Question

In: Statistics and Probability

A large retai lchain purchases a certain kind of electronic device from a manufacturer. The manufacturer...

A large retai lchain purchases a certain kind of electronic device from a manufacturer. The manufacturer indicates that the defective rate of the device is 3%. The inspector at the retailer randomly picks 100 items from a shipment.

(a) What is the expected value of the number of defective electronic devices?

(b) What is the variance of the number of defective electronic devices?

Solutions

Expert Solution

Solution:

n = 100

p = 0.03

binomial probability distribution

Formula:

P_{(k out of n )}= n!*p^k * q^n-k / k! *(n - k)!

( a )

Expected value = 0(P(0)) + 1(P(1))+2(P(2)) + 3(P(3))+............+ 99(P(99))+100(P(100))

= 0(0.0476)+1(0.1471)+2(0.2252)+3(0.2275)+4(0.1706)+5(0.1013)+6(0.0496)

+7(0.0206)+8(0.0074)+9(0.0023)+10(0.0007)+11(0.0002)+...........

+.................+99(0.0000)+100(0.0000)

= 2.9992

( b )

varience = =

= 0²(P(0)) + 1²(P(1))+2²(P(2)) + 3²(P(3))+............+ 99²(P(99))+100²(P(100))

=0²(0.0476)+1²(0.1471)+2²(0.2252)+3²(0.2275)+4²(0.1706)+5²(0.1013)+6²(0.0496)

+7²(0.0206)+8²(0.0074)+9²(0.0023)+10²(0.0007)+11²(0.0002)+...........

+.................+99²(0.0000)+100²(0.0000)

= 11.903

= = 11.903 - 2.9992² = 2.9077

( c )

standard deviation = =

= 0²(P(0)) + 1²(P(1))+2²(P(2)) + 3²(P(3))+............+ 99²(P(99))+100²(P(100))

=0²(0.0476)+1²(0.1471)+2²(0.2252)+3²(0.2275)+4²(0.1706)+5²(0.1013)+6²(0.0496)

+7²(0.0206)+8²(0.0074)+9²(0.0023)+10²(0.0007)+11²(0.0002)+...........

+.................+99²(0.0000)+100²(0.0000)

= 11.903

= = = = 1.7052

orchestra answered 1 year ago

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