Business Statistics
Second Exam

Instructions: Answer the following four questions. Write legibly and show all work. Begin each numbered question on a fresh page. Unsupported answers will receive zero points. You must work independently.

Due: Tuesday 6/12, 11:59pm

1. Sears rates its salespersons according to their sales ability and their potential for advancement. They sampled 500 salespeople with following data:

(a) Calculate the probability that a randomly selected Sear's salesperson has above average sales ability and is an excellent potential for advancement?

(b) Calculate the probability that a randomly selected Sear's salesperson will have average sales ability an and good potential for advancement?

(c) Calculate the probability that a randomly selected Sear's salesperson will have below average sales ability and fair potential for advancement?

(d) Calculate the probability that a randomly selected Sear's salesperson will have an excellent potential for advancement given they also have above average sales ability?

(e) Calculate the probability that a randomly selected Sear's salesperson will have an excellent potential for advancement given they also have average sales ability?

2.A study by the Information Technology department at WPU revealed company employees receive an average of four e-mails per hour. Assume the arrival of these e-mails is approximated by the Poisson distribution.

(a) What is the probability , Prof. Smith, received exactly one e-mail between 4pm and 5pm yesterday?

(b) What is the probability he did not receive any e-mail during this period?

(c) What is the probability he received ten or more e-mails during the same period?
3. A recent study in NJ showed that 50% of all patients will return to the same dentist. Suppose nine patients are selected at random, what is the probability that:

(a) Exactly five of the patients will return?

(b) All nine will return?

(c) At least eight will return?

(d) At least one will return?

(e) How many patients would be expected to return to the same dentist, i.e., what is the mean of the distribution?

4. A recent study of long distance phone calls made from WPU, showed that the length of the calls follows the normal probability distribution with a mean of 3.2 minutes per call and a standard deviation of 0.50 minutes.

(a) What fraction of the calls last between 3.2 and 4 minutes?

(b) What fraction of the calls last more than 4 minutes?

(c) What fraction of the calls last between 4 and 4.5 minutes?

(d) What fraction of the calls last between 3 and 4.5 minutes?

Answer

Submitted by StatSolver on Tue, 2012-06-12 21:13

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Please see the attachment for solution. If you need any clarification please ask me. Thanks.

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Please xxx the xxxxxxxxxx for xxxxxxxxx xx you xxxx xxx xxxxxxxxxxxxx please xxx xxx Thanks.

file1.doc preview (832 words)

1. Sears xxxxx xxx xxxxxxxxxxxx according xx xxxxx xxxxx xxxxxxx xxx xxxxx xxxxxxxxx for xxxxxxxxxxxx xxxx sampled xxx salespeople with following xxxxxx

xxxxxxxxxxxxx

		Fair	xxxx	xxxxxxxxx	Total
xxxxx average	16	xx	xx	xx
stability xxxxxxx	xx	xx	45	150
below xxxxxxx	93	xx	135	300
xxxxx		154	144	xxx	500

xxx xxxxxxxxx xxx xxxxxxxxxxx that a randomly selected Sear's salesperson xxx xxxxx xxxxxxx xxxxx xxxxxxx xxx is xx excellent xxxxxxxxx for xxxxxxxxxxxx

P(Above xxxxxxx and Excellent) = 22/500 = xxxxx

(b) xxxxxxxxx the probability xxxx x randomly selected xxxxxx salesperson will xxxx average sales xxxxxxx an xxx xxxx xxxxxxxxx for advancement?

P(Stability xxxxxxx xxx Good) = 60/500 = 0.12

(c) Calculate the probability xxxx x randomly selected xxxxxx xxxxxxxxxxx will have xxxxx xxxxxxx xxxxx xxxxxxx xxx xxxx xxxxxxxxx for advancement?

xxxxxxxx xxxxxxx xxx xxxxx = xxxxxx x xxxxx

(d) Calculate xxx xxxxxxxxxxx that a randomly xxxxxxxx xxxxxx salesperson xxxx

- - - more text follows - - -

file2.xls preview (69 words)

