hw5sol - 2 /12) Note: Large sample is not required because...

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Page 1 of 2 The Hong Kong University of Science and Technology IS O M 111 - Business statistics Lecture Section 5 (Prof Inchi HU) Assignment 5 Solution (Problem Sheet 3: #1, #11 & #12) Solutions 1. (a) μ = E( x ) = 19.3 or $19300 E ( x 2 ) = 397.8 σ = 2 2 ) ( μ x = 5.0309 or $5030.9 (b) x μ σ 2 /n ) P( x > 20.5) = P( 8 / 0309 . 5 3 . 19 5 . 20 > z ) = P( z > 1.91) = 0.5 – 0.4719 = 0.0281 11. (a) Let x be the monthly sales of Dark-Man Toothpaste. x ~ N (1000, 150 2 ) P( x > 1100) = P( z > 0.67) = 0.5 – 0.2486 = 0.2514 (b) Let c be the required stock for the monthly demand. Then P( x < c) = 0.95 P( z < 150 1000 c ) = 0.95 P(0 < z < 150 1000 c ) = 0.45 150 1000 c = 1.645 c = 1246.75 (c) Let x i be the sales of the i th month in the next year, i = 1, 2, …, 12 P( = 12 1 i i x > 12600) = P( x > 1050) where x = = 12 1 12 1 i i x
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Unformatted text preview: 2 /12) Note: Large sample is not required because we are sampling from a normal population. ∴ P(Total sales &gt; 12600) = P( x &gt; 1050) = P( z &gt; 12 150 1000 1050 − ) = P( z &gt; 1.15) = 0.5 – 0.3749 = 0.1251 Page 2 of 2 12. n = 300, N = 10,000, p = 0.019 As n/N = 300 / 10000 = 0.03 &lt; 0.05 We can use the Binomial distribution to approximate the Hypergeometric Distribution. (a) E( p ˆ ) = p = 0.019 or 1.9% Var( p ˆ ) = p(1-p) / n = 0.019 (1 – 0.019) / 300 = 6.213 x 10-5 (b) P( p ˆ ≥ 0.03) = P( z &gt; 5 10 213 . 6 019 . 03 . − × − ) = P( z &gt; 1.40) = 0.5 – 0.4192 = 0.0808 (c) P( x = 2) = 5 2 C p 2 (1 – p) 3 = 10 (0.0808) 2 (1 – 0.0808) 3 = 0.050705...
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This note was uploaded on 09/18/2010 for the course BBA ISOM111 taught by Professor Hu during the Fall '08 term at HKUST.

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hw5sol - 2 /12) Note: Large sample is not required because...

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