Question Completion Status A manufacturer of a consumer electronics products exp

Question

Question Completion Status A manufacturer of a consumer electronics products expects 2% of units to fail during the warranty period. A sample of 500 independent units is tracked for warranty performance. (a) What is the probability that none fail during the warranty period? (b) What is the expected number of failures during the warranty period? (c) What is the probability that more than 2 units fail during the warranty period? Round your answer to four decimal places (e.g. 98.7654 (b) (c) C Round your answer to the nearest integer. Round your answer to four decimal places (e.q.98.7654). > Moving to another question wil save this response. Question 3 of 16

Explanation / Answer

P = 0.02

n = 500

mean = n * p = 500 * 0.02 = 10

Standard deviation = sqrt(n * p * (1 - p)) = sqrt(500 * 0.02 * 0.98) = 3.13

A) P(-0.5 < X < 0.5) = P((-0.5 - 10) / 3.13 < Z <( 0.5 - 10)/3.13)

= P(-3.35 < Z < -3.04)

= P(Z < -3.04) - P(Z < -3.35)

= 0.0012 - 0.0004 = 0.0008

B) expected value = n * p = 500 * 0.02 = 10

C) P(X > 2.5) = P((X - mean) / SD > (2.5 - 10)/3.31)

= P(Z > -2.27)

= 1 - P(Z < -2.27) = 1 - 0.0116 = 0.9884

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