MBOSE Class 12 Sample Question Paper 2026 Statistics

Meghalaya Board of School Education

SAMPLE
PAPERS

STATISTICS
CLASS- XII
SUBJECT: Time: 3 Hours.
Maximum Marks: 80
General Instructions:
Read the following instructions very carefully and strictly follow them:
(i) Write all answers in the answer script
(ii) Attempt Part-A questions serially
(iii) Attempt all parts of a question together at one place

(PART-A: OBJECTIVE)
(Marks:32)
SECTION-I
(Marks:16)
1. Choose and write the Correct Answer: 1x8=8
(i) If 𝑋𝑋 is a random variable, then the variance of 𝑋𝑋 denoted by 𝑉𝑉𝑉𝑉𝑉𝑉(𝑋𝑋)𝑜𝑜𝑜𝑜 𝑉𝑉(𝑋𝑋) is defined by
(A) 𝑉𝑉(𝑋𝑋) = 𝐸𝐸 2 (𝑋𝑋) − 𝐸𝐸(𝑋𝑋 2 )
(B) 𝑉𝑉(𝑋𝑋) = 𝐸𝐸(𝑋𝑋 2 ) − {𝐸𝐸(𝑋𝑋)}2
(C) 𝑉𝑉(𝑋𝑋) = {𝐸𝐸(𝑋𝑋 2 )}2 − 𝐸𝐸(𝑋𝑋 2 )
(D) 𝑉𝑉(𝑋𝑋) = {𝐸𝐸(𝑋𝑋)}2 − 𝐸𝐸(𝑋𝑋 2 )
(ii) If 𝑋𝑋 is a random variable, and 𝑎𝑎, 𝑏𝑏 are constants, then:
(A) 𝑉𝑉(𝑎𝑎𝑎𝑎 + 𝑏𝑏) = 𝑎𝑎2 𝑉𝑉(𝑋𝑋) + 𝑏𝑏
(B) 𝑉𝑉(𝑎𝑎𝑎𝑎 + 𝑏𝑏) = 𝑎𝑎𝑎𝑎(𝑋𝑋) + 𝑏𝑏
(C) 𝑉𝑉(𝑎𝑎𝑎𝑎 + 𝑏𝑏) = (𝑎𝑎 + 𝑏𝑏)𝑉𝑉(𝑋𝑋)
(D) None of the above
(iii) If 𝐸𝐸(3𝑋𝑋 + 5) = 12, then the value of 𝐸𝐸(𝑋𝑋) is:
(A) 7/3
(B) 3/7
(C) 5
(D) 12
(iv) Which of the following is not correct?
(A) 𝐸𝐸(𝑎𝑎𝑎𝑎) = 𝑎𝑎𝑎𝑎(𝑋𝑋)
(B) 𝐸𝐸(𝑎𝑎𝑎𝑎) = 𝑋𝑋𝑋𝑋(𝑎𝑎)
(C) 𝐸𝐸(𝑎𝑎) = 𝑎𝑎
(D) 𝐸𝐸(𝑎𝑎𝑎𝑎 + 𝑏𝑏𝑏𝑏) = 𝑎𝑎𝑎𝑎(𝑋𝑋) + 𝑏𝑏𝑏𝑏(𝑌𝑌)
(v) If 𝑛𝑛 = 32 𝑎𝑎𝑎𝑎𝑎𝑎 𝑝𝑝 = 𝑞𝑞 = 1/2, then standard of deviation of binomial distribution is:
(A) 2√2
(B) 2
(C) 4
(D) 36
(vi) For binomial distribution, the mean is:
(A) Greater than the variance
(B) Lesser than the variance

(C) Equal to the variance
(D) None of the above
(vii) Binomial distribution is symmetrical if:
(A) 𝑝𝑝 = 𝑞𝑞 = 1
1
(B) 𝑝𝑝 = 𝑞𝑞 =
2
1
(C) 𝑝𝑝 = 1, 𝑞𝑞 =
2
1
(D) 𝑝𝑝 = , 𝑞𝑞 = 1
2
(viii) In SRSWR, the sample mean 𝑥𝑥̅ is an unbiased estimator of:
(A) Population means 𝑋𝑋�
(B) Population variance 𝜎𝜎 2
(C) Sample mean square 𝑆𝑆 2
(D) None of the above

2. Fill in the blanks 1x4=4
(a) Normal distribution is …….
(b) In a sampling distribution, a finite population of 𝑁𝑁 units samples of size 𝑛𝑛 can be selected
as ……
(c) Time reversal test is satisfied if ……….
(d) In case of SRSWR 𝑉𝑉(𝑥𝑥̅ ) is equal to ………

3. State whether the following statements are True or False 1x4=4
(a) The term ‘Statistic’ is used to denote the characteristics of the population
(b) Moving average method can be used to determine only trend
(c) Cyclical variations refer to the movements which occur after time interval of one year
(d) Sampling which provides for a known non-zero equal chance of selection is called quota
sampling