Sheet1

xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx

Poisson distribution	Binomial distribution	xxxxxx xxxxxxxxxxxx
x	xxxxxx	x	xxxxxx	x	x	P(Z < xx
0	xxxxxx	x	xxxxxx	xxxx	0	xxxxxxxxxxxx
x	xxxxxx	1	0.0176	x	xxx	0.9452
x	xxxxxx	x	xxxxxx	4.5	2.6	xxxxxx
3	xxxxxx	3	xxxxxx	3	-0.4	xxxxxx
4	xxxxxx	4	xxxxxx
x	xxxxxx	5	0.2461
6	0.1042	6	xxxxxx
7	xxxxxx	x	0.0703
x	xxxxxx	x	0.0176
9	0.0132	x	xxxxxx
xxxx	Good	xxxxxxxxx	xxxxx
Above average	16	12	xx	50
xxxxxxxxx average	xx	xx	xx	150
xxxxx average	93	72	135	300
xxxxx	154	144	xxx	500

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Answer

Submitted by Kumail Raza on Wed, 2012-06-13 09:13

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Answer file with formula and step by step calculation with printable format is attached. Feel free to contact for any further assistance.

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xx

xxxxxxxxxxxxxxxxxxxxxxxxxxxxx

xx	xxxxx	xxxxxxxxx xxx advancement
xxxxx xxxxxxx	Fair	xxxx	xxxxxxxxx	xxxxx
F	G	x	I
xxxxx xxxxxxx	5	xx	xx	xx	xx
average	6	xx	60	xx	xxx
xxxxx average	x	xx	xx	xxx	xxx
xxxxx	x	xxx	xxx	202	500
xxxxxxxxx
xxx	xxxxxxxxx the probability that x xxxxxxxx selected Sear’s salesperson has above average xxxxx xxxxxxx and xx xx xxxxxxxxx xxxxxxxxx xxx xxxxxxxxxxxx
xxxxxx	x	0.27
(b)	Calculate xxx probability that a xxxxxxxx xxxxxxxx xxxx’s salesperson xxxx have average xxxxx ability an and xxxx xxxxxxxxx for advancement?
=G6/I8	=	0.12
(c)	xxxxxxxxx xxx xxxxxxxxxxx xxxx a randomly xxxxxxxx xxxx’s xxxxxxxxxxx xxxx have xxxxx average xxxxx xxxxxxx xxx fair potential xxx advancement?
=F5/I8	x	xxxx
(d)	Calculate xxx xxxxxxxxxxx that x xxxxxxxx xxxxxxxx Sear’x - - - more text follows - - -

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Answer

Submitted by shahimermaid on Wed, 2012-06-13 05:13

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xxxxx xxxxx xxx salespersons xxxxxxxxx xx xxxxx sales ability xxx xxxxx xxxxxxxxx xxx advancement. They sampled 500 xxxxxxxxxxx with xxxxxxxxx data:

xxx Calculate xxx probability xxxx x randomly selected Sear's xxxxxxxxxxx xxx xxxxx xxxxxxx xxxxx ability and is an excellent potential for advancement? xxxxxxxx =0.27

(b) Calculate xxx probability that a randomly selected xxxxxx xxxxxxxxxxx will have average xxxxx xxxxxxx and good potential for xxxxxxxxxxxx =60/500=0.12

(c) Calculate xxx xxxxxxxxxxx that x xxxxxxxx xxxxxxxx xxxxxx xxxxxxxxxxx xxxx xxxx xxxxx average sales xxxxxxx and fair xxxxxxxxx for advancement? =16/500=0.032

xxx xxxxxxxxx xxx probability that x randomly selected Sear's salesperson will xxxx an xxxxxxxxx xxxxxxxxx xxx xxxxxxxxxxx xxxxx xxxx xxxx have above xxxxxxx sales ability? =135/(93+72+135)=0.45

xxx xxxxxxxxx xxx xxxxxxxxxxx xxxx a xxxxxxxx selected Sear's salesperson will xxxx xx xxxxxxxxx potential xxx xxxxxxxxxxx given xxxx also xxxx xxxxxxx xxxxx xxxxxxxx =

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