The median time a customer waits to be served at a large retail company is 20 minutes. On a day when 6 customers pitch up, what is the probability that more than half will have to wait more than 20 minutes. (Answer: 1-0.3438)
I think that $\theta = 0.5$ since half of the customers will have a waiting time from 20 mins upwards.
so $1 - \theta = 0.5$
Let X denote the number of customers that will have to wait more than 20 minutes.
$P(X>3) = 1 -P(X \leq 3) = 1 - [P(X=0) + P(X=1) + P(X=2) + P(X=3)] = 1 - [{6 \choose 0}(0.5)^0(0.5)^6 + {6 \choose 1}(0.5)^1(0.5)^5 + {6 \choose 2}(0.5)^2(0.5)^4 + {6 \choose 3}(0.5)^3(0.5)^3] = 1 - 0.6563$
I know that if the answer was $P(X \geq 3) = 1 - P(X<3)$ then it would be their answer but the question stated more than half. Not half or more...
$1 - P(X < 3) = 1 - [P(X=0) + P(X=1) + P(X=2)] = 1 -[{6 \choose 0}(0.5)^0(0.5)^6 + {6 \choose 1}(0.5)^1(0.5)^5 + {6 \choose 2}(0.5)^2(0.5)^4] = 1-0.3438$