The help desk for a software company logs the lengths of times of each call that

Question

The help desk for a software company logs the lengths of times of each call that technicians have with customers. Historical records based on the past 3 years indicate that call length (L) is a normally distributed variable with an average () of 7.25 minutes and a standard deviation () of 1.15 min. For each of the problems below, find the appropriate probability.
Note: For problems A, B, and C round your answers to four decimal places.

A) P(L 7.73)   

B) P(L > 5.92)   

C) P(6.48 L < 9.75)   

D) The call times that enclose the central 88% of the distribution are:
    Note: Round your answers to two decimal places for this problem.
    Low:  to High:

Explanation / Answer

=Mean is 7.25 and s is 1.15. z is given as (x-mean)/s

a) P(x<7.73) = P(z<(7.73-7.25)/1.15)=P(z<0.42), from normal distribution table we get 0.6628

b) P(x>5.9) =P(z>(5.92-7.25)/1.15)=P(z>-1.1.6) or P(z<1.16) , from normal table it is 0.8770

c) P(6.48<x<9.75) =P((6.48-7.25)/1.15<z<(9.75-7.25)/1.15)=P(-0.67<z<2.17) this is P(z<2.17)-P(z<-0.67) or 0.9850-(1-0.7486)=0.7336

d) P(-X<x<X)=0.88

for 0.88, the z value for top and bottom can be calculated as P = .88 a = .12 & a/2 = .06 z = ± 1.5548

Thus lower value is 7.25-1.5548*1.15 =5.46198

and upper value is 7.25+1.5548*1.15=9.03802

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