A small repair shop orders parts for its repairs. Its supplier claims that only 2% of its parts are defective. In order of 500 parts, an employee takes a sample of 100 parts. Determine the likelihood that the employee finds three or more parts are defective.

Correct answer:

p =  0.3233

Step-by-step explanation:

r=2%=1002=501=0.02 n=500 n2=100  p0=(0n2)r0(1r)n20=(0100)0.020(10.02)1000=10.0200.981000.1326 p1=(1n2)r1(1r)n21=(1100)0.021(10.02)1001=1000.0210.98990.2707 p2=(2n2)r2(1r)n22=(2100)0.022(10.02)1002=49500.0220.98980.2734  P=p0+p1+p2=0.1326+0.2707+0.27340.6767  p=1P=10.6767=0.3233

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