SECTION-II
(Marks: 16)
4. Answer the following questions (Any Eight) 2 x 8 = 16
(a) “The mean of a binomial distribution is 5 and standard deviation is 3”. State whether the
statement is true or false with proper justification.
(b) In a binomial distribution, prove that mean > variance
(c) If 𝑋𝑋 follows a binomial distribution with mean 4 and variance 2, find 𝑃𝑃(𝑋𝑋 = 5)
(d) If 𝑋𝑋 and 𝑌𝑌 are independent random variables, then prove that 𝐸𝐸[{𝑋𝑋 − 𝐸𝐸(𝑋𝑋)}{𝑌𝑌 − 𝐸𝐸(𝑌𝑌)}] =
0
1 1
(e) If 𝑓𝑓(𝑥𝑥) = , 𝑥𝑥 = 2,4,8,16, find 𝐸𝐸 � � where 𝑓𝑓(𝑥𝑥) = 𝑃𝑃(𝑋𝑋 = 𝑥𝑥)
4 𝑋𝑋
(f) Find the mean of a Poisson distribution
(g) A player rolls one fair die. If the die shows a prime number, the player wins the value that
appears on the die, else loses that value that appears on it. Find whether the game is
favourable to the player.
(h) “Index numbers are economic barometers”. Explain.
(i) Distinguish between seasonal variation and cyclical variation.
(j) Mention the three methods of sampling.

PART-B: DESCRIPTIVE
(Marks: 48)
Answer any Four Questions, taking at least One from each Group
GROUP-A
5. (a) Given 𝐸𝐸(𝑋𝑋 + 4) = 10 𝑎𝑎𝑎𝑎𝑎𝑎 𝐸𝐸[(𝑋𝑋 + 4)2 ] = 116. Determine 𝐸𝐸(𝑋𝑋 2 ) 𝑎𝑎𝑎𝑎𝑎𝑎 𝑉𝑉𝑉𝑉𝑉𝑉(𝑋𝑋)
OR
The random variable 𝑋𝑋 has the following distribution:

𝑋𝑋 1 2 3
𝑝𝑝(𝑥𝑥) 1/6 2/6 3/6

Find Var (X) and 𝑃𝑃(𝑋𝑋 ≥ 2) 4
(b) If the sum of the mean and variance of a binomial distribution for 5 trials is 1.8, then find
the distribution. 4
(c) Define the following:
(i) Mathematical Expectation.
(ii) Discrete random variable.
(iii) Continuous random variable.
(iv) Variance of a random variable 1+1+1+1=4

6. (a) Define Poisson distribution and binomial distribution. When does a binomial distribution or
a Poisson distribution tend to a normal distribution.
(b) A traffic engineer records the number of bicycle riders that use a particular cycle track. He
records that an average of 3.2 bicycle riders use the cycle track every hour. Given that the
number of bicycles that use the cycle track follow a Poisson distribution, what is the probability
that 2 or less bicycle riders will use the cycle track within an hour? Also find the mean
expectation and variance for the random variable. (Given 𝑒𝑒−3.2 = 0.041) 4
(c) If 1% residents of a city are colour-blind, find the probability that out of 100 persons
selected at random, at most one is colour-blind. 4

GROUP-B

7. (a) What are the components of a time series? Discuss each of them with suitable examples 4
(b) Explain briefly how Fischer’s ideal index number is constructed and justify it being called
“ideal” 4
(c) The quarterly profits of a small-scale industry (₹ in thousands) are as follows:

YEAR QUARTER 1 QUARTER 2 QUARTER 3 QUARTER
4
2020 39 47 20 56
2021 68 59 66 72
2022 88 60 60 67

Calculate 4-Quarterly moving averages 4

8. (a) Show that Laspeyres’ price index number and Paasche’s index number do not satisfy Factor
Reversal Test. 6
(b) Calculate the Cost-of-Living Index Number from the following data and interpret your result:

Article Weight Price Relative
Food 24.9 1.02
Fuel 27.5 1.28
Clothing 9.4 0.83
House Rent 10.8 1.14
Transportation 5.3 1.00
Miscellaneous 22.1 1.18
6

GROUP-C
9. (a) Prove that in Simple Random Sampling, sample mean is an unbiased estimate of the
population 4
𝜎𝜎 2 𝑁𝑁−𝑛𝑛
(b) Show that 𝑉𝑉(𝑥𝑥̅ )𝑆𝑆𝑆𝑆𝑆𝑆𝑆𝑆𝑆𝑆𝑆𝑆 = 𝑛𝑛 . where 𝑥𝑥̅ and 𝜎𝜎 have usual meanings 5
𝑁𝑁
(c) Write short notes on the following: (Any three) 1x3=3
(i) Sampling error
(ii) SRSWOR
(iii) SRSWR
(iv) Judgement sampling

