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 6.50 minutes and a standard deviation () of 0.95 min. For each of the problems below, find the appropriate probability.

My Notes Ask Your Teacher 3. 05 points I Previous Answers 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 ca length (L) is a normally distributed variable with an average (Au) of 6.50 minutes and a standard deviation (a) of 0.95 min. For each of the problems below, find the appropriate probability. Note: For problems Ar B, and C round your answers to four decimal places. A) PCL s 7.63) 0.3409 x B) PCL 6.38 C) P(4.47 S L 7.61) D) The call times that enclose the central 80% of the distribution are: Note: Round your answers to two decimal places for this problem. to High: Low 4. -1 points My Notes Ask Your Teacher Counselors at an exclusive private college look carefully at the applications from high school students seeking admission to the college. One of their criteria is that these students must score in the top 2.25% of all students who took the required entrance exam. The exam is constructed such that the scores are normally distributed with an average of 1200 and a standard deviation of 95 The minimum exam score necessary for applicants to be further considered for admission is: Hint: I know you guys hate to do this because you think it's a waste of your time but y'a should really draw an appropriate diagram Note: Round all answers to the nearest integer.

Explanation / Answer

mean = 6.50 . s = 0.95

3)

A)

P(L< 7.63)

z = ( L - mean) / ( s /sqrt(n))

= ( 7.63 - 6.50) / (0.95)

= 0.3767

now, we need to find P(Z <0.3767)

P(L < 7.63) = P( z <2.06) = 0.8829

b)

P(L >6.38)

z = ( L - mean) / ( s )

= ( 6.38 - 6.50) / (0.95)

= -0.04

now, we need to find P(Z > -0.04)

P(L > 6.38) = P( z >-0.04) = 0. 5503

c) P(4.47 < L < 7.61)

P(L <4.47)

z = ( L - mean) / ( s)

= ( 4.47 - 6.50) / (0.95)

= -0.676

P(L <7.61)

z = ( L - mean) / ( s )

= ( 7.61 - 6.50) / (0.95)

= 0.37

P(4.41 < L < 7.61) = P(-0.676 <z < 0.37) = 0.8648

4) z value at 2.25% = 2.004

mean = 1200, s =95

x bar = mean + z *s

= 1200 + 2.004 * 95

= 1390.38 = 1340

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