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ACCOUNTING – XII

2016 – Regular & PRIVATE Solved Paper

Compiled & Solved by: Sameer Hussain

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Compiled & Solved by: Sameer Hussain www.a4accounting.weebly.com

www.facebook.com/a4accounting.net www.twitter.com/a4accounting2

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Statistics – XII – 2016 Regular & Private Solution 1

Time: 20 Minutes Max. Marks: 20 SECTION “A” (MULTIPLE CHOICE QUESTIONS)

Note: (i) This section contains of 10 part questions and all are to be answered. Each question carries equal marks. (ii) Do not copy down the part question in your answer book. Write only the answer in full against the proper number of the question and its part. (iv) The code of your question paper must be mentioned in bold letters in the beginning. Q.No.1 Choose the correct answer for each from the given options:

(i) Data obtained from newspaper is called: * Primary * Secondary * Continuous * None of these

(ii) If �̅� = 50 and y = 3x – 10, then mean of y is:

* 140 * 0 * 100 * 200

(iii) If a card is drawn from a pack of playing cards, then the probability of getting a face card is: * 4/52 * 13/52 * 12/52 * 1/52

(iv) The blood group of the students of a class is:

* Quantitative variable * Qualitative variable * Continuous variable * Discrete variable

(v) The number of days in the month of December is: * Variable * Constant * Sample * parameter

(vi) If, in a moderately skewed distribution, mean = 35, median = 30, then mode is:

* 10 * 20 * 5 * 65

(vii) The mode of the data 1, 7, 2, 3, 4, 5, 5, 7, 4, 4 is: * 4 * 5 * 7 * No mode

(viii) If Laspeyre’s index number = 132.5%, Paasche’s index number = 135.75%, then

Fisher’s index number is: * 154.11% * 144.11% * 134.11% * 164.11%

(ix) For a certain distribution, if ∑(𝒙 − 𝟏𝟓) = 𝟓 𝐚𝐧𝐝 ∑(𝒙 − 𝟏𝟖) = 𝟎, then mean is:

* 0 * 15 * 18 * 5

(x) If two fair coins are tossed, then the probability of one head is: * 0.25 * 0.50 * 0.75 * 1



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Time: 1 Hours 45 Minutes Max. Marks: 40 SECTION “B” (SHORT – ANSWER QUESTIONS)

Note: Attempt any FIVE questions. All questions carry equal marks. The use of calculator is allowed.

Q:2 (i) Describe the sources of the collection of Primary Data. SOLUTION 2 (i) Sources of Collection of Primary Data: 1. Direct Personal observation: Here the investigator directly contacts the informants, solicits their cooperation and enumerates the data. The information are collected by direct personal interviews. 2. Indirect Oral Interviews: Here information are not collected directly from the source but by interviewing persons closely related with the problem. 3. Mailed Questionnaire Method: Here information are collected through a set of questionnaire. A questionnaire is a document prepared by the investigator containing a set of questions. These questions relate to the problem of enquiry directly or indirectly. 4. Schedule Method: Here the questionnaires are sent through the enumerators to collect information. Enumerators are persons appointed by the investigator for the purpose. They directly meet the informants with the questionnaire. 5. From Local Agents: Sometimes primary data are collected from local agents or correspondents. These agents are appointed by the sponsoring authorities. They are well conversant with the local conditions like language, communication, food habits, traditions etc. Being on the spot and well acquainted with the nature of the enquiry they are capable of furnishing reliable information. Q:2 (ii) Draw a pie diagram in your answer script of the following data:

Category A B C D E F Frequency 9 12 57 24 10 8

SOLUTION 2 (ii)

Category Frequency Degree A 9 9/120 x 360 = 27 B 12 12/120 x 360 = 36 C 57 57/120 x 360 = 171 D 24 24/120 x 360 = 72 E 10 10/120 x 360 = 30 F 8 8/120 x 360 = 24

Total 120 360



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Q:2 (iii) From the given data, calculate price index numbers, taking 1967 as base

year using Simple Aggregative Method:

Item Price

1967 1970 1973 A 12.00 15.00 15.60 B 3.00 3.60 3.60 C 5.00 6.00 9.70

SOLUTION 2 (iii)

