Statistics: Definition, Importance, and Computations

Part 1: Definition of Statistics (2 marks)

Statistics is the branch of science that deals with the collection, organization, presentation, analysis, and interpretation of numerical data in order to draw valid conclusions and make rational decisions under conditions of uncertainty.

Part 2: Importance of Statistics in Computational Sciences (3 marks)

1. Fast computation: Computers perform statistical operations (mean, variance, ANOVA, regression) rapidly and reliably, removing tedious hand calculation.
2. Handling large/multivariate data: Complex methods like multivariate analysis, clustering, and linear programming become feasible.
3. Basis of modern techniques: Data mining, machine learning, pattern recognition, bootstrap methods, and image analysis are inherently statistical and computational.
4. Research and reliability: Enables reproducible, high-speed experimental research and simulation.
5. Practical CS/IT problem solving: Reliability testing (e.g., expected lifetime of hardware), performance benchmarking, and predictive modeling all depend on statistics.

Part 3: Statistical Computations (5 marks)

Given data (n = 20)

$$15, 10, 2, 17, 5, 8, 3, 10, 2, 9, 5, 9, 13, 1, 10, 12, 5, 10, 8, 4$$

Sorted: $$1, 2, 2, 3, 4, 5, 5, 5, 8, 8, 9, 9, 10, 10, 10, 10, 12, 13, 15, 17$$

Mean

$$\sum X = 158$$ $$\bar{X} = \frac{158}{20} = 7.9 \text{ minutes}$$

Median (n even → average of 10th and 11th values)

10th = 8, 11th = 9 $$\text{Median} = \frac{8+9}{2} = 8.5 \text{ minutes}$$

Mode

Value 10 occurs 4 times (highest frequency). $$\text{Mode} = 10 \text{ minutes}$$

Variance and Standard Deviation

$$\sum X^2 = 1626$$

$$\sigma^2 = \frac{\sum X^2}{n} - \bar{X}^2 = \frac{1626}{20} - (7.9)^2$$ $$= 81.3 - 62.41 = 18.89$$

$$\sigma = \sqrt{18.89} = 4.3463 \approx 4.35 \text{ minutes}$$

(Note: if the sample formula with $n-1$ is used: $$s^2 = \frac{\sum X^2 - n\bar{X}^2}{n-1} = \frac{1626 - 1248.2}{19} = \frac{377.8}{19} = 19.8842$$ $$s = 4.4592 \text{ minutes})$$

Coefficient of Variation

$$CV = \frac{\sigma}{\bar{X}} \times 100 = \frac{4.3463}{7.9} \times 100 = 55.02%$$

(Using sample SD: $CV = \frac{4.4592}{7.9}\times100 = 56.45%$)

Final Summary

Normal Distribution: Definition, Characteristics, and Numerical

Definition

A normal distribution is a continuous probability distribution, symmetrical and bell-shaped about its mean, defined by two parameters, mean $\mu$ and standard deviation $\sigma$. Its probability density function is:

$$f(x) = \frac{1}{\sigma\sqrt{2\pi}} e^{-\frac{1}{2}\left(\frac{x-\mu}{\sigma}\right)^2}, \quad -\infty < x < \infty$$

Main Characteristics

1. It is a continuous distribution with parameters $\mu$ (mean) and $\sigma^2$ (variance).
2. The curve is bell-shaped and symmetrical about the mean $\mu$.
3. Mean = Median = Mode (all coincide at the centre).
4. Maximum ordinate occurs at $x = \mu$, equal to $\frac{1}{\sigma\sqrt{2\pi}}$.
5. Skewness $\beta_1 = 0$ (perfectly symmetric).
6. Kurtosis $\beta_2 = 3$ (mesokurtic).
7. Empirical area rule:
   • $P(\mu-\sigma < X < \mu+\sigma) = 0.6826$
   • $P(\mu-2\sigma < X < \mu+2\sigma) = 0.9544$
   • $P(\mu-3\sigma < X < \mu+3\sigma) = 0.9974$
8. Total area under the curve = 1.
9. Asymptotic: the curve never touches the X-axis.

Given Data

• Mean $\mu = 5$ inches
• Standard deviation $\sigma = 0.05$ inches
• $X \sim N(5, 0.05^2)$
• Standard normal variate: $Z = \dfrac{X - 5}{0.05}$

Part (i): Proportion exceeding tolerance limits 4.9 to 5.1 inches

Z-scores:

$$Z_1 = \frac{4.9 - 5}{0.05} = -2, \qquad Z_2 = \frac{5.1 - 5}{0.05} = +2$$

Proportion within limits:

$$P(4.9 < X < 5.1) = P(-2 < Z < 2) = 2 \times P(0 < Z < 2) = 2 \times 0.4772 = 0.9544$$

Proportion exceeding limits:

$$P(X < 4.9 \text{ or } X > 5.1) = 1 - 0.9544 = \boxed{0.0456 \ (4.56%)}$$

Part (ii): Proportion having length greater than 6.5 inches

Z-score:

$$Z = \frac{6.5 - 5}{0.05} = \frac{1.5}{0.05} = 30$$

Since $Z = 30$ lies far beyond the standard normal table (which effectively ends near $Z \approx 4$):

$$P(X > 6.5) = P(Z > 30) \approx \boxed{0}$$

A rod exceeding 6.5 inches is 30 standard deviations above the mean, so the proportion is practically zero.

