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A study of long distance phone calls made from the corporate offices ...

By Sarah Gibson,2014-07-15 19:52
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A study of long distance phone calls made from the corporate offices ...

    A study of long distance phone calls made from the corporate offices of the Pepsi Bottling Group, Inc., in Somers, New York, showed the calls follow the normal distribution. The mean length of time per call was 4.2 minutes and the standard deviation

    was 0.60 minutes.

    a. What fraction of the calls last between 4.2 and 5 minutes?

     Mean=μ= 4.2 minutes

    Standard deviation =σ= 0.6 minutes

    x1= 4.2 minutes

    x2= 5.0 minutes

    z1=(x1-μ)/σ= 0 =(4.2-4.2)/0.6

    z2=(x2-μ)/σ= 1.3333 =(5-4.2)/0.6

    Cumulative Probability corresponding to z1= 0 is= 0.5 0r= 50.00% Cumulative Probability corresponding to z2= 1.3333 is= 0.9088 0r= 90.88%

    Therefore probability that the value of x will be between x1= 4.2 and x2= 5.0 is = 40.88% =90.88%-50.%

    Fraction of the call between 4.2 and 5.0 minutes= 0.4088 or 40.88%

b. What fraction of the calls last more than 5 minutes?

Mean=μ= 4.2 minutes

    Standard deviation =σ= 0.6 minutes

    x= 5

    z=(x-μ)/σ= 1.3333 =(5-4.2)/0.6

    Cumulative Probability corresponding to z= 1.3333 is= 0.9088 Therefore probability corresponding to x> 5.00 is 1-Prob(Z)= 0.0912 =1-0.9088 0r= 9.12%

    Fraction of the calls greater than 5.0 minutes= 0.0912 or 9.12%

    c. What fraction of the calls last between 5 and 6 minutes?

Mean=μ= 4.2 minutes

    Standard deviation =σ= 0.6 minutes

    x1= 5.0 minutes

    x2= 6.0 minutes

    z1=(x1-μ)/σ= 1.3333 =(5-4.2)/0.6

    z2=(x2-μ)/σ= 3 =(6-4.2)/0.6

    Cumulative Probability corresponding to z1= 1.3333 is= 0.9088 0r= 90.88% Cumulative Probability corresponding to z2= 3 is= 0.9987 0r= 99.87%

    Therefore probability that the value of x will be between x1= 5.0 and x2= 6.0 is = 8.99% =99.87%-90.88%

    Fraction of the call between 5.0 and 6.0 minutes= 0.0899 or 8.99%

d. What fraction of the calls last between 4 and 6 minutes?

Mean=μ= 4.2 minutes

    Standard deviation =σ= 0.6 minutes

    x1= 4.0 minutes

    x2= 6.0 minutes

    z1=(x1-μ)/σ= -0.3333 =(4-4.2)/0.6

    z2=(x2-μ)/σ= 3 =(6-4.2)/0.6

    Cumulative Probability corresponding to z1= -0.3333 is= 0.3695 0r= 36.95% Cumulative Probability corresponding to z2= 3 is= 0.9987 0r= 99.87%

    Therefore probability that the value of x will be between x1= 4.0 and x2= 6.0 is = 62.92% =99.87%-36.95%

    Fraction of the call between 4.0 and 6.0 minutes= 0.6292 or 62.92%

    e. As part of her report to the president, the Director of Communications would like to report the length of the longest (in duration) 4 percent of the calls. What is this time?

Z value corresponding to 4% is 1.7507

Mean=μ= 4.2 minutes

    Standard deviation =σ= 0.6 minutes

    z=(x-μ)/σ= 1.7507

    x=μ+zσ 5.25 minutes =4.2+(1.7507*0.6)

    Length of the longest (in duration) 4% of calls is greater than 5.25 minutes

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