10. (a) Write a brief note on the errors that creep up in sample survey 6
(b) What is sampling? Give its objective and name the laws which form the basis of sampling
6

MARKING SCHEME
CLASS XII
STATISTICS
SECTION: A (Solution of MCQs of 1 Mark each)

Q. No. Answer Hints/Solution
1 (a) B 𝑉𝑉(𝑋𝑋) = 𝐸𝐸[𝑋𝑋 − 𝐸𝐸(𝑋𝑋)]2 = 𝐸𝐸[𝑋𝑋 2 − 2𝑋𝑋𝑋𝑋(𝑋𝑋) + {𝐸𝐸(𝑋𝑋)}2 ]
= 𝐸𝐸(𝑋𝑋 2 ) − 2𝐸𝐸(𝑋𝑋)𝐸𝐸(𝑋𝑋) + {𝐸𝐸(𝑋𝑋)}2
= 𝐸𝐸(𝑋𝑋 2 ) − 2{𝐸𝐸(𝑋𝑋)}2 + {𝐸𝐸(𝑋𝑋)}2
= 𝐸𝐸(𝑋𝑋 2 ) − {𝐸𝐸(𝑋𝑋)}2
1 (b) A 𝑉𝑉(𝑎𝑎𝑎𝑎 + 𝑏𝑏) = 𝐸𝐸[(𝑎𝑎𝑎𝑎 + 𝑏𝑏) − 𝐸𝐸(𝑎𝑎𝑎𝑎 + 𝑏𝑏)]2
= 𝐸𝐸[𝑎𝑎𝑎𝑎 + 𝑏𝑏 − 𝑎𝑎𝑎𝑎(𝑋𝑋) − 𝑏𝑏]2
= 𝐸𝐸[𝑎𝑎{𝑋𝑋 − 𝐸𝐸(𝑋𝑋)}]2
= 𝑎𝑎2 𝐸𝐸{𝑋𝑋 − 𝐸𝐸(𝑋𝑋)}2
= 𝑎𝑎2 𝑉𝑉(𝑋𝑋)
1 (c) A 𝐸𝐸(3𝑋𝑋 + 5) = 12
⟹ 3𝐸𝐸(𝑋𝑋) + 5 = 12
⟹ 3𝐸𝐸(𝑋𝑋) = 7
⟹ 𝐸𝐸(𝑋𝑋) = 7/3
1 (d) B 𝐸𝐸(𝑎𝑎𝑎𝑎) = ∑𝑥𝑥 𝑎𝑎 𝑥𝑥 𝑝𝑝(𝑥𝑥) = 𝑎𝑎 ∑𝑥𝑥 𝑥𝑥𝑥𝑥(𝑥𝑥) = 𝑎𝑎𝑎𝑎(𝑋𝑋)
1 (e) A 1 1
𝑆𝑆𝑆𝑆 = �𝑉𝑉(𝑋𝑋) = �𝑛𝑛𝑛𝑛𝑛𝑛 = �32 × 2 × 2 = √8 = 2√2
1 (f) A Greater than variance
1
1 (g) B 𝑝𝑝 = 𝑞𝑞 = 2
1 1 1 𝑛𝑛𝑋𝑋�
1 (h) A 𝐸𝐸(𝑥𝑥̅ ) = 𝐸𝐸 �𝑛𝑛 ∑𝑛𝑛𝑖𝑖=1 𝑥𝑥𝑖𝑖 � = 𝑛𝑛 ∑𝑛𝑛𝑖𝑖=1 𝐸𝐸(𝑥𝑥𝑖𝑖 ) = 𝑛𝑛 ∑𝑛𝑛𝑖𝑖=1 𝑋𝑋� = 𝑛𝑛 = 𝑋𝑋�
2 (a) Bi-
parametric
2 (b) 𝑁𝑁 𝑛𝑛 The first unit can be drawn from 𝑁𝑁 units in 𝑁𝑁𝐶𝐶 1 = 𝑁𝑁 ways, the second unit
can be drawn in 𝑁𝑁 ways and so on. Hence total ways is 𝑁𝑁. 𝑁𝑁. 𝑁𝑁 … . . 𝑁𝑁 = 𝑁𝑁 𝑛𝑛
2 (c) 𝐼𝐼𝑜𝑜𝑜𝑜 × 𝐼𝐼𝑛𝑛𝑛𝑛 This test requires that if in an index number formula base and current year
=1 be interchanged, then one should be reciprocal of the other i.e. their product
should be unity
2 1
2 (d) 𝜎𝜎 𝑉𝑉(𝑥𝑥̅ ) = 𝑉𝑉 � ∑𝑛𝑛𝑖𝑖=1 𝑥𝑥𝑖𝑖 �
𝑛𝑛
𝑛𝑛 1
= 2 𝑉𝑉(∑𝑛𝑛𝑖𝑖=1 𝑥𝑥𝑖𝑖 )
𝑛𝑛
1
= ∑𝑛𝑛 𝑉𝑉(𝑥𝑥𝑖𝑖 )
𝑛𝑛2 𝑖𝑖=1
1 1 𝜎𝜎 2
= 𝑛𝑛2 ∑𝑛𝑛𝑖𝑖=1 𝜎𝜎 2 = 𝑛𝑛2 . 𝑛𝑛. 𝜎𝜎 2 = 𝑛𝑛
3 (a) False Sample Data
3 (b) True Moving average values plotted against time give the trend line
3 (c) False Cyclical variations refer to the movements which occur after time intervals
of more than one year
3 (d) False Systematic Sampling as it is a type of probability sampling while others are
types of non-probability sampling.
(When selection of objects from the population is random, then objects of
the
population have an equal probability i.e., has a known non-zero equal chance
of selection. In other words, in probability sampling, sample units are
selected at random.)