Summary

Difference Between Correlation and Regression Analysis

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Given Data

$n = 10$

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Preliminary Calculations

$$\bar{X}=\frac{446}{10}=44.6,\qquad \bar{Y}=\frac{112}{10}=11.2$$

$$\Sigma X^2-n\bar X^2 = 20834-10(44.6)^2=20834-19891.6=942.4$$ $$\Sigma Y^2-n\bar Y^2 = 1318-10(11.2)^2=1318-1254.4=63.6$$ $$\Sigma XY-n\bar X\bar Y = 5210-10(44.6)(11.2)=5210-4995.2=214.8$$

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Part (i): Correlation Coefficient

$$r=\frac{\Sigma XY-n\bar X\bar Y}{\sqrt{(\Sigma X^2-n\bar X^2)(\Sigma Y^2-n\bar Y^2)}}=\frac{214.8}{\sqrt{942.4\times63.6}}$$

$$r=\frac{214.8}{\sqrt{59916.64}}=\frac{214.8}{244.78}\approx \boxed{0.877}$$

Interpretation: $r=0.877$ shows a strong positive linear relationship. As humidity increases, moisture content also increases.

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Part (ii): Regression Equation of Y on X

$$b_{YX}=\frac{\Sigma XY-n\bar X\bar Y}{\Sigma X^2-n\bar X^2}=\frac{214.8}{942.4}=0.2279\approx 0.228$$

$$a=\bar Y-b\bar X=11.2-0.228(44.6)=11.2-10.169=1.031$$

$$\boxed{\hat Y = 1.03 + 0.228X}$$

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Part (iii): Estimate Y when X = 45

$$\hat Y = 1.03 + 0.228(45)=1.03+10.26=\boxed{11.29%}$$

When humidity is 45%, estimated moisture content ≈ 11.29%.

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Part (iv): Interpretation of Regression Coefficient

$b=0.228$ means that for every 1 percentage point increase in relative humidity, the moisture content increases on average by about 0.228 percentage points. The positive sign confirms a direct relationship: higher storage humidity leads to higher moisture in the raw materials.

Sampling: Definition, Purpose, and Stratified Sampling

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1. What is Sampling?

Sampling is the process of selecting a subset (called a sample) of units from a larger group (called the population) in order to draw conclusions or make inferences about the entire population. Instead of studying every unit in the population (census), sampling allows us to study only a representative portion, saving time, cost, and effort.

The difference between the value of the sample statistic obtained from a sample and the value of the corresponding population parameter is called the sampling error.

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2. Main Purpose of Sampling

The main purposes of sampling are:

• Economy: Studying the entire population is costly; sampling reduces expenditure of money, time, and manpower.
• Feasibility: In many cases, examining every unit is practically impossible (e.g., testing every light bulb for lifespan).
• Speed: Results can be obtained more quickly from a sample than from a complete census.
• Accuracy: A carefully conducted sample survey can sometimes be more accurate than a full census, since more attention can be given to each selected unit.
• Defining survey objectives: The first step in sampling is to clearly define the objective of the survey, commensurate with available resources in terms of money, manpower, and time.

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3. Stratified Random Sampling

Definition

When the units in the population are not similar in nature, the population is first divided into sub-groups called strata, and then a simple random sample is drawn from each stratum in proportion to its size. This technique is known as Stratified Random Sampling.

Conditions for Stratification

The strata must satisfy the following conditions:

• The strata should be non-overlapping and together should comprise the whole population.
• The strata should be as homogeneous within groups (similar units inside each stratum) and heterogeneous between groups (different units across strata) as possible.

Purpose of Stratification

According to the notes, the purposes of stratification are:

1. To make the sample more representative of the population.
2. For greater accuracy in the results.
3. For administrative convenience in conducting the survey.

Example

Suppose we want to study the income level of people in a city. The population can be stratified into groups such as:

• Low income group
• Middle income group
• High income group

A random sample is then drawn from each group separately, ensuring all income levels are properly represented in the final sample.

Advantage over Simple Random Sampling

For a given level of precision, stratified random sampling usually requires a smaller sample size as compared to simple random sampling, making it more efficient.

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Summary Table:

Partition Values and Computation

Definition

Partition values are values that divide a distribution (arranged in order) into a number of equal parts. Common ones:

• Quartiles ($Q$): divide data into 4 equal parts
• Deciles ($D$): divide data into 10 equal parts
• Percentiles ($P$): divide data into 100 equal parts

Given Data (Cumulative Frequency Table)

$N = 200$, class width $h = 10$.

