৪৯তম বিসিএস ⎯ পরিসংখ্যান [৯৮১]

প্রশ্ন:


সমাধান:

প্রশ্ন: প্রশ্নবোধক স্থানে কোন সংখ্যাটি বসবে?

সমাধান:
(২য় কলাম × ৩য় কলাম) - ১ম কলাম = ৪র্থ কলাম

(6 × 10) - 2 = 60 - 2 = 58
(7 × 11) - 3 = 77 - 3 = 74
(8 × 12) - 4 = 96 - 4 = 92

সুতরাং, প্রশ্নবোধক স্থানে 92 সংখ্যাটি বসবে।

Answer: D (Mutually exclusive events are generally not independent, except in special cases).

Mutually exclusive events cannot happen at the same time. If AAA and BBB are mutually exclusive, then P(A∩B)=0
Independent events are such that the occurrence of one does not affect the probability of the other. For A and B independent, P(A∩B)=P(A)⋅P(B).
Now, if two events are mutually exclusive and both have non-zero probability, then:

P(A∩B)=0 ≠P(A)⋅P(B)>0
So they cannot be independent.

Conclusion: Mutually exclusive events with non-zero probabilities are not independent.

now
P(A∪B)=P(A)+P(B)−P(A∩B)
0.7=0.5+0.3−P(A∩B)
so, P(A∩B)=0.1
since, P(A∩B)>0 → They are not mutually exclusive
P(A∩B)≠P(A)P(B) → They are not independent

here, kurtosis, β₂ = µ₄ /µ₂²
=5 > 3

so, it is leptokurtic

Proportions or percentages – Pie charts show the relative size of each category compared to the whole.

There are abr total observations. Each of the ab treatment combinations has r replicates, so Total df is- ABr-1

source- statistics and probabiliy: an introductory analysis by siddiqur rahman, fundamentals of probability and probability distribution by manindra kumar roy, statistics by Md abdul aziz

If we fail to reject Ho when it is actually false, we are making: Type II error

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Answer. C) x² test
Explanation: The chi-square test for independence is used in contingency
tables (e.g., gender vs. preference).

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the probabilities that all seeds are germinated will be, P(x=6) = f(6; 6 , .7) = 6C₆ * 0.7⁶ * 0.3⁰
=0.1176
=11.76%

WE have to wait untill first success, so it follows geometric distribution,

now P(X=3) = p * (1-p)^3-1
= 0.2 * 0.8²
=0.128

since in uniform distribution , β₁ = 0 and β₂ < 3

so, the characteristics / shape of uniform distribution is - symmetrical and platykurtic

in type I error, null hypothesis is rejected when it is true

in hypothesis testing using the criminal trial analogy:

Null hypothesis (H₀): The defendant is innocent.
Alternative hypothesis (H₁): The defendant is guilty.

Rejecting H₀ when it is actually true → convicting an innocent person.

for outside range -extrapolation
for within range -interpolation

Explanation:

Regression uses observed data to model the relationship between a predictor (independent) variable X and a response (dependent) variable Y.
Interpolation predicts Y for X within the range of observed X values.
Extrapolation predicts Y for X outside the observed range of X values.
Caution: Extrapolation can be risky because the relationship may not hold outside the observed data range.

3 fertilizers → more than one treatment.
Applied on several plots → this is Replication.
Applied randomly → this is Randomization.
Across fields with different fertility levels → fertility is a known source of heterogeneity, and controlling this by comparing treatments within fertility groups is Local Control (Blocking).

So, all three principles are applied.

Reject H₀​ → at least one treatment mean differs

If F-calculated > F-critical:

The observed treatment variation is larger than expected by random error.
We reject H₀
Interpretation: At least one treatment mean is significantly different from the others.

Answer: b) Histogram

The data are continuous and grouped into class intervals (e.g., 50–59, 60–69).
A histogram shows frequencies for such intervals using adjacent bars (no gaps), because the classes are continuous.

Answer: c) Number of heads in 5 times coin tosses because its associated with chance

A random variable (RV) is a variable that takes numerical values based on the outcome of a random experiment.

