MGMT2262 Midterm practice solutions

 

Question 1

a)      The stem and leaf plot is:

2  | 0 5 5

3  | 2 8

4  | 7 9

10| 1 9

12 | 0 5 7

13 | 3 3

14 | 4

16 | 2

b)     There are 2 modes: 25 and 133.

c)      The lowest value is 20, the highest 162. The range is 162 – 20 = 142.

d)     The median is the average of the 8th and 9th values which are 101 and 109 to get 105. The mean is 86.875 and the standard deviation is 50.7055 which you should get from your calculator. The coefficient of skewness is 3(86.875 – 105)/50.7055 = -1.07. The interpretation of the coefficient is that the data is moderately skewed left.

e)      The coefficient of variation is 50.7055/86.875 = 58.37%. Since the other data set has a lower coefficient of variation, it means that it has less variation relative to its mean. This, in turn, means that the second data set is more consistent.

f)       Q1 = average of 4th and 5th values = (32 + 38)/2 = 35

Q3 = average of 12th and 13th values = (127 + 133)/2 = 130

IQR = 95

Lower inner fence = 35 – 1.5(95) = -107.5

Upper inner fence = 130 + 1.5(95) = 272.5

Since all the values are inside the fences, there are no outliers.

 

Question 2

First we create a crosstab:

 

Under 30

30+

Total

CSI

12

39

51

not CSI

36

13

49

Total

48

52

100

a)      P(30+ or not CSI) = P(30+) + P(not CSI) – P(30+ and not CSI) = 52% + 49% - 13% = 88%

b)     P(under 30 and not CSI) = 36% from the crosstab

c)      P(under 30 and not CSI) + (30+ and CSI) = 36% + 39% = 75%

d)     P(under 30 and CSI) = 12%; P(under 30)*P(CSI) = (48%)(51%) = 24.48%; since the two sides are not equal, the events are not independent. Thus, watching CSI depends on the person’s age.

 

Question 3

a)      P(medium and not in Calgary) = (46 + 82)/500 = 128/500 = 25.6%

b)     P(low/high or not in RD) = P(low/high) + P(not in RD) – P(low/high and not in RD)

= (118 + 176)/500 + (225 + 200)/500 – (55 + 44 + 92 + 74)/500

= 294/500 + 425/500 – 265/500 = 454/500 = 90.8%

c)      P(not in RD | medium/high) = (78 + 92 + 82 + 74)/(206 + 176) = 326/382 = 85.34%

d)     P(low and in RD) = 19/500 = 3.8%; P(low)*P(RD) = (118/500)(75/500) = 3.54%; since the two sides are not equal, being low income depends on living in Red Deer.

 

Question 4

a)      (10C3)(5C3)(4C3)(8C3) = 120 x 10 x 4 x 56 = 268,800

b)     (30C5)(25C8)(17C12) = 142,506 x 1,081,575 x 6,188 = 9.5 x 1014. Note that with each successive term, the number of people to choose from decreases since no person can serve on more than 1 committee.

c)      One solution is 25 x 10 x 25 x 9 x 24 x 8 x 23 x 7 = 1,808,352,000. The other solution is 26P4 x 10P4 = 358,800 x 5,040 = 1,808,352,000

 

Question 5

a)      P(fewer than 7) = P(no more than 6). Using Table A.2, we see the cumulative percentage for n=12, p=0.4 and x=6 is 0.842.

b)     P(more than 8) = P(at least 9) = 100% - P(no more than 8). Using Table A.2, we see that the cumulative percentage for n=17, p=0.4 and x=8 is 0.801 = 80.1%. Then P(more than 8) = 100% - 80.1% = 19.9%

c)      Using Table A.1, we find the percentage for n=9, p=0.4 and x=4 is 0.251 = 25.1%

d)     m = 1200(0.4) = 480; s = sqrt(480*0.6) = 16.97. The coefficient of variation is 16.97/480 = 3.54%. Thus, this town has a lower coefficient of variation.

e)      The lower limit is m-3s = 480 – 3(16.97) = 480 – 50.91 = 429 after rounding to the nearest whole number. The upper limit is 480 + 50.91 = 531 after rounding to the nearest whole number.

 

Question 6

a)      Z = (15 – 16.7)/(3.9/sqrt(40)) = -2.76

P(xbar > 15) = P(Z > -2.76) = 0.5 + 0.4971 = 0.9971 = 99.71%

b)     We want P(xbar > 18) = 0.0606. The first thing is to find the appropriate Z value which we will call p. Then P(Z > p) = 0.0606. That means P(0 < Z < p) = 0.4394. Working backwards in the Z table, we find p = 1.55. The next step is to solve the problem: 1.55 = (18 – 16.7)/( s/sqrt(40)). Then s = (18 – 16.7)/(1.55/sqrt(40)) = 5.3045.

c)      Let k represent the number of standard deviations a value is from the mean. We have 2 equations:

11.24 = 16.7 – 3.9k (equation 1)

22.16 = 16.7 + 3.9k (equation 2)

Using either equation, we get 3.9k = 5.46. Then k = 1.4. So we know that 11.24 and 22.16 are both 1.4 standard deviations from the mean. Then, the minimum percentage between these values is 1 – 1/(1.4)2 = 48.98%

 

Question 7

a)      m = 50,000/10,000 = 5. P(more than 1) = P(at least 2) = 100% - P(no more than 1).

P(no more than 1) = P(0) + P(1) = 0.0067 + 0.0337 = 0.0404 = 4.04%

P(more than 1) = 100% - 4.04% = 95.96%

It should be noted that P(0) and P(1) are obtained from Table A.3.

b)     m = 75,000/10,000 = 7.5. P(no more than 4) = P(0) + P(1) + P(2) + P(3) + P(4) = 0.0006 + 0.0041 + 0.0156 + 0.0389 + 0.0729 = 0.1321 = 13.21%. Once again, these values are obtained from Table A.3.

c)      We already have m = 7.5. Then s = sqrt(7.5) = 2.7; Then m-3s = -0.8 which we round to 0 (since this is the lowest value in the Poisson distribution) and m+3s = 15.7 which we round to 16.

 

Question 8

a)      Z = (550 – 532)/89 = 0.2. Then P(X > 550) = P(Z > 0.2) = 0.5 – 0.0793 = 0.4207 = 42.07%

b)     Z = (700 – 532)/89 = 1.89. Then P(X < 700) = P(Z < 1.89) = 0.5 + 0.4706 = 0.9706 = 97.06%

c)      Z1 = (600 – 532)/89 = 0.76; Z2 = (650 – 532)/89 = 1.33; Then P(600 < X < 650) = P(0.76 < Z < 1.33) = 0.4082 – 0.2704 = 0.1318 = 13.18%

d)     Let the percentile be p. Then P(Z > p) = 0.0708. Then P(0 < Z < p) = 0.4292. Then p = 1.47 = (x-532)/89. Solving for x, x = 532 + 1.47(89) = 662.83.

e)      Let the Z value be p. Then P(-p < Z < p) = 0.4778. This means P(0 < Z < p) = 0.2389. Working backwards, we find p = 0.64. So, the upper Z value is 0.64 while the lower Z value is –0.64. The first equation to solve is 0.64 = (x – 532)/89 from which we get x = 588.96. The second equation to solve is –0.64 = (x – 532)/89 from which we get x = 475.04. So, 47.78% of the contributions are between $475.04 and $588.96.