a) As one in five customers received, the coupon,
therefore
P(received coupon) = 1/5 = 0.2
Therefore out of 6 customers, the number of customers who received the coupon here is modelled as:
Therefore this is a binomial distribution here. This is because the probability of any customer receiving the coupon is same and equal to 0.2. Also the number of customers here is constant n = 6 here.
b) The mean and standard deviation of X here is computed as:
c) P(X >= 2) = 1 - P(X = 0) - P(X = 1) = 1 - 0.86 - 6*0.85*0.2 = 0.34464
As the probability here is 0.34464 > 0.05, therefore the given that two of the six customers received the coupon the event is not unusual here. Therefore we cannot reject the claims here.
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