Answer: b) All numbers are equal, in this time AM, GM, HM are equal


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Answer: b) Arithmetic Mean, it is most affected by outliers


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Answer: B) Long-term oscillations around a trend due to business cycles

A cyclical variation is a long-term, wave-like fluctuation in a time series that repeats over periods longer than a year.

These are usually related to business or economic cycles, such as booms and recessions.
The duration is often 2–10 years.
It is different from seasonal variation, which repeats within a year.

When an interviewer intentionally records wrong data, the error does not arise due to the process of sampling itself — it’s caused by human bias or misconduct during data collection.

This type of error is called a non-sampling error.

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Answer: C) Parametric
Explanation: Parametric tests (t, z, ANOVA) depend on mean and variance.

source- statistics and probabiliy: an introductory analysis by siddiqur rahman, fundamentals of probability and probability distribution by manindra kumar roy, statistics by Md abdul aziz and class lecture

Answer: B) Median
Explanation: Median remains stable even with extreme values, unlike the mean.

source- statistics and probabiliy: an introductory analysis by siddiqur rahman, fundamentals of probability and probability distribution by manindra kumar roy, statistics by Md abdul aziz and class lecture

Answer: C) Time series trends use arithmetic mean
Explanation: Trend methods like moving averages are based on arithmetic mean.

In correlation, regression, and time series analysis — we primarily use the Arithmetic Mean.

Answer: A) Dataset A
Explanation: we know, CV = SD/Mean * 100
CV_A=5/50×100
=10%,

CV_B=10/200×100
=5%

So data set A is more variable relative to its mean.

The data set is listed in increasing order of magnitude.

The number of observations, n = 20, and (n + 1)/2 = (20+1)/2 = 10.5
Hence, Me = half way between 10th and 11th observations in the ordered list.
= (8+8)/2
=8.

The first half of the entire data set below median is 1 3 5 5 7 8 8 8 8 8

Hence, first quartile,
Q1= halfway between 5th observation and 6th observation in the first half

(7 + 8) / 2 = 7.5

The last half of the entire data set below median is 8 9 9 9 9 9 10 10 10 10

Hence, third quartile,
Q3= halfway between 5th observation and 6th observation in the last half

(9 + 9) / 2 = 9

since, Q3- Me > Me - Q1, the data set is skewed to the right

A distribution with light tails and flat peak has- β2 <3 (platykurtic)

if the correlation coefficient r is belongs to 0.10 to 0.29 than it is weak positive correlation
for 0.30 to 0.49 it is medium and
for 0.50 to 0.79 it is moderate positive correlation
for 0.80 to 0.99 it is strong positive correlation 

Multiple correlation coefficient can not be less than simple correlation coefficient

it is one of the properties of Multiple correlation

Products in a production batch should be checked randomly to detect defects.
Systematic sampling (e.g., every 10th item) is also efficient in production lines.

both sampling method Ensures representative inspection without checking all items.

source- statistics and probabiliy: an introductory analysis by siddiqur rahman, fundamentals of probability and probability distribution by manindra kumar roy, statistics by Md abdul aziz and class lecture

it follows geometric distribution because : The geometric distribution is a discrete probability distribution that models the number of trials required to get the first success in a sequence of independent Bernoulli trials (each trial has only two outcomes: success or failure).
so,
E(X)=1/p
=1/0.2
=5

Laspayer`s index number has-
i) upward bias and it is unit free

A trend variation shows the long-term, smooth, and consistent movement in a time series over several years.

In this case, the steady growth of national income over time reflects a long-term upward trend — not short-term or random changes.

if p < 1/2 then the shape of the binomial distribution is positively skewed and for p > 1/2 it is negatively skewed

HERE, P(A∪B) = P(A) + P(B) - P(A∩B)
P(A∩B) = 3/20
now,
P(A/B‾) = P(A∩B‾) / P(B‾)
= [P(A)- P(A∩B)] /[ 1- P(B)]
= 1/10 / 3/5
=1/6

Two or more events are non-mutually exclusive if they can occur at the same time, meaning they have common elements.
that is, P(A∩B)≠0

here,
Events can overlap.
The occurrence of one event does not prevent the other from happening.

