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