SECTION: II (Solution of SA type Questions of 2 Marks each)

Q. Answer/Hints Marking
No Scheme
4 If 𝑋𝑋~𝐵𝐵(𝑛𝑛, 𝑝𝑝), then
(a) 𝑛𝑛𝑛𝑛 = 5 𝑎𝑎𝑎𝑎𝑎𝑎 �𝑛𝑛𝑛𝑛𝑛𝑛 = 3
𝟏𝟏
⟹ 𝑛𝑛𝑛𝑛𝑛𝑛 = 9 1
𝟐𝟐
𝑛𝑛𝑛𝑛𝑛𝑛 9
⟹ =
𝑛𝑛𝑛𝑛 5
⟹ 𝑞𝑞 > 1 which is impossible.
Hence the statement is wrong
½
4 𝑉𝑉𝑉𝑉𝑉𝑉𝑉𝑉𝑉𝑉𝑉𝑉𝑉𝑉𝑉𝑉 𝑛𝑛𝑛𝑛𝑛𝑛
𝑀𝑀𝑀𝑀𝑀𝑀𝑀𝑀
= 𝑛𝑛𝑛𝑛
(b) 𝟏𝟏
𝑉𝑉𝑉𝑉𝑉𝑉𝑉𝑉𝑉𝑉𝑉𝑉𝑉𝑉𝑉𝑉 1
⟹ = 𝑞𝑞 < 1 𝟐𝟐
𝑀𝑀𝑀𝑀𝑀𝑀𝑀𝑀
⟹ 𝑉𝑉𝑉𝑉𝑉𝑉𝑉𝑉𝑉𝑉𝑉𝑉𝑉𝑉𝑉𝑉 < 𝑀𝑀𝑀𝑀𝑀𝑀𝑀𝑀 ½
⟹ 𝑀𝑀𝑀𝑀𝑀𝑀𝑀𝑀 > 𝑣𝑣𝑣𝑣𝑣𝑣𝑣𝑣𝑣𝑣𝑣𝑣𝑣𝑣𝑣𝑣
4 𝑀𝑀𝑀𝑀𝑀𝑀𝑀𝑀 = 𝑛𝑛𝑛𝑛 = 4
(c) 𝑛𝑛𝑛𝑛𝑛𝑛 2 1 1 1 𝟏𝟏
𝑉𝑉𝑉𝑉𝑉𝑉𝑉𝑉𝑉𝑉𝑉𝑉𝑉𝑉𝑉𝑉 = 𝑛𝑛𝑛𝑛𝑛𝑛 = 2 ⟹ = ⟹ 𝑞𝑞 = ⟹ 𝑝𝑝 = 1 − = 1𝟐𝟐
𝑛𝑛𝑛𝑛 4 2 2 2
1
Now, 𝑛𝑛𝑛𝑛 = 4 ⟹ 𝑛𝑛 = 4 ⟹ 𝑛𝑛 = 8
2
1 1 8! 1 1 14 7
Hence, 𝑃𝑃(𝑋𝑋 = 5) = 8𝐶𝐶 5 (2)5 (2)8−5 = 5!3! . 32 . 8 = 32 = 16 𝟏𝟏
𝟐𝟐
4 𝐸𝐸[{𝑋𝑋 − 𝐸𝐸(𝑋𝑋)}{𝑌𝑌 − 𝐸𝐸(𝑌𝑌)}] = 𝐸𝐸[𝑋𝑋𝑋𝑋 − 𝑋𝑋𝑋𝑋(𝑌𝑌) − 𝑌𝑌𝑌𝑌(𝑋𝑋) + 𝐸𝐸(𝑋𝑋)𝐸𝐸(𝑌𝑌)]
(d) = 𝐸𝐸(𝑋𝑋𝑋𝑋) − 𝐸𝐸(𝑌𝑌)𝐸𝐸(𝑋𝑋) − 𝐸𝐸(𝑋𝑋)𝐸𝐸(𝑌𝑌) +
𝐸𝐸(𝑋𝑋)𝐸𝐸(𝑌𝑌) 𝟏𝟏
1
𝟐𝟐
= 𝐸𝐸(𝑋𝑋𝑋𝑋) − 𝐸𝐸(𝑋𝑋)𝐸𝐸(𝑌𝑌)
Since 𝑋𝑋 𝑎𝑎𝑎𝑎𝑎𝑎 𝑌𝑌 are independent, 𝐸𝐸(𝑋𝑋𝑋𝑋) = 𝐸𝐸(𝑋𝑋)𝐸𝐸(𝑌𝑌) 𝟏𝟏
Hence, 𝐸𝐸[{𝑋𝑋 − 𝐸𝐸(𝑋𝑋)}{𝑌𝑌 − 𝐸𝐸(𝑌𝑌)}] = 0 𝟐𝟐
4 1 1 𝟏𝟏
𝐸𝐸 � � = ∑ . 𝑃𝑃(𝑋𝑋 = 𝑥𝑥)
𝑋𝑋 𝑥𝑥 𝟐𝟐
(e) 1 1 1 1 1 1 1 1
= . + . + . + .
2 4 4 4 8 4 16 4
8+4+2+1 15
= 64 = 64 𝟏𝟏
1𝟐𝟐
4 𝑒𝑒 −𝜆𝜆 𝜆𝜆𝑥𝑥
𝑀𝑀𝑀𝑀𝑀𝑀𝑀𝑀 = 𝐸𝐸(𝑋𝑋) = ∑ 𝑥𝑥 𝑝𝑝(𝑥𝑥) = ∑∞
𝑥𝑥=0 𝑥𝑥. 𝑥𝑥!
(f)
∞ 𝑒𝑒 −𝜆𝜆 𝜆𝜆𝑥𝑥
∑
= 𝑥𝑥=1 𝑥𝑥.
𝑥𝑥(𝑥𝑥−1)!
𝜆𝜆𝑥𝑥−1
1
−𝜆𝜆 ∑∞
= 𝜆𝜆𝑒𝑒 𝑥𝑥=1 (𝑥𝑥−1)!
𝜆𝜆2 𝜆𝜆3
= 𝜆𝜆𝑒𝑒 −𝜆𝜆 �1 + 𝜆𝜆 + 2! + 3! + ⋯ . �
−𝜆𝜆 𝜆𝜆
= 𝜆𝜆𝑒𝑒 𝑒𝑒 = 𝜆𝜆 1

