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Suppose 1.5 percent of the antennas on new Nokia cell phones are defective. For a random sample of 200 antennas, find the probability that: a. None of the antennas is defective. b. Three or more of the antennas are defective. a. None of the antennas is defective. = (1 - 0.015)^200 ~ 0.0486 b. Three or more of the antennas are defective. Probability of choosing 3 or more defective = 1 –{ probability of choosing at least 2 defective antennas} = 1 – [(1 - 0.015)^200 + 200(0.015)^1(1-0.015)^(199) + 19900(0.015)^2(1-0.015)^(198)] ~ 0.5785 IS THIS RIGHT???? If not, please correct for me.

Submitted: 14 days and 5 hours ago.

Category: Math

Value: $9

Status: CLOSED

Level/Year: Graduate
Subject: Statistics

Already Tried:
To figure it out by the textbook

Accepted Answer

Hi,

Your answers are correct. Just a little rounding error on the first one.

a. P(x = 0) = (200C0)(0.015^0)(1-0.015)^200 = 0.0487

b.

P(x ≥ 0) = 1 - P(x = 0) - P(x = 1) - P(x = 2)

= 1 - (200C0)(0.015^0)(1-0.015)^200 - (200C1)(0.015^1)(1-0.015)^199 - (200C2)(0.015^2)(1-0.015)^198

= 0.5785

Accepts:
Expert:	abozer
Pos. Feedback:	99.9 %
Answered:	11/8/2009

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Level/Year: GraduateSubject: Statistics Already Tried: To figure it out by the textbook

Accepted Answer

Hi,Your answers are correct. Just a little rounding error on the first one.a. P(x = 0) = (200C0)(0.015^0)(1-0.015)^200 = 0.0487b. P(x ≥ 0) = 1 - P(x = 0) - P(x = 1) - P(x = 2) = 1 - (200C0)(0.015^0)(1-0.015)^200 - (200C1)(0.015^1)(1-0.015)^199 - (200C2)(0.015^2)(1-0.015)^198 = 0.5785

Related Math Questions

Level/Year: Graduate
Subject: Statistics

Already Tried:
To figure it out by the textbook

Hi,

Your answers are correct. Just a little rounding error on the first one.

a. P(x = 0) = (200C0)(0.015^0)(1-0.015)^200 = 0.0487

b.

P(x ≥ 0) = 1 - P(x = 0) - P(x = 1) - P(x = 2)

= 1 - (200C0)(0.015^0)(1-0.015)^200 - (200C1)(0.015^1)(1-0.015)^199 - (200C2)(0.015^2)(1-0.015)^198

= 0.5785