SOLUTION: Please help with explanation:
The time spent on customer service calls in normally distributed with a mean of 4.5 minutes.
1) You are given that the probability that a call
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-> SOLUTION: Please help with explanation:
The time spent on customer service calls in normally distributed with a mean of 4.5 minutes.
1) You are given that the probability that a call
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Question 773759: Please help with explanation:
The time spent on customer service calls in normally distributed with a mean of 4.5 minutes.
1) You are given that the probability that a call will take less that 5 minutes is equal to 60%. What is the probability that a call will take between 4.5 and 5 minutes.
2) You are given that the a call will take more than 3 minutes is equal to 80%. What is the probability what a call will take more than 6 minutes.
Struggling with this.
Thank you. Answer by rothauserc(4718) (Show Source):
You can put this solution on YOUR website! 1) the probability that a call will take less that 5 minutes is equal to 60%. What is the probability that a call will take between 4.5 and 5 minutes
Pr(4.54.5)
by definition of the mean Pr(X>4.5) is 50%
Pr(4.5
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2) You are given that the a call will take more than 3 minutes is equal to 80%. What is the probability what a call will take more than 6 minutes.
Pr(X>3) = 1 - Pr(X<3) = 80% and Pr(X<3) = 20%
need to calculate std dev
z-value = (3 - 4.5) / std. dev
consult Z table for 20% probability = -.84
-.84 = -1.5 / std. dev
std. dev = 1.5/.84 = 1.78
now
Pr(X>6) = 1 - Pr(X<6)
z-value = (6 -4.5) / 1.78 = .84
Pr(X<6) = 80%
Pr(X>6) = 1 - Pr(X<6)= 1 -.80 = 20%
