Answer to Question #2460 in Statistics and Probability for mudassar
A. Find the probability that the time taken for a randomly chosen customer to be dealt with by a cashier is more than 180s.
B. One of the cashier deals with two customers, one straight after the other.
Then Z = (X-150)/45is a random variable which has standard normal.
A. P(X>180) =
= P(& (X-150)/45>(180-150)/45 ) =
= P(& Z>30/45& ) = P(Z>0.6667) =
= 1 -& P(Z<0.6667) = 1 - f(0.6667) =
= 1 -& 0.747518 = 0.252482
Question B is not completely formulated
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