A shift supervisor at a technical support call center claims that the true mean

Question

A shift supervisor at a technical support call center claims that the true mean waiting time for customers to speak to a live representative is 8.8 minutes, with a standard deviation of 5.6 minutes. The district manager examines some data and finds that, in a random sample of 37 phone calls, the sample mean waiting time was 10.67 minutes. If the shift supervisor is correct about the true mean waiting time, what is the probability that the district manager would observe a sample mean of 10.67 minutes or more? Hint: this is a sampling distribution problem.

Explanation / Answer

Mean ( u ) =8.8
Standard Deviation ( sd )=5.6
Number ( n ) = 37
Normal Distribution = Z= X- u / (sd/Sqrt(n) ~ N(0,1)                  
P(X < 10.67) = (10.67-8.8)/5.6/ Sqrt ( 37 )
= 1.87/0.9206= 2.0312
= P ( Z <2.0312) From Standard NOrmal Table
= 0.9789                  
P(X > = 10.67) = 1 - P(X < 10.67)
= 1 - 0.9789 = 0.0211

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