The word “Statistics” is derived from the italian word- statista।

source- statistics and probabiliy: an introductory analysis by siddiqur rahman, fundamentals of probability and probability distribution by manindra kumar roy, statistics by Md abdul aziz and class lecture

number of childrens is variable where rest of is constant for everytime.

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Two positive numbers: a and b
a+b=20
√ab = 8
ab = 64

from this we calcule a= 16
b= 4

given, f(x) = x/8; 0 < x < 4 ,
E(x²) = ∫₀⁴ x²f(x)dx
=8 by cal

so, E(2x² +3) = 2*8 + 3
=16+3
=19

we know, GM² = AM * HM
⇒ GM² = 9
GM = 3

Explanation:
CV = (SD / Mean) × 100 → ratio → dimensionless.

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Explanation:
Values between −1.96 and +1.96 → accept H₀.

The Region of Acceptance is the range of values of the test statistic for which we do not reject the null hypothesis (H₀).

In other words:

It is the area under the sampling distribution where the observed test statistic is not extreme enough to reject H₀.

here, total outcomes 36

the total of the numbers on the dice is greater than 8 where first dice face is 5 is 3 [(5,4), (5,5),(5,6)]

so, the probability is- 3/36
= 1/12

Outliers can dramatically inflate or deflate the Pearson correlation coefficient, giving a misleading measure of the linear relationship.

Homoscedasticity and linearity are assumptions; violating them affects interpretation but does not distort r as strongly as outliers.

Small sample size affects the stability and significance of r, but it doesn’t distort the actual value as directly as outliers do

Two non null events together can not be mutually exclusive and independents

Cause if they independents than A intersection B is not equal 0, so they are not mutually exclusive

And if they are mutually exclusive then P(A).P(B) is Greater than 0, so they are not independent

The probability that the randomly selected consumer will use at least one of the soap brands is P(A ∪ B ∪ C)= P(A) + P(B) + P(C) -P(A ∩ B)-P(B ∩ C)-P(A ∩ C)+P(A ∩ B ∩ C) = 0.5 + 0.45 + 0.4 - 0.25 - 0.1 - 0.16 + 0.08
= 0.92

So 92% use at least one of the brands,
So the chance not use any of the brand is 8%

this can solve by venn diagram also

Events A and B are independent If P(A/B)= P(A)

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Fail to reject H₀ does not mean H₀ is true.
It only means that the sample data does not provide strong enough evidence to conclude that H₀ is false.
In hypothesis testing terms:

If the test statistic lies in the region of acceptance
Or if the p-value > α
→ We fail to reject H₀.

It follows hypergeometric distribution

Formula for Hypergeometric probability:
P(X=x)=(^KC_x)(^N−KC_n−x)/(^NC_n)

Here, Total population N=20
Number of defectives K=5
Sample size n=4
We want P(X=1) probability that exactly 1 defective is chosen.

By putting values, we get
p(x=1) = 0.469
= 46%

here, P(A∩B)=2/52 (Red aces)
P(B)=26/52 (red card)
=1/2

P(A∣B)=2/52 / 26/52
=2/26
=1/13

the range of multiple correlation coefficient value is -0≤R≤1

The multiple correlation coefficient (R) measures how strongly a dependent variable (Y) is linearly related to two or more independent variables (X₁, X₂, …).
R is always non-negative (0 ≤ R ≤ 1).
Direction of the relationship (positive or negative) is not provided by R; for direction, we look at regression coefficients.

In Systematic Sampling, we first randomly select a starting point from the population list.
Then, we select every k-th member at fixed intervals.

Example: If k = 5, after picking a random start, we select every 5th member.
This method is easier and faster than Simple Random Sampling for large populations.

Every member has a known and equal chance of being selected only if the population doesn’t have a periodic pattern matching the interval.
Advantage: Saves time and effort.
Limitation: May introduce bias if there is a hidden pattern in the population.

