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A brand name has a 50% recognition rate. Assume the owner of the brand wants to verify that rate by beginning with a small sample of 7 randomly selected consumers.

A) What is the probability that exactly 6 of the selected consumers recognize the brand name is ______? (Round to 3 decimal places)
B) What is the probabilty that all the selected consumers recognize the brand _______?
C) What is the probability that 3 of the selected consumers recognize the brand _______?
D) If 4 consumers are randomly selected is 3 an unusually high number of consumers that recognize the brand name ______?

User Loyal
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Answer:

0.0547,0.0078,0.25

Explanation:

Any random customer selected probability to recognize = 50% = 0.50 which remains the same for each customer. Also there are only two outcomes. Hence X the no of customers who recognize brand name is Binomial

X is Bin (7,0.50)

a)
P(x=6) = 7C6 (0.5)^7 = 0.0547

b)
P(X=7) = (0.5)^7 = 0.0078

c)
P(X=3) = 7C3 (0.5)^7\\= 0.2734

d) Here n changes from 7 to 4. But p probability for success remains the same.

Hence
P(X=3) = 4C3 (0.5)^4 = 0.25

This probability is not unusually high.

If 4 consumers are randomly selected is 3 an unusually high number of consumers that recognize the brand name is 0.25

User Zenvelope
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