Suppose the daily customer volume at a call center has a normal distribution with mean 4,200 and standard deviation 850.
What is the probability that the call center will get between 3,600 and 3,800 calls in a day?
Please specify your answer in decimal terms and round your answer to the nearest hundredth (e.g., enter 12 percent as 0.12).
Solution :
Given that ,
mean = = 4200
standard deviation = = 850
P( 3600< x < 3800 ) = P[(3600-4200)/ 850) < (x - ) / < (3800-4200) / 850) ]
= P(-0.71 < z < -0.47)
= P(z <-0.47 ) - P(z <-0.71 )
Using standard normal table
= 0.3192 - 0.2389 = 0.0803
Probability = 0.08
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