Question

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?

(c) What is the standard deviation of the number of defective electronic devices?

Answer 1

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