To compare two or more distributions, polygon is the most appropriate graph Where histogram generally used in single distribution

Explanation:

The distribution function of a random variable (usually called the cumulative distribution function, or CDF) describes the probability that the variable takes a value less than or equal to a given number.
For a random variable XXX, the distribution function F(x) is defined as:

F(x)=P(X≤x),
That is — F(x) gives the total probability accumulated up to x.

here,
F(1)=P(X ≤ 1)
P(X ≤ 1) = P(X=1) + P(X=0)
= 1/2 + 1/8
= 5/8

Empirical probability (also called experimental probability) is based on actual observation or experimentation.

Toss a coin 100 times and heads appear 56 times →
P(Heads)=56/100
=0.56

This is empirical probability because it comes from an actual experiment.

These two are the properties of mean deviation

When a data set contains an outlier, we want a measure of dispersion that is not affected by extreme values.

Standard deviation (SD) → greatly affected by outliers

Coefficient of variation (CV) → depends on SD, also affected

Mean deviation → also influenced by extreme values

Interquartile Range (IQR) → based only on the middle 50% of data (Q3 – Q1), not affected by outliers

IQR (Interquartile Range) is the most appropriate and robust measure of dispersion when outliers are present.

Explanation: A continuous random variable is a type of random variable that can take any value within a given range or interval.
Its possible values are uncountably infinite, meaning they can include fractions and decimals.

Except life of an electric fan, remaining all takes finite values. Life of an electric fan takes infinite values so it is a continuous random variable.

Weighted average formula
Average return=(wA×rA)+(wB×rB)+(wC×rC)

Substitute:

=(0.25×0.05)+(0.25×0.06)+(0.50×0.02)
=0.0125+0.015+0.010
=0.0375
=3.75%

Type I error (α) = Rejecting a true null hypothesis (false positive).
By decreasing α (e.g., from 0.05 to 0.01), we make it harder to reject H₀, so the chance of committing a Type I error decreases.

note: Decreasing α reduces Type I error, but it may increase Type II error (β) because now we are more conservative and may fail to reject H₀ even when it is false.

Answer: b
Explanation: Bernoulli trial has only two possible outcomes and is mutually exclusive. Those two outcomes are ‘success’ and ‘failure’. So, it is also called as a ‘yes’ or ‘no’ question.

Arrange in ascending order
0,8,9,12,13,16,10000,
There are n=7 observations.

The first quartile (Q1) position is: n+1/ 4
= 2
Position 2 corresponds to the value 8.
So, Q1= 8

Given:

Y=20+0.8X
X = 10 + 0.9Y

So,

bY∣X=0.8
bX∣Y=0.9

Formula for correlation coefficient:

r=±√(bY∣X×bX∣Y)

Compute:

r=√(0.8×0.9)
=0.8485

Since both slopes are positive, r is also positive.

Let f(x,y) be a joint joint density function. The marginal density function of X is-
f(x)= ∫ f(x,y) dy for - ∞ < x <∞

Let f(x,y) be a joint joint density function. The marginal density function of y is-
f(y)= ∫ f(x,y) dx for - ∞ < y <∞

Climate change involves long-term rise in temperature, sea level, and salinity.
These are persistent changes over decades, not just seasonal or random fluctuations.

So, Climate change of coastal areas reflects the Trend (T) component of a time series.

A trend line is drawn by hand through the data points.
The curve is subjective, so different persons may draw slightly different curves for the same data.

Partial correlation measures the linear relationship between two variables while controlling for the effect of one or more other variables.
It essentially “removes” the influence of the controlling variable(s) to see the direct association between the two primary variables.

And partial correlation can be smaller or larger than the simple correlation, depending on how the controlling variable affects the relationship—but the main point is that it isolates the direct effect, not always larger.

Since we have several independent trials (missiles) and more than two possible outcomes (3 targets), this follows a multinomial distribution.

If:

X1= number of missiles hitting target 1
X2= number of missiles hitting target 2
X3= number of missiles hitting target 3

Each missile is equally likely to hit any target (probability 1/3 for each).
So, each missile’s hit can be represented as a random variable taking one of three outcomes:
→ Target 1 with probability 1/3​,
→ Target 2 with prob 1/3,
→ Target 3 with prob 1/3.

All missiles are independent.