4 The possible outcomes 𝑥𝑥𝑖𝑖 of the game with their corresponding probabilities
(g) 𝑝𝑝(𝑥𝑥𝑖𝑖 ) are as follows:
𝑥𝑥 2 3 5 -1 -4 -6
𝑝𝑝(𝑥𝑥) 1/6 1/6 1/6 1/6 1/6 1/6
The negative values -1, -4, -6 correspond to the fact that the player loses if a 1
non prime number occur
1 1 1 ½
The expected value of the game is 𝐸𝐸(𝑋𝑋) = ∑ 𝑥𝑥 𝑝𝑝(𝑥𝑥) = 2. + 3. + 5. −
6 6 6
1 1 1 1
1. 6 − 4. 6 − 6. 6 = − 6 ½
Thus the game is not favourable to the player.

4 Indices of price, output, foreign exchange, bank deposits etc. acts as
(h) barometers to find out ups and downs in the general economic conditions of
a country. Index numbers measure the pressure of economic and business 2
behaviour
4 Seasonal variations occur regularly and therefore they are definite and can be 1
(i) forecasted with a little effort.
Cyclical variations though more or less regular are not uniformly periodic 1
4 Simple Random, Stratified and Cluster Samplings 2
(j)

PART-B: DESCRIPTIVE (Solutions of Long Answer Type Questions)
Q. Answer/Hints Marking
No Scheme
5 𝑉𝑉(𝑋𝑋) = 𝐸𝐸(𝑋𝑋 2 ) − {𝐸𝐸(𝑋𝑋)}2 1
1 2 3 1+8+27
(a) Now, 𝐸𝐸(𝑋𝑋 2 ) = ∑ 𝑥𝑥 2 𝑝𝑝(𝑥𝑥) = 12 . 6 + 22 . 6 + 32 . 6 = 6 = 6
1 2 3 1+4+9 7 1
𝐸𝐸(𝑋𝑋) = ∑ 𝑥𝑥 𝑝𝑝(𝑥𝑥) = 1. + 2. + 3. = =
6 6 6 6 3
7 2 5
Hence, 𝑉𝑉(𝑋𝑋) = 6 − � � =
3 9 1
𝑃𝑃(𝑋𝑋 ≥ 2) = 𝑃𝑃(𝑋𝑋 = 2) + 𝑃𝑃(𝑋𝑋 = 3)
2 3 5
=6+6=6 1
OR 𝑉𝑉(𝑋𝑋) = 𝐸𝐸(𝑋𝑋 2 ) − {𝐸𝐸(𝑋𝑋)}2 ½
1 2 3 1+8+27
Now, 𝐸𝐸(𝑋𝑋 2 ) = ∑ 𝑥𝑥 2 𝑝𝑝(𝑥𝑥) = 12 . 6 + 22 . 6 + 32 . 6 = 6 = 6 1
1 2 3 1+4+9 7 1
𝐸𝐸(𝑋𝑋) = ∑ 𝑥𝑥 𝑝𝑝(𝑥𝑥) = 1. + 2. + 3. = = 1
6 6 6 6 3
7 2 5
Hence, 𝑉𝑉(𝑋𝑋) = 6 − � � =
3 9
𝑃𝑃(𝑋𝑋 ≥ 2) = 𝑃𝑃(𝑋𝑋 = 2) + 𝑃𝑃(𝑋𝑋 = 3)
2 3 5 𝟏𝟏
=6+6=6
𝟐𝟐
5 Here 𝑛𝑛 = 5 and 𝑛𝑛𝑛𝑛 + 𝑛𝑛𝑛𝑛𝑛𝑛 = 1.8
(b) ⟹ 5𝑝𝑝 + 5𝑝𝑝𝑝𝑝 = 1.8 ½
⟹ 5𝑝𝑝 + 5𝑝𝑝(1 − 𝑝𝑝) = 1.8
⟹𝑝𝑝 + 𝑝𝑝(1 − 𝑝𝑝) = 0.36
36
⟹𝑝𝑝 + 𝑝𝑝 − 𝑝𝑝2 =
100
36
⟹𝑝𝑝2 − 2𝑝𝑝 + = 0.
100
2
⟹100𝑝𝑝 − 200𝑝𝑝 + 36 = 0.
2
⟹25𝑝𝑝 − 50𝑝𝑝 + 9 = 0.
⟹25𝑝𝑝2 − 45𝑝𝑝 − 5𝑝𝑝 + 9 = 0
⟹5𝑝𝑝(5𝑝𝑝 − 9) − (5𝑝𝑝 − 9) = 0
⟹(5𝑝𝑝 − 9)(5𝑝𝑝 − 1) = 0.
9 1 2
∴ 𝑝𝑝 = (rejected) or 𝑝𝑝 = = 0.2
5 5
⟹ 𝑞𝑞 = 0.8 𝟏𝟏
1𝟐𝟐
∴ 𝑃𝑃(𝑋𝑋 = 𝑟𝑟) = 5Cr (0.2) r (0.8)5-r where r = 0,1,2,3,4,5.
5 Definitions 1 Mark each
(c)
6 A discrete random variable is said to follow Poisson distribution if its p. m.
(a) f. is given by
𝑒𝑒 −𝜆𝜆 𝜆𝜆𝑥𝑥
𝑃𝑃(𝑋𝑋 = 𝑥𝑥) = 𝑝𝑝(𝑥𝑥) = ; 𝑥𝑥 = 0,1,2,3, … . 𝜆𝜆 > 0 2
𝑥𝑥!

A discrete random variable is said to follow binomial distribution if its p. m.
f. is given by
𝑃𝑃(𝑋𝑋 = 𝑥𝑥) = 𝑛𝑛𝐶𝐶 𝑥𝑥 𝑝𝑝 𝑥𝑥 𝑞𝑞 𝑛𝑛−𝑥𝑥 ; 𝑥𝑥 = 0,1,2,0000, 𝑛𝑛; 𝑝𝑝 + 𝑞𝑞 = 1 2

