Starbucks manager at city wants to find a the probability of

Starbucks manager at city wants to find:

(a) the probability of the total average cost less than or equal to $1000 per day. He is only concerned about the cost of waiting time of customers. Customers spend time at the checkout register with mean of 4 minutes and standard deviation of 4 minutes. On average, he got 100 customers per day. 1 minute of waiting time is considered as cost of $0.5.

(b) the probability of the total expected revenue greater than or equal to $1500 per day. On average, each customer spends $4 at a time with standard deviation of $1.

Solution

a)

P( x < 1000)

P( z < [1000 - 4*400*0.5 ] / [ 4 *0.5 * 100 ] )

P( z <1 ) = 1 - P( z > 1)

P( z < 1 ) = 1 - 0.1587 = 0.8413

b)

P( x > 1500)

P( z > [1500 - 4 *0.5*100 ] / [ 1 * 4 *100 ])

P ( z > 3.25 ) = 0.00058