Item Price

1967 1970 1973 A 12.00 15.00 15.60 B 3.00 3.60 3.60 C 5.00 6.00 9.70

Total 20.00 24.60 28.90

𝐼𝑜𝑛 =∑𝑃𝑛∑𝑃𝑜

𝑥100

𝐼1967,1970 =24.60

20.00 x 100

𝑰𝟏𝟗𝟔𝟕,𝟏𝟗𝟕𝟎 = 𝟏𝟐𝟑%

𝐼1967,1973 =28.90

20.00 x 100

𝑰𝟏𝟗𝟔𝟕,𝟏𝟗𝟕𝟑 = 𝟏𝟒𝟒. 𝟓%

A

B

C

D

E

F

Pie - Diagram

A B C D E F



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Q:2 (iv) The ages of employees in a department are classified as: Age in Years 16 – 20 21 – 25 26 – 30 31 – 35 36 – 40

Number of Employees 14 25 30 25 4 Draw a histogram and find the mode graphically. SOLUTION 2 (iv)

Class Interval Frequency Class Boundaries 16 – 20 14 15.5 – 20.5 21 – 25 25 20.5 – 25.5 26 – 30 30 25.5 – 30.5 31 – 35 25 30.5 – 35.5 36 – 40 4 35.5 – 40.5

30

29 28 27 26 25

24 23 22 21 20 19 18 17 16 15 14

13 12 11 10

9 8 7 6 5 4

3 2 1 0

2.5 7.5 12.5 17.5 22.5 27.5 Mode



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Q:2 (v) The data given below shows the daily wages paid to workers in a factory. Classify this data taking a class interval of 18 and starting with 100. Also calculate relative frequencies:

240 165 175 185 210 117 100 140 175 165 103 145 179 104 176 179 214 126 147 169 175 210 217 225 217 222 217 179 165 145 175 107 172 124 134 146 157 165 179 145 175 156

SOLUTION 2 (v)

Class Interval Tally Marks Frequency Relative Frequency 100 – 117 //// 5 5/42 = 0.12 118 – 135 /// 3 3/42 = 0.07 136 – 153 //// / 6 6/42 = 0.14 154 – 171 //// // 7 7/42 = 0.17 172 – 189 //// //// // 12 12/42 = 0.29 190 – 207 0 0/42 = 0.00 208 – 225 //// /// 8 8/42 = 0.19 226 – 243 / 1 1/42 = 0.02

∑𝑓 = 42 ∑= 1.00

Q:2 (vi) For a set of 15 observations, the mean came out to be 12. Later on checking it

is discovered that an observation 20 was incorrectly recorded whereas the correct value was 02. Calculate the correct mean.

SOLUTION 2 (vi) 𝑛 = 15, �̅� = 12

�̅� =∑𝑥

𝑛

12 =∑𝑥

15

∑𝑥 = 12 x 15

∑𝑥 = 180

∑𝑥 = 180 − 20 + 2

∑𝑥 = 162

�̅� =∑𝑥

𝑛

�̅� =162

15

�̅� = 𝟏𝟎. 𝟖

Q:2 (vii) A card is drawn from a deck of 52 playing cards. What is the probability that

it is: (a) Diamond? (b) Face card?



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SOLUTION 2 (vii) 𝑛 = 52 𝑟 = 1

𝑛𝐶𝑟 = 𝑛!

𝑟! (𝑛 − 𝑟)!

52𝐶1 = 52!

1! (52 − 1)!

52𝐶1 = 52 𝑤𝑎𝑦𝑠 Diamond Card: 𝑛 = 13 𝑟 = 1

13𝐶1 = 13!

1! (13 − 1)!

13𝐶1 = 13 ways

𝑃 =𝑛(𝐴)

𝑛(𝑆)

𝑃 =13

52

𝑷 =𝟏

𝟒 𝒐𝒓 𝑷 = 𝟎. 𝟐𝟓

Face Card: 𝑛 = 12 𝑟 = 1

12𝐶1 = 12!

1! (12 − 1)!