6 Given Mean = 𝜆𝜆 = 3.2 ½
(b) Let 𝑋𝑋 be the number of bicycle riders which use the cycle track
Required Probability 𝑃𝑃(𝑋𝑋 ≤ 2) = 𝑃𝑃(𝑋𝑋 = 2) + 𝑃𝑃(𝑋𝑋 = 1) + 𝑃𝑃(𝑋𝑋 = 0) 1
𝑒𝑒 −3.2 (3.2)2 𝑒𝑒 −3.2 (3.2)1 𝑒𝑒 −3.2 (3.2)0
= + +
2! 1! 0! 2
−3.2 (1
= 𝑒𝑒 + 3.2 + 5.12)
= 0.041 × 9.32 = 0.618
Also, mean expectation=variance of 𝑋𝑋 = 𝜆𝜆 = 3.2 ½
6 Given, n =100
(c) 1 ½
p = probability of colour-blind residents = 100.
Since, n is sufficiently large and p is very small, we can
apply Poisson distribution.
𝜆𝜆 = 𝑛𝑛𝑛𝑛 = 1
½
Let X be the number of colour-blind residents,
𝑒𝑒 −1 1𝑥𝑥
then 𝑃𝑃(𝑋𝑋 = 𝑥𝑥) = ; x = 0, 1, 2, ………… ½
𝑥𝑥!
The probability that at most one person is colour-blind is
P(𝑋𝑋 ≤ 1) = 𝑃𝑃(𝑋𝑋 = 0) + 𝑃𝑃(𝑋𝑋 = 1)
𝑒𝑒 −1 10 𝑒𝑒 −1 11
= + = 0.368 +0.368 𝟏𝟏
0! 1! 𝟐𝟐 𝟐𝟐
= 0.736.
7 Components of time series are; 1+1+1+1
(a) 1. Secular Trend or Trend
2. Seasonal Variations
3. Cyclical Variations
Random or Irregular Variations
7 Fisher’s Index number is the GM of Laspeyre’s and Paasche’s index numbers
(b) given by
∑ 𝑝𝑝1 𝑞𝑞0 ∑ 𝑝𝑝1 𝑞𝑞1
𝑃𝑃01 = � × × 100 2
∑ 𝑝𝑝0 𝑞𝑞0 ∑ 𝑝𝑝0 𝑞𝑞1
It is called ideal number because it is the GM of Laspeyre’s and Paasche’s 1
index numbers, effectively balancing the biases of both.
It also satisfies both time reversal and factor reversal tests making it reliable 1
and accurate measure of price and quantity changes over time
7 Yearly/ Small 4- 4- 4-year 𝟏𝟏
𝟐𝟐
(c) Quarterly scale quarterly quarterly centered 𝟐𝟐
