Answer:
Hence, the probability that seven or more of them used their phones for guidance on purchasing decisions is 0.7887
Step-by-step explanation:
We are given the following information in the question:
We treat people using phones in the store for guidance as a success.
P(People use phone for guidance) = 57% = 0.57
Then the number of people follows a binomial distribution, where
Formula:
[tex]P(X=x) = \binom{n}{x}.p^x.(1-p)^{n-x}[/tex]
where n is the total number of observations, x is the number of success, p is the probability of success.
Now, we are given n = 14 and x = 7
We have to evaluate:
[tex]P(x \geq 7) = P(x = 7) + P(x = 8) + P(x = 9) + P(x = 10) + P(x =11) + P(x = 12) + P(x = 13) + P(x = 14)\\= \binom{14}{7}(0.57)^7(1-0.57)^7 + \binom{14}{8}(0.57)^8(1-0.57)^6 + \binom{14}{7}(0.57)^9(1-0.57)^5 + \binom{14}{7}(0.57)^{10}(1-0.57)^4 + \binom{14}{7}(0.57)^{11}(1-0.79)^3 + \binom{14}{7}(0.57)^{12}(1-0.57)^2 + \binom{14}{7}(0.57)^{13}(1-0.57)^1 + \binom{14}{7}(0.57)^{14}(1-0.57)^0\\= 0.7887 = 78.67\%[/tex]
Hence, the probability that seven or more of them used their phones for guidance on purchasing decisions is 0.7887