For a multinomial distribution,

P(X1=x1,X2=x2,X3=x3)=n!/(x1! x2! x3!) × p1^x1 p2^x2 p3^x3

where x1+x2+x3=n.
By putting values we get,
= 20/243

If you take a large number of independent random samples of size n from any population with mean μ and variance σ2, then the sampling distribution of the sample mean Xˉ will be approximately normal for large n, even if the population itself is not normal.

The sum of squares of deviations of the observations about arithmetic mean is- least

This is one of the properties of AM

Given:
Treatments t=5 → Fertilizers A, B, C, D, E
Blocks r=4 → Soil types

Design: Randomized Block Design (RBD),

Error degrees of freedom - (t-1)(r-1)
= 4*3
=12

Total- tr-1
= 5*4- 1
=19

Bowley’s coefficient of skewness (based on quartiles) is:

Sk=(Q3+Q1−2Me)/(Q3−Q1)

Where:

Q1=4, Q3=16, Me=6

So, Sk = 8/12
= 0.67

Incorrect values: 580 and 590
Correct values: 850 and 950
Now, adjust the total sum ?

Correct total=100000−(580+590)+(850+950)
=100000−1170+1800
=100000+630
=100630

Correct average 1006.30

Latin Square Design (LSD) is used when there are two known sources of nuisance variation that need to be controlled.
In this example:

Row factor = soil type
Column factor = slope
Treatments (fertilizers) are applied so that each treatment appears once in each row and once in each column.
This design reduces experimental error due to both sources simultaneously.

Why not the others?

A) Lab mice are homogeneous → CRD is sufficient.
B) Uniform soil → No need for LSD; CRD or RBD works.
D) One classroom → Only one source of variation; LSD unnecessary.

Parameters (like μ, σ, p) are fixed values that describe the entire population.
Statistics (like x̄, s, p̂) are calculated from a sample and used to estimate the corresponding population parameters.

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the regression coefficient of Y on X:

bᵧₓ = r × (σᵧ / σₓ)
= 0.8 × (10 / 5)
= 1.6

Number of nuisance factors
Rows (soil type) → 1 nuisance factor
Columns (slope) → 1 nuisance factor
Total nuisance factors = 2

Error degrees of freedom formula in LSD
Error d.f.=(t−1)(t−2)

Here, t=5
So, 4*3
= 12

Non-parametric tests do not assume a specific distribution for the data.
They are preferred when:

The data is ordinal or ranked (e.g., Likert scale responses)
The data does not follow a normal distribution
Sample size is small and distribution is unknown

Examples of non-parametric tests:

Mann-Whitney U test
Wilcoxon signed-rank test
Chi-square test

When all numbers are equal, AM = GM = HM = that number.
So here, AM = GM = HM = 70.

Explanation:
In a hypergeometric distribution, we sample without replacement from a finite population.
Because each draw changes the composition of the population, the probability of success changes after each trial.
Hence, the trials are dependent.
Example:

Population: 20 items, 5 defective
Draw 4 items without replacement → probability of drawing a defective item changes after each draw.

The waiting time until the next event follows an Exponential distribution with rate λ=3 per hour.

We know mean and SD are same in exponentiall distribution that is- 1/λ

Here 1/3 hours
= 20 minutes

Explanation:
Statistic is derived from a sample, e.g., sample mean (x̄).

Statistic are computed from samples, while parameters describe the whole population.

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Here, this series is first 7 natural numbers
Again second central moment is variance.

We know the variance of the first n natural number is- n²-1/12
=7²-1/12
=4

Here, X̄ = 100, Ȳ = 200, slope b = 0.5, intercept a = 200- 100*0.5
= 150

predicted Y when X = 120?
Y= 150+ 120*0.5
=150+60
=210

Non-sampling errors are mistakes not caused by sampling. They arise from:

Human errors (e.g., recording, calculation mistakes)
Respondent errors (e.g., lying, misunderstanding questions)
Biases in the survey or questionnaire

we know ,
E(X) = ∑xf(x)
= 0*0.4 -4*0.2 +1*0.3+5*0.1
=-0.8+0.8
=0

The type of trial in this case is a Bernoulli trial.

