questions 1,2,3. thank you so much. 8 Due 03/02/18 in Assignments (Click Assessm

Question

questions 1,2,3. thank you so much.

8 Due 03/02/18 in Assignments (Click Assessments on D2L) Math 1213 Project 1 Direction: Write your work on a blank paper, scan it and make sure it is readable. Show major steps and use correct notation to get full eredit Problem 1 (8 points) Use the grouped data formulas (my note in D2L Content) to obtain (a) the mean and (b) standard deviation for the following quiz grade data for a class of 50 students, then (c) give the Chebyshev's interval with k-2. Use the defining formula and show x Grade Frequency 12 26 Problem 2 (8 points) In a sales effectiveness seminar, a group of sales representatives tried two approaches to selling a customer a new automobile: the aggressive approach and the passive approach. For 1160 customers, the following record was kept as shown below: Sale No Sale Row Total Aggressive 289 291 580 Passive 494 86 580 Column Total 783 377 1160 Suppose a customer is selected at random from the 1160 participating customers. Let us use the following notation for events: A-aggressive approach, Pa-passive approach, S-sale, N= no sale. So, PIA) is the probability that an aggressive approach was used, and so on. (Enter your answers as fractions.) a) Compute P(S), P(S|A), and P(S | Pa) b) Compute P(N) and P(VIA). Are N and A independent? Explain using a formula c) Compute P(A or S). Problem 3 (4 points) The qualified applicant pool for five management trainee positions consists of eight women and four men a) How many different groups of applicants can be selected for the positions? b) How many different groups of trainees would consist entirely of women? c) Probability extension: If the applicants are equally qualified and the trainee positions are selected by drawing the names at random so that all groups of five are equally likely, what is the probability that the trainee class will consist entirely of women? (Round your answer to four decimal places.)

Explanation / Answer

1)mean=12.5 Standard deviation=9.433981

chebyshev's interval when k=2 is (mean-2(standard deviation),mean+2(standard deviation)) i.e (-6.38,31.38)

2)P(S)=783/1160 P(S/A)=289/580 P(S/pa)=494/580

P(N)=377/1160 P(N/A)=291/580  

P(NA)=P(N)*P(A) then N and A are independent.

P(NA)=291/1160

P(N)*P(A)=(377/1160)*(289*1160)

this is not equal. therefore, N & A are not independent.

P(A or S)=P(A)+P(S)-P(AS)=(580/1160)+(783/1160)-(289/1160)=(1074/1160)=0.9258

3) a)C12,5 =(12!/(8!*5!))

b)5 women can be selected from 8 =(8!/(5!*3!))

c)(8C5)/(12C5)=56/792=0.07

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