A company just opened a new massage station at the local airport. The company ha

Question

A company just opened a new massage station at the local airport. The company has a stand that offers massages to travelers. Customers can select a length of massage between 5 and 20 minutes and there is a unique rate of $30 independently of the length selected by customers. The average length of massage requested by customers is of 15 minutes with standard deviation of 10 minutes. There are three employees delivering massages. The average number of potential customers requesting a massage is of 15 per hour. The inter-arrival times are assumed to be exponentially distributed. If spot is available when the customer arrives, she leaves in order not to risk missing her/his flight. What is average number of customers being serviced per hour at the massage station? (Fractions are possible. Round your answer to one digit after the decimal place. For example, if you calculate 10.63?, type 10.6) Consider a process that has 3 stations, ordered in sequence: A, B and C. At each station, two consecutive tasks are performed one after the other. The time (in seconds per unit) it takes for a single person to perform each task is given in the table below (e.g., task A2 takes 67 seconds per unit):

Explanation / Answer

Given Information

Mean (µ) = 8 cm and Std deviation () = 0.04 cm

Find out the prob (X > 8.08) = ?

Z = X - µ/

Z = 8.08 – 8 / 0.04

Z = 2

The prob for this Z = 2 will be 0.4772 (from the Table)

Therefore prob (X > 8.08) = 0.5 – 0.4772 = 0.0228

Or prob = 2.28 %

Find out prob (X < 7.08) = ?

Z = X - µ/

Z = 7.08 – 8/ 0.04

Z = - 0.05

The corresponding prob will be 0.0199

Therefore prob (X < 7.08) = 0.5 – 0.0199

                                             = 0.4801

Or 48.01 %

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