Explanation Given: The mean of the distribution is 60 and the standard deviation is 10 . Concept used: The probability associated with the distribution x = a is shown below. P ( X > a ) = P ( Z > a − μ σ ) … … ( 1 ) Calculation: Consider X be the normal random variable that denotes the score in an examination. The probability that an applicant score 70 or more than 70 is P ( X ≥ 70 ) . The value of a is 70 , the mean μ is 60 and the standard deviation is 10 . Substitute 70 for a , 60 for μ and 10 for σ in equation (1) as follows: P ( X ≥ 70 ) = P ( X > 70 ) = P ( Z > 70 − 60 10 ) = P ( Z > 10 10 ) = P

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Author: Tan

Publisher: INTER CENG

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Chapter 10.3, Problem 30E

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To find: The percentage of applicants that passed the written examination.

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R₂ = X2 2) slots per pole per phase = 3/31 B-180 60 msl kd Kas Sin () 2 I sin (6) sin(30) Sin (30) اذا مريد شرح الكتب بس 0 بالفراغ 3 Cos (30) 0.866 4) Rotating ined sove in peaper 5) Synchronous speed s 120×50 6 s = 1000-950 1000 Copper losses 5kw Rotor input 5 0.05 6) 1 loo kw اذا ميريد شرح الكتب فقط Look 7) rotov DC I need a detailed solution on paper please 0 64 Solve the following equations: 0 Q1// Find the solution of: ( y • with y(0) = 1. dx x²+y²

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The percentage of applicants that passed the written examination.

Solution Summary: The author explains how the probability associated with the distribution x=a is shown.

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Chapter 10 Solutions