12𝐶1 = 12 ways

𝑃 =𝑛(𝐴)

𝑛(𝑆)

𝑃 =12

52

𝑷 =𝟑

𝟏𝟑 𝒐𝒓 𝑷 = 𝟎. 𝟐𝟑

Q:2 (viii) How many permutations can be formed from the word COMMITTEE? SOLUTION 2 (viii) COMMITTEE

C = 1 = 𝑛1 𝑃 =

𝑛!

𝑛1! x 𝑛2! x 𝑛3! x 𝑛4! x 𝑛5! x 𝑛6

𝑃 = 9!

1! x 1! x 2! x 1! x 2! x 2!

𝑃 = 362,880

1 x 1 x 2 x 1 x 2 x 2

𝑃 = 362,880

8

𝑷 = 45,360 ways

O = 1 = 𝑛2 M = 2 = 𝑛3 I = 1 = 𝑛4 T 2 = 𝑛5 E = 2 = 𝑛6 9 = 𝑛



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SECTION “C” (DETAILED – ANSWER QUESTIONS) (15) Note: Attempt any Two questions from this section. Q:3 Calculate the mean, median and mode of the following data:

Group 0 – 10 10 – 20 20 – 30 30 – 40 40 – 50 50 – 60 Frequency 12 18 27 20 17 06

SOLUTION 3

C – I F X Fx CF 0 – 10 12 5 60 12

10 – 20 18 15 270 30 20 – 30 27 25 675 57 30 – 40 20 35 700 77 40 – 50 17 45 765 94 50 – 60 06 55 330 100

∑𝑓 = 100 ∑𝑓𝑥 = 2800

Mean: Median: Mode:

�̅� =∑𝑓𝑥

∑𝑓

�̅� =2800

100

�̅� = 𝟐𝟖. 𝟎𝟎

𝑀𝑒𝑑𝑖𝑎𝑛 𝑔𝑟𝑜𝑢𝑝 = (∑𝑓

2)

𝑡ℎ 𝑣𝑎𝑙𝑢𝑒

𝑀𝑒𝑑𝑖𝑎𝑛 𝑔𝑟𝑜𝑢𝑝 = (100

2)𝑡ℎ 𝑣𝑎𝑙𝑢𝑒

𝑀𝑒𝑑𝑖𝑎𝑛 𝑔𝑟𝑜𝑢𝑝 = (50)𝑡ℎ 𝑣𝑎𝑙𝑢𝑒

�̃� = 𝑙 +ℎ

𝑓(∑𝑓

2− 𝐶𝐹)

�̃� = 20 +10

27 (100

2− 30)

�̃� = 20 +10

27 (50 − 30)

�̃� = 20 +10

27 (20)

�̃� = 20 +200

27

�̃� = 20 + 7.41 �̃� = 𝟐𝟕. 𝟒𝟏

𝑥 = 𝑙 + ℎ (𝑓𝑚 − 𝑓1

2𝑓𝑚 − 𝑓1 − 𝑓2)

𝑥 = 20 + 10 (27 − 18

2(27) − 18 − 20)

𝑥 = 20 + 10 (9

16)

𝑥 = 20 + (90

16)

𝑥 = 20 + 5.625 �̂� = 𝟐𝟓. 𝟔𝟐𝟓



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Q:4 (a) A fair coin is tossed three times. What is the probability of getting at least one head?

SOLUTION 4 (a)

𝑆. 𝑆. = {𝐻𝐻𝐻,𝐻𝐻𝑇,𝐻𝑇𝐻, 𝑇𝐻𝐻𝑇𝑇𝑇, 𝑇𝑇𝐻, 𝑇𝐻𝑇,𝐻𝑇𝑇

}

At Least One Head:

𝑆. 𝑆. = {𝐻𝐻𝐻,𝐻𝐻𝑇,𝐻𝑇𝐻, 𝑇𝐻𝐻

𝑇𝑇𝐻, 𝑇𝐻𝑇,𝐻𝑇𝑇}

𝑃 =𝑛(𝐴)

𝑛(𝑆)

𝑃 =7

8

𝑷 =𝟕

𝟖 𝒐𝒓 𝑷 = 𝟎. 𝟖𝟕𝟓

Q:4 (b) What is the probability of getting a total of 7 or 11 when a pair of dice is

tossed? SOLUTION 4 (b)

