problem set 3 answers - 7.94 = 24 minutes = 7 minutes and n...

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7.94 μ = 24 minutes, σ = 7 minutes, and n = 100 μ x = μ = 24 minutes and σ x = σ n = 7 100 = . 7 minutes The sampling distribution of x is approximately normal because n > 30 7.96 μ = 24 minutes, σ = 7 minutes, and n = 100 μ x = μ = 24 minutes and σ x = σ n = 7 100 = . 7 minutes a. For x = 22 : z = ( x μ ) σ x = (22 24) . 7 = 2 . 86 P ( x < 22) = P ( z < 2 . 86) = . 0021 b. For x = 23 : z = ( x μ ) σ x = (23 24) . 7 = 1 . 43 For x = 26 : z = ( x μ ) σ x = (26 24) . 7 = 2 . 86 P (23 < x < 26) = P ( 1 . 43 < z < 2 . 86) = P ( z < 2 . 86) P ( z < 1 . 43) = . 9979 . 0764 = . 9215 c. P ( x within 1 minute of μ ) = P (23 x 25) For x = 23 : z = ( x μ ) σ x = (23 24) . 7 = 1 . 43 For x = 25 : z = ( x μ ) σ x = (25 24) . 7 = 1 . 43 P (23 x 25) = P ( 1 . 43 z 1 . 43) = P ( z 1 . 43) P ( z ≤ − 1 . 43) = . 9236 . 0764 = . 8472 d. P ( x greater than μ by 2 minutes or more ) = P ( x 26) For x = 26 : z = ( x μ ) σ x = (26 24) . 7 = 2 . 86 1
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P ( x 26) = P ( z 2 . 86) = 1 P ( z 2 . 86) = 1 . 99979 = . 0021 7.99 p = . 88 , q = 1 p = 1 . 88 = . 12 , and n = 80 μ e p = p = . 88 , and σ e p = r pq n = r ( . 88)( . 12) 80 = . 03633180 np = (80) ( . 88) = 70 . 4 > 5 , nq = (80)(
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