly earnings for employees is at least approximately normal.

> 460) = P(Z > 2.0) = 0.5000 - 0.4772 = 0.0228
(E) The sample size is large; n = 100 is greater than 30, so the distribution of the average weekly earnings for employees is at least approximately normal.

(A) Find the Mean and Standard Deviation of the Sampling

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(A) Find the Mean and Standard Deviation of the Sampling

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Correct Answer:VerifiedShow Answer

Correct Answer:
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