industry moving moving moving Marks for
correct table
total average average 𝟏𝟏
2020 I 39 &1
𝟐𝟐
II 47 162 40.5 Marks for
last column
III 20 44.125
191 47.75
IV 56 49.25
203 50.75
2021 I 68 56.5
249 62.25
II 59 64.25

265 66.25
III 66 68.75
285 71.25
IV 72 71.375
286 71.5
2022 I 88 70.75
280 70.0
II 60 69.375
275 68.75
III 60
IV 67
8 Laspeyre’s price index number
(a) ∑ 𝑝𝑝1 𝑞𝑞0
𝐿𝐿
𝑃𝑃01 =
∑ 𝑝𝑝0 𝑞𝑞0
Interchanging base (0) and current year (1) we get
𝐿𝐿
∑ 𝑝𝑝0 𝑞𝑞1
𝑃𝑃10 =
∑ 𝑝𝑝1 𝑞𝑞1
𝐿𝐿 𝐿𝐿
Hence, 𝑃𝑃01 × 𝑃𝑃10 ≠1 3
Hence Laspeyre’s index number does not satisfy time reversal
test
Also, Paasche’s price index number
𝑝𝑝 ∑ 𝑝𝑝1 𝑞𝑞1
𝑃𝑃01 =
∑ 𝑝𝑝0 𝑞𝑞1
And
𝑝𝑝 ∑ 𝑝𝑝0 𝑞𝑞0
𝑃𝑃10 =
∑ 𝑝𝑝1 𝑞𝑞0
𝑝𝑝 𝑝𝑝 3
Hence, 𝑃𝑃01 × 𝑃𝑃10 ≠ 1
Hence Paasche’s index number does not satisfy time reversal
test
8 Article Weight W 𝑃𝑃1 𝐼𝐼𝐼𝐼 𝟒𝟒
(b) 𝐼𝐼 = × 100 Marks for
𝑃𝑃0
Food 24.9 correct table
102 2539.8
&2
Fuel 27.5 128 3520.0 Marks for
Clothing 9.4 83 780.2 interpretation
House Rent 10.8 114 1231.2
Transportation 5.3 100 530.0
Miscellaneous 22.1 118 2607.8
� 𝑊𝑊 = 100 � 𝐼𝐼𝐼𝐼
= 11209

∑ 𝐼𝐼𝐼𝐼 11209
CLIN = ∑ 𝑊𝑊 = 100 = 112.09
Interpretation: There is an increase of 12.09% in the cost of living
in the current year compared to the base year
1
9 𝐸𝐸(𝑥𝑥̅ ) = 𝐸𝐸 �𝑛𝑛 ∑𝑛𝑛𝑖𝑖=1 𝑥𝑥𝑖𝑖 �
(a) 1
1
= 𝑛𝑛 ∑𝑛𝑛𝑖𝑖=1 𝐸𝐸(𝑥𝑥𝑖𝑖 )
2

1 1
𝐴𝐴𝐴𝐴 𝐸𝐸(𝑥𝑥𝑖𝑖 ) = ∑𝑛𝑛𝑖𝑖=1 𝑥𝑥𝑖𝑖 𝑝𝑝𝑖𝑖 = ∑𝑛𝑛𝑖𝑖=1 𝑥𝑥𝑖𝑖 . 𝑁𝑁 = 𝑁𝑁 . ∑𝑛𝑛𝑖𝑖=1 𝑥𝑥𝑖𝑖 = 𝑋𝑋�
1 𝑛𝑛𝑋𝑋 �
Hence, 𝐸𝐸(𝑥𝑥̅ ) = 𝑛𝑛 ∑𝑛𝑛𝑖𝑖=1 𝑋𝑋� = 𝑛𝑛 = 𝑋𝑋� 1

