Aug 14, 2014

(1)

The Council of Community Colleges of Jamaica Page 1

THE COUNCIL OF COMMUNITY COLLEGES OF JAMAICA

ASSOCIATE OF SCIENCE EXAMINATION

SUMMER/SUPPLEMENTAL – 2014 AUGUST

PROGRAMMES: SOCIAL WORK

COURSE NAME: STATISTICS FOR SOCIAL SCIENCES

CODE : (MATH1108)

YEAR GROUP: ONE

DATE: WEDNESDAY, 2014 AUGUST 13

TIME: 9:00 A.M. – 12:00 NOON

DURATION: 3 HOURS

EXAMINATION TYPE: FINAL

This Examination paper has 6 pages

INSTRUCTIONS:

SECTION A: ANSWER ALL QUESTIONS IN THIS SECTION.

SECTION B: ANSWER ANY THREE (3) QUESTIONS FROM THIS SECTION.

(2)

The Council of Community Colleges of Jamaica Page 2 SECTION A

Instructions: On the computerized answer sheet provided, shade the letter that corresponds with the most appropriate response for each of the following.

1. The midpoint of a class is obtained by:

A. adding the lower limit to the upper limit of the previous class B. subtracting the lower limit from the upper limit

C. dividing the sum of the two class limits by 2 D. multiplying the frequency by 0.1

2. The prices of five books are $129, $183, $151, $102and $99 .The median price in this case is :

A. $129 B. $102 C. $99 D. $151

3. The data that occurs the most in a given set of values is known as the:

A. median B. mode C. mean

D. standard deviation

4. The standard deviation of a set of numbers is 0. Which of the following must be true?

i. all the numbers are the same ii. the mean is equal to 0

iii. all the numbers are equal to 0

A. i only B. ii only C. iii only D. i and ii only

(3)

The Council of Community Colleges of Jamaica Page 3 5. If the mean is greater than the median for a given distribution. The distribution is said to

be.

A. any kind of distribution

B. a skewed to the right distribution C. a skewed to the left distribution D. a normal distribution

The data in Table below refers to Questions 6 - 9 .The following shows results of a survey of time taken by a group of children to complete a reading assignment.

6. What is the upper class boundary of the class ‘40-49”?

A. 49 B. 40 C. 49.5 D. 40.5

7. What is the modal class?

A. 30-39 B. 50 -59 C. 20-29 D. 40 -49

8. What is the lower class limit of the class “50 -59” ?

A. 50 B. 59.5 C. 49.5 D. 55

Time( Min) # of students

10-19 2

20-29 6

30-39 12

40-49 16

50-59 8

60-69 4

(4)

The Council of Community Colleges of Jamaica Page 4 9. How many students took part in the survey?

A. 49 B. 40 C. 50 D. 55

The following information refers to questions 10 – 13.

A ransom variable X has the probability distribution shown in the table below:

X 0 `1 2 4

P(X)) 0.2 d 0.3 0.2

10. The value of d is:

A. 0.1 B. 0.4 C. 3 D. 0.3

11. The expected value of X is:

A. 1.67 B. 5 C. 1.7 D. 1.6

12. The variance of X is:

2.14 4.6 1.2 1.81

13. The standard deviation of X is:

(5)

The Council of Community Colleges of Jamaica Page 5 14. Which of the following measures of centre is most affected by outliers

A. Mode B. Median

C. Standard deviation D. Mean

15. Which of the following statements is TURE?

A. Quantitative variable is said to have no numerical significant. B. Quantitative variable is said to have numerical significant. C. Qualitative variable is said to have numerical significant. D. Qualitative variable can be counted or measured

16. Which of the follow is a continuous variable?

A. Number of students in the class B. The time taken to complete a test

C. The number of books a store sold in a day D. The number of tables in the class

17. If the probability that it will rain today is 0.4. What is the probability that it will not rain today?

A. 0.4 B. 0.8 C. 0.6 D. -0.6

Question 18 to 21 relates to the following information:

For a given binomial distribution, n=8 and p=0.6.

18. Determine the value of q.

A. 0.6 B. 7 C. 0.4 D. 2

19. Find the expected value E(X) of the distribution.

(6)

The Council of Community Colleges of Jamaica Page 6 20. Find VAR(X).

A. 1.82 B. 1.9 C. 4 D. 1.92

21. Find the standard deviation.

A. 1.39 B. 3.69 C. 3.54 D. 2

22. The data value that is in the middle when the data set is arranged in order of size is known as the:

A. range B. mode C. median D. mean

23. The mean of the following numbers 12, 23, 8, 16, 34 is:

A. 18.6 B. 93 C. 21.4 D. 19.6

24. If an increase in x results in a decrease in y then the relationship between x and y is?

A. a positive relationship B. a negative relationship C. skewed to the left D. skewed to the right

25. The coefficient of correlation is 0.94 then the coefficient of determination is:

A. 0.94 B. 0.88 C. 0.06

D. 0.49 (Total 25 marks)

(7)

THE COUNCIL OF COMMUNITY COLLEGES OF JAMAICA

ASSOCIATE OF SCIENCE EXAMINATION

SUMMER/SUPPLEMENTAL – 2014 AUGUST

CODE : (MATH1108)

TIME: 9:00 A.M. – 12:00 NOON

This Examination paper has 4 pages

INSTRUCTIONS:

(8)

The Council of Community Colleges of Jamaica Page 8 SECTION B

Instructions: Answer any THREE (3) questions from this section.