𝑆. 𝑆. =

{

(1,1)(1,2)(1,3)(1,4)(1,5)(1,6)(2,1)(2,2)(2,3)(2,4)(2,5)(2,6)(3,1)(3,2)(3,3)(3,4)(3,5)(3,6)(4,1)(4,2)(4,3)(4,4)(4,5)(4,6)(5,1)(5,2)(5,3)(5,4)(5,5)(5,6)(6,1)(6,2)(6,3)(6,4)(6,5)(6,6)}

Events – A: Sum is 7: 𝑆. 𝑆. = {(1,6)(2,5)(3,4)(4,3)(5,2)(6,1)}

𝑃 =𝑛(𝐴)

𝑛(𝑆)

𝑃 =6

36

𝑷 =𝟏

𝟔 𝒐𝒓 𝑷 = 𝟎. 𝟏𝟕

Event – B: Sum is 11: 𝑆. 𝑆. = {(5,6)(6,5)}

𝑃 =𝑛(𝐴)

𝑛(𝑆)

𝑃 =2

36

𝑷 =𝟏

𝟏𝟖 𝒐𝒓 𝑷 = 𝟎. 𝟎𝟔

Mutually Exclusive Events: 𝑃(𝐴 ∪ 𝐵) = 𝑃(𝐴) + 𝑃(𝐵) 𝑃(𝐴 ∪ 𝐵) = 0.17 + 0.06 𝑷(𝑨 ∪ 𝑩) = 𝟎. 𝟐𝟑



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Q:4 (c) The probability that a student will pass in accounting is 6/9 and the probability that he will pass in statistics is 8/13, what is the probability that he will pass in both the subjects?

SOLUTION 4 (c)

Event − A: Pass in Accounting =6

9= 0.67

Event − B: Pass in Statistics =8

13= 0.62

𝑃(𝐴 ∩ 𝐵) = 𝑃(𝐴) . 𝑃(𝐵) 𝑃(𝐴 ∩ 𝐵) = 0.67 x 0.62 𝑷(𝑨 ∩ 𝑩) = 𝟎. 𝟒𝟐 Q:5 (a) A bag contains 3 red, 4 white and 3 green balls. If a sample of 3 balls is

selected at random, what is the probability that: (a) All are of different colours. (b) Two are of white colour.

SOLUTION 5 (a) 𝑛 = 10 𝑟 = 3

𝑛𝐶𝑟 = 𝑛!

𝑟! (𝑛 − 𝑟)!

10𝐶3 = 10!

3! (10 − 3)!

10𝐶3 = 120 𝑤𝑎𝑦𝑠 All are of different colours:

3𝐶1 x 4C1 x 3C1 = 3!

1! (3 − 1)! x

4!

1! (4 − 1)! x

3!

1! (3 − 1)!

3𝐶1 x 4C1 x 3C1 = 3 x 4 x 3 3𝐶1 x 4C1 x 3C1 = 36 𝑤𝑎𝑦𝑠

𝑃 =𝑛(𝐴)

𝑛(𝑆)

𝑃 =36

120

𝑷 =𝟑

𝟏𝟎 𝒐𝒓 𝑷 = 𝟎. 𝟑𝟎



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2 White Balls:

6𝐶1 x 4C2 = 6!

1! (6 − 1)! x

4!

2! (4 − 2)!

6𝐶1 x 4C2 = 6 x 6 6𝐶1 x 4C2 = 36

𝑃 =𝑛(𝐴)

𝑛(𝑆)

𝑃 =36

120

𝑷 =𝟑

𝟏𝟎 𝒐𝒓 𝑷 = 𝟎. 𝟑𝟎

Q:5 (b) The probability that A will alive for 30 years is 0.4 and probability that B will

be alive for 30 years is 0.8. What is the probability that both will be alive for 30 years?

SOLUTION 5 (b) Event − A: A will alive for 3 years = 0.4 Event − B: B will alive for 3 years = 0.8 𝑃(𝐴 ∩ 𝐵) = 𝑃(𝐴) . 𝑃(𝐵) 𝑃(𝐴 ∩ 𝐵) = 0.4 x 0.8 𝑷(𝑨 ∩ 𝑩) = 𝟎. 𝟑𝟐

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