A manufacturer of a consumer electronics products expects 2% of units to fail du

Question

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) Round your answer to the nearest integer. Round your answer to four decimal places (e.g. 98.7654)

Explanation / Answer

The probability that a unit fails during the warranty period = 0.02

The probability that a unit does not fail during the warranty period = 1 - 0.02 = 0.98

(a) Probability that none fail during the warranty period = (0.98)500 = 0.000041.

Note: Four decimal places gives 0.0000! You may try giving 0.0001 in case that fails.

(b) Expected number of failures

= 500 * 0.02 = 10

(c) If one unit fails, it can be any of the 500 units.

Probability of exactly one unit failing = 500 * 0.02 * 0.98499 = 0.0004

If two units fail, they can come in 500C2 = 124750 ways.

Probability of exactly two units failing = 124570 * 0.022 * 0.98498 = 0.0021

=> Probability that two or less units fail = 0.0000 + 0.0004 + 0.0021 = 0.0025

Probability that more than 2 units fail = 1 - 0.0025 = 0.9975.

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