Question 1

The following table gives the frequency distribution of the number of orders received each day at the office of a mail-ordering company.

a. Calculate the mean (4 marks) b. Calculate the Mode (4 marks) c. Calculate the median (5 marks) d. Calculate the variance and standard deviation (7 marks)

e. Calculate the skewness and interpret the results (5 marks)

(Total 25 marks) Question 2

a. Let X be the random variable with the probability distribution given by:

i. Find the value of C (2 marks) ii. Find E(X) (3 marks) iii. Calculate the variance and the standard deviation (8 marks) iv. Determine P(2≤X≤5) (3 marks)

b. Two Events A and B are such that P(A) = 0.65 and P(B1) = 0.7. If A and B are Independent events , find:

i. P(A∩B) (5 marks)

ii. P(AUB) (4 marks)

(Total 25 marks) Number of Orders Frequency (days)

10-19 7

20-29 13

30-39 18

40-49 24

50-59 14

60-69 4

X 1 2 3 4 5

(9)

The Council of Community Colleges of Jamaica Page 9 Question3

The data below shows the following information:

X 11 13 14 16 18 19 21 24

Y 11 12 14 14 15 17 18 19

a. Calculate and interpret r and r2 _{(12 marks)}

b. Calculate the value of the slope of the regression line. (4 marks)

c. Calculate the value of y intercept of the regression line. (5 marks)

d. Using the regression equation to predict the value of x when y =16 (4 marks)

(Total 25 marks)

Question 4

a. The probability that a given grain of corn planted in the soil will germinate is 0.3. Determine the probability that out of ten grains of corn:

i. none will germinate (3 marks)

ii. more than two (2) will germinate (4 marks)

iii. At least four (4) will germinate (5 marks)

iv. The expected number of grains of corn that will germinate (2 marks)

b. The average number of accidents on a stretch of road per week is three (3). What is the probability of :

i. Exactly six (6) accidents per week (2 marks)

ii. At least two (2) accidents per fortnight (5 marks)

iii. Exactly three (3) accidents per month (4 marks)

(10)

The Council of Community Colleges of Jamaica Page 10 Question 5

The following shows results of a survey of time taken by a group of children to complete a reading assignment.

a. Construct a less than cumulative frequency table (3 marks)

b. Using a scale of 1cm: 5 units on both axis draw a less that CF curve (10 marks)

c. Using our graph to determine the median (3 marks)

d. Determine form the graph the quartile deviation (5 marks)

e.

Calculate the probability that a given student will complete a reading assignment in at least 40 minutes. (4 marks)

(Total 25 marks)

END OF EXAMINATION Time( Min) # of students

10-19 2

20-29 6

30-39 12

40-49 16

50-59 8

60-69 4

(11)

THE COUNCIL OF COMMUNITY COLLEGES OF JAMAICA

ASSOCIATE OF SCIENCE EXAMINATION

SEMESTER II – 2013 MAY

CODE : (MATH1108)

TIME: 9:00 A.M. – 12:00 NOON

SOLUTIONS SECTION A

1 C

11 C

21 A

31

41

2 B

12 D

22 C

32

42

3 B

13 A

23 A

33

43

4 A

14 D

24 B

34

44

5 B

15 B

25 B

35

45

6 C

16 B

26

36

46

7 D

17 C

27

37

47

8 A

18 C

28

38

48

9 C

19 B

29

39

49

(12)

The Council of Community Colleges of Jamaica Page 12 SECTION B

Question 1

Mean = x = ∑fx = 148 = 39.35 (2marks) ∑f 80

Mode = L1 + D1_ C

D1 + D2

= 39.5 +( 6/6+10 ) 10 =39.5 + (6/10) 10

= 39.5 + 3.75 = 43.25

MEDIAN= D1 + (N/2 -∑F1/fm) C

= 39.5 + (80/2 -38/24) 10 = 39.5 + (40-38/24) 10 =39.5 + 0.83

= 40.33

Variance= ∑fx2_/_∑_{f – x}2 _{= 136450/ 80 – (39.35)}2

= 1705.625 – 1548.42 = 157.2

Standard deviation= √∑fx2/∑f– x2 =√157.2 = 12.54

Skewness = 3(mean – median/standard deviation) = 3(39.35- 40.33)/12.54

=3(-0.98)/12.54 = - 2.94/12.54 =-0.234

The deviation has a slight degree of negative skewness. X

Midpoint

F

Frequency

Fx X2 Fx2

14.5 7 101.5 210.25 1471.75

24.5 13 318.5 600.25 7803.25

34.5 18 639 1190.25 21424.5

44.5 24 1068 1980.25 47,526

54.5 14 763 2970.25 41,583.5

64.5 4 258 4160.25 16,641

(13)