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Part (i): Minimum Score of Top 10% Students

Top 10% lie above $P_{90}$, so the minimum score of top 10% = $P_{90}$.

$$\frac{90N}{100} = \frac{90 \times 200}{100} = 180\text{th item}$$

Since $169 < 180 \le 189$, the $P_{90}$ class is 70-80.

• $L = 70,\ cf = 169,\ f = 20,\ h = 10$

$$P_{90} = 70 + \frac{180 - 169}{20} \times 10 = 70 + \frac{11}{20}\times 10 = 70 + 5.5 = \boxed{75.5}$$

Minimum score of top 10% students = 75.5

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Part (ii): Range of Middle 60% Students

Middle 60% lies between $P_{20}$ and $P_{80}$.

Calculation of $P_{20}$:

$$\frac{20 \times 200}{100} = 40\text{th item}$$

Since $14 < 40 \le 64$, $P_{20}$ class is 40-50.

$$P_{20} = 40 + \frac{40 - 14}{50}\times 10 = 40 + \frac{26}{50}\times 10 = 40 + 5.2 = 45.2$$

Calculation of $P_{80}$:

$$\frac{80 \times 200}{100} = 160\text{th item}$$

Since $124 < 160 \le 169$, $P_{80}$ class is 60-70.

$$P_{80} = 60 + \frac{160 - 124}{45}\times 10 = 60 + \frac{36}{45}\times 10 = 60 + 8 = 68$$

Range of middle 60%:

$$P_{80} - P_{20} = 68 - 45.2 = \boxed{22.8}$$

Range of the middle 60% of students = 22.8 marks

Bayes' Theorem - Definition and Application

STEP 1: Given Data

Resignations equally likely among all employees. Observed event: the resigning employee is a woman. Find $P(\text{Store B} \mid \text{Woman})$.

Definition of Bayes' Theorem

If $B_1, B_2, \dots, B_n$ are mutually exclusive and exhaustive events with $P(B_i) > 0$, and $A$ is any event with $P(A) > 0$, then:

$$P(B_i \mid A) = \frac{P(B_i), P(A \mid B_i)}{\sum_{j=1}^{n} P(B_j), P(A \mid B_j)}$$

• $P(B_i)$ = prior probability
• $P(A \mid B_i)$ = likelihood
• $P(B_i \mid A)$ = posterior probability

STEP 2: Solution

Define Events

• $S_A, S_B, S_C$ = employee works in Store A, B, C
• $W$ = resigning employee is a woman

Prior Probabilities (equal likelihood over 225 employees)

$$P(S_A) = \frac{100}{225}, \quad P(S_B) = \frac{75}{225}, \quad P(S_C) = \frac{50}{225}$$

Likelihoods

$$P(W \mid S_A) = 0.70, \quad P(W \mid S_B) = 0.60, \quad P(W \mid S_C) = 0.50$$

Total Probability of a Woman Resigning

$$P(W) = \frac{100}{225}(0.70) + \frac{75}{225}(0.60) + \frac{50}{225}(0.50)$$

$$P(W) = \frac{70 + 45 + 25}{225} = \frac{140}{225}$$

Apply Bayes' Theorem

$$P(S_B \mid W) = \frac{P(S_B), P(W \mid S_B)}{P(W)} = \frac{\dfrac{75}{225}(0.60)}{\dfrac{140}{225}} = \frac{45}{140}$$

$$\boxed{P(S_B \mid W) = \frac{45}{140} = \frac{9}{28} \approx 0.3214 \approx 32.14%}$$

Conclusion

The probability that the resigning woman works in Store B is approximately 0.3214 (32.14%).

Poisson Probability Distribution

Definition

The Poisson distribution is a discrete probability distribution that describes the number of events occurring in a fixed interval of time (or space), given that these events occur with a known constant average rate $\lambda$ and independently of each other. It is a limiting case of the binomial distribution when $n \to \infty$, $p \to 0$, while $np = \lambda$ stays finite.

Probability mass function:

$$P(X = x) = \frac{e^{-\lambda},\lambda^x}{x!}, \quad x = 0, 1, 2, 3, \ldots$$

where $\lambda$ is the mean number of occurrences, and $\text{mean} = \text{variance} = \lambda$.

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Given Data

• Average arrival rate = $3$ cars per minute
• $\lambda = 3$ (per minute)

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(i) No car arrives during a particular minute

$x = 0$, $\lambda = 3$:

$$P(X = 0) = \frac{e^{-3},3^0}{0!} = e^{-3} \approx 0.0498$$

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(ii) At least one car arrives during a particular minute

$$P(X \geq 1) = 1 - P(X = 0) = 1 - e^{-3}$$

$$P(X \geq 1) = 1 - 0.0498 \approx 0.9502$$

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(iii) Four cars arrive in any 2 minutes

For a 2-minute interval, the mean scales:

$$\lambda' = 3 \times 2 = 6$$

With $x = 4$:

$$P(X = 4) = \frac{e^{-6},6^4}{4!} = \frac{e^{-6}\times 1296}{24}$$

$$= \frac{0.00247875 \times 1296}{24} = \frac{3.2124}{24} \approx 0.1339$$

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Summary