Explanation:
Each donor can have two possible outcomes:

O+ (success)
Not O+ (failure)
A trial with exactly two outcomes is called a Bernoulli trial.
If you check multiple donors and count the number of O+ donors, that would follow a Binomial distribution.

Geometric – Discrete (counts number of trials until first success)
Negative Binomial – Discrete (counts trials until r-th success)
Exponential – Continuous (models time between events)

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r² = proportion of variance in Y explained by one independent variable (simple correlation).
R² = proportion of variance in Y explained by all independent variables together (multiple correlation).
When you add more predictors:

The explained variance can only stay the same or increase, never decrease.
Therefore, r² ≤ R² always.

Adding more predictors in multiple regression increases R², so it is never less than r² from a single predictor.

Let X denote the number of pregnant in the sample. Sample size, n = 100. Probability to be pregnant of a woman during family planning period is p=5/100
=0.05.
here, it follows poission distribution,
here, parameter, λ = np
=5
Probability that at least three women will be pregnant in the sample is P(X≥3)=1-P(X≤2)
=1-[f(0;5)+f(1; 5)+f(2; 5)]
=1- [ e^-55⁰/0! + e^-55¹/1! + e^-55²/2!]

= 1-[ e⁻⁵+5e⁻⁵+12.5e⁻⁵]
= 1- 18.5e⁻⁵.

Poisson distribution focuses on counting occurrences over intervals, not restricting the total number of outcomes. That’s why the sample space does not need to be finite.

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sample of size n = 5. Probability of suffering a person from gastric ulcer, p=0.24.
it follows binomial distribution,
so mean = np
=1.2

Red marbles = 3
Blue marbles = 4
Total marbles = 7
Two marbles are drawn without replacement.

after removing blue ball first, their are 6 in total.
so, getting red in 2nd ball = 3/6
=1/2

second central moment of binomial distribution describes- variance
1st central moment of binomial distribution is mean,

The second central moment measures how much the values of the random variable deviate from the mean on average.
In a Binomial distribution, it quantifies the spread of the number of successes around the expected number of successes.
So, the second central moment of a binomial distribution describes the variance of the number of successes.

Since f(x, y) is a density function,

∫₀¹∫₀¹ f(x, y) dxdy =1
⇒ ∫∫ k(x² + 2xy) dxdy =1
> k ∫ [∫ k(x2 + 2xy) dy]dx = 1
> k ∫ [x²y + xy²]₀¹dx =1
>k ∫ [x2 + x]dx =1
>k [x³/3 + x²/2]₀¹ = 1
>k * 5/6 = 1
>k = 6/5

Solution : Let us define the following events be as:
B: The selected item selected is defective
A1: The selected item was produced by machine M₁
A2: The selected item was produced by machine M₂
A3: The selected item was produced by machine M3.

We have,
P(A1) = 0.25, P(B/A1) = 5/100=0.05
P(A2) = 0.35, P(B/A2) = 4/100=0.04
P(A3) = 0.40, P(B/A3) =2/100= 0.02

NOW,

The probability that the selected defective item produced by machine M1 or M2

P( A1 ∪ A2| B)=P(A1 |B)+P(A2|B)
= P(A1) P(B|A 1 )+P(A2 )P(B|A2) / [P(A1)P(B|A1)+P(A2)P(B|A2 )+P(A3)P(B|A3)]

The probability = [(0.25)(0.05) + (0.35)(0.04)] / [(0.25)(0.05) + (0.35)(0.04) + (0.4)(0.02)]
= 0.0265/0.0345 = 0.7681

Answer: d
here, 5 red, 4 blue, and 3 green ball,
total = 12 ball

p(no red) = 7/12

total outcomes =10c3 = 120
Ways to choose 3 blue balls = 4c3 = 4

probability of getting 3 blue = 4/120
=1/30

Answer: c
Explanation: The probability of drawing a four = 4/52
The probability of drawing a queen = 4/52

Total probability = 4/52×4/52
=1/169

Answer: c
Explanation: Probability of pink = 1/4
Probability of choosing pink pair of gloves twice = 1/4×1/4
=1/16