∑P(X=x)=1

1=0.1+0.2+C+0.15+0.25 1=0.7+C

1-0.7=C 0.3=C

E(X)=∑PiXi=(1*0.1)+(2*0.2)+3(0.3)+(4*0.15)+(5*0.25) =0.1+0.4+0.9+0.6+1.25

=3.25

X P(X=x) PiXi X2 _PiX2_i

1 0.1 0.1 1 0.1

2 0.2 0.4 4 0.8

3 0.3 0.9 9 2.7

4 0.15 0.6 16 2.4

5 0.25 1.25 25 6.25

∑PiXi=3.25 ∑Pix2i=12.25

ﬞ◌2_=vanance=_∑_PiX2_-(E(x))2

12.25-3.252

12.25-10.5625 =1.6875 =1.69

ﬞ◌= √ﬞ◌2 =√VAR(X)= √∑PiX2i-(E(x))2= 1.69 =1.3

P(25X ≤ 5)=P(X=2)+P(X=3)+P(X=4)+P(X=5) =0.2+0.3+0.15+0.25

=0.9

B. P(A∩ B)=P(A)* P(B)

P(B)=1-P(B1₎

=1-0.7 =0.3

P(A∩B) = P(A)*P(B) =0.6 *0.3

(14)

The Council of Community Colleges of Jamaica Page 14 P(AU B)=P(A)+P(B)-P(A∩ B)

=0.6+0.3-0.18 =0.72

Question 3

X Y XY X2 Y2

11 11 121 121 121

13 12 156 169 144

14 14 196 196 196

16 14 224 256 196

18 15 270 324 225

19 17 323 361 289

21 18 378 441 324

24 19 456 576 361

∑X= 136 ∑Y=120 ∑XY= 2124 ∑X2_{= 2444} _∑_Y2_{= 1856}

= 8(2124) – (136) (120)

√8(2444) – (136)2_√_{8(1856) – (120)}2

= 16992 - 16320

√19552- 18496 √14848- 14400

= 672

√1056 √448

= 672 32.5 x 21.17

= 672/687.9 = 0.977

r2 _{= (0.977)}2

(15)

The Council of Community Colleges of Jamaica Page 15 b= n (∑XY) – (∑X) (∑Y)

n(∑X2) - (∑X)2

= 8(2124) – (136) (120) 8(2444) – (136)2

= 16992 - 16320 19552- 18496

= 672 1056

= 0.64

A = ∑Y - b ∑X n n

= 120/8 - 0.64(136)/8

= 15 - 0.64(17)

= 15 – 10.88

= 4.12

Y = A + bX

Y = 4.12 + 0.64X

16 = 4.12 + 0.6X

16 – 4.12 = 0.64x 11.88 = .64x

(16)

P(X= ×) = ACX PX qn-x

P=0.3 N=10 q= 1-p

= 1-0.3 q= 0.7

i. P(X=0) = 10_C

0 (0.3)0(0.7)10

=1*1*0.02825 = 0.02825

ii. P(X>2) = 1- [ P(X=0)+ P(X=1)+P(X=2)]

=1-(0.02825 + 10C1 (0.3)1 (0.7)9 +10C2 (0.3)3 (0.7)8]

=1-(0.02825+0.12106+0.2335 =1- 0.382

=0.6718

iii. P (X≥4) = 1- [P(x=0) + P(x=1) + P(x=2) + P(x=3)] =1-[0.02825 + 0.12106 + 0.2335+ 10_C

3 (0.3)3 (0.3)7]

=1- [0.382 + 0.2668] =1- 0.6488

=0.3512

Iv. E(X) =np = 10 x 0.3 = 3 (B.)

I P(X=x) = e-λλx X!

P(X=x) = e-3₃6_{= 0.05}

6!

ii) λ= 3x2 =6

P(X≥2) = 1 - { P(X=0) +P(X=1)} = 1 – { e-6 60 + e-6 61 }

o! 1!

= 1- ( 0.002478 + 0.01487) = 1 – 0.0135

= 0.983

ii) λ= 3x4 =12

P(X=x) = + e-12₁₂3_}

(17)

(i) less than cf table

Time Cf

<9.5 0

<19.5 2

<29.5 8

<39.5 20

<49.5 36

<59.5 44

<69.5 48

<79.5 50

(iii) Median position= n+1 = 50+1 = 20.5th

2 2

From graph median = 43mins

(iv)Quartile deviation = Q3 - Q1

Q3 Position=3(n+1)= (50+1)=38.25th

4 4

From graph Q3= 51.5 mins

Q1 position=n+1 = 50+1=12.75th

4 4 From graph Q1=34

Quartile deviation=Q3-Q1=51.5-34

(18)