9 2
𝑉𝑉(𝑥𝑥̅ ) = 𝐸𝐸�𝑥𝑥̅ − 𝐸𝐸(𝑥𝑥̅ )�
(b)
= 𝐸𝐸(𝑥𝑥̅ − 𝑋𝑋�)2
1 2
= 𝐸𝐸 �𝑛𝑛 ∑𝑛𝑛𝑖𝑖−1(𝑥𝑥𝑖𝑖 − 𝑋𝑋�)�
1
= 𝑛𝑛2 𝐸𝐸(∑𝑛𝑛𝑖𝑖−1(𝑥𝑥𝑖𝑖 − 𝑋𝑋�))2
1
= 𝑛𝑛2 𝐸𝐸(∑𝑛𝑛𝑖𝑖−1(𝑥𝑥𝑖𝑖 − 𝑋𝑋�)2 + ∑𝑛𝑛𝑗𝑗≠1(𝑥𝑥𝑖𝑖 − 𝑋𝑋�)�𝑥𝑥𝑗𝑗 − 𝑋𝑋��
𝑎𝑎𝑎𝑎 (∑𝑛𝑛𝑖𝑖=1 𝑥𝑥𝑖𝑖 )2 = ∑𝑛𝑛𝑖𝑖=1 𝑥𝑥𝑖𝑖2 + ∑𝑛𝑛𝑗𝑗≠1 𝑥𝑥𝑖𝑖 𝑥𝑥𝑗𝑗 1
𝟏𝟏
𝟐𝟐
Again, 𝐸𝐸(𝑥𝑥𝑖𝑖 − 𝑋𝑋�)2 = ∑𝑁𝑁 � 2
𝑖𝑖=1(𝑥𝑥𝑖𝑖 − 𝑋𝑋 ) . 𝑝𝑝𝑖𝑖
= ∑𝑁𝑁 � 2 1
𝑖𝑖=1(𝑥𝑥𝑖𝑖 − 𝑋𝑋 ) . 𝑁𝑁 = 𝜎𝜎
2 1
1
Also, 𝐸𝐸(𝑥𝑥𝑖𝑖 − 𝑋𝑋�)�𝑥𝑥𝑗𝑗 − 𝑋𝑋�� = ∑𝑁𝑁 � �
𝑖𝑖≠𝑗𝑗(𝑥𝑥𝑖𝑖 − 𝑋𝑋 )�𝑥𝑥𝑗𝑗 − 𝑋𝑋 �.
𝑁𝑁(𝑁𝑁−1)
1 𝑁𝑁 �
= 𝑁𝑁(𝑁𝑁−1) [{∑𝑖𝑖=1(𝑥𝑥𝑖𝑖 − 𝑋𝑋)} − ∑𝑁𝑁
2 � 2
𝑖𝑖=1(𝑥𝑥𝑖𝑖 − 𝑋𝑋 ) ]
1
= 𝑁𝑁(𝑁𝑁−1) [0 − ∑𝑁𝑁 � 2
𝑖𝑖=1(𝑥𝑥𝑖𝑖 − 𝑋𝑋 ) ]
1 1
= − 𝑁𝑁−1 . 𝑁𝑁 ∑𝑁𝑁 � 2
𝑖𝑖=1(𝑥𝑥𝑖𝑖 − 𝑋𝑋 )
1
= − 𝑁𝑁−1 𝜎𝜎 2 𝟏𝟏
1 1 1
Therefore, 𝑉𝑉(𝑥𝑥̅ ) = 𝑛𝑛2 �∑𝑛𝑛𝑖𝑖=1 𝜎𝜎 2 + ∑𝑛𝑛𝑖𝑖≠𝑗𝑗 − 𝑁𝑁−1 𝜎𝜎 2 � 𝟐𝟐

1 2 𝑛𝑛(𝑛𝑛−1) 2
= 𝑛𝑛2 �𝑛𝑛𝜎𝜎 − 𝜎𝜎 �
𝑁𝑁−1
1 𝑁𝑁−1−𝑛𝑛+1 𝑁𝑁−1
= 𝑁𝑁 �𝑁𝑁−1
�.
𝑛𝑛
𝜎𝜎 2
1 𝑁𝑁−𝑛𝑛
= 𝑛𝑛 � 𝑁𝑁 � 𝜎𝜎 2 1
9 Short Notes of each 1 Mark for
(c) each
10 Sampling & Non-Sampling Error 3 Marks each
(a)
10 Definition of Sampling, Objective of Sampling & Principle of Sample 1 Mark for
(b) survey Definition &
2 Marks each
for objective
& principle

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