A brand name has a 60% recognition rate. Assume the owner of the brand wants to

Question

A brand name has a 60% recognition rate. Assume the owner of the brand wants to verify that rate by beginning with a small sample of 66 randomly selected consumers. Complete parts (a) through (d) below. a. What is the probability that exactly 5 of the selected consumers recognize the brand name? The probability that exactly 5 of the 6 consumers recognize the brand name is nothing. b. What is the probability that all of the selected consumers recognize the brand name? The probability that all of the selected consumers recognize the brand name is nothing

Explanation / Answer

Here, X is random variable which represents number of customers recognize the brand name out of 66 customers & probability of customers will recognize the brand name is 60/100=0.6 .

Clearly , X ~ Binomoial (n=66 , p=0.6)

Therefore , probability mass function of X is ,

P(X=x) = nCx px (1-p)n-x =66Cx 0.6x 0.466-x ; x=0,1,.......,n=66

(a) We have to find here the probability that exactly 5 of the selected consumers recognize the brand name i.e. P(X=5).

Consider,

P(X=5) =66C5 0.65 0.466-5   

={66! / [5!*(66-5)!]}*0.65 * 0.461

={66*65*64*63*62*61! / [5*4*3*2*1*61!]}* 0.07776* 5.316912e-25

={66*65*64*63*62/[5*4*3*2*1]} *4.134431e-26

=8936928 *4.134431e-26

=3.694911e-19

Thus, the probability that exactly 5 of the selected consumers recognize the brand name is 3.694911e-19 .

(b) We have to find here the probability that all of the selected consumers recognize the brand name i.e. P(X=66).

Consider,

P(X=66) =66C66 0.666 0.466-66

={66! / [66!*(66-66)!]}*0.666 * 0.40

={66! / [66!*0!]}*0.666

=0.666    ..............Since 0!=1

=2.28025e-15

Thus, the probability that all of the selected consumers recognize the brand name is 2.28025e-15 .

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