Regression is a statistical technique used to model and analyze the relationship between a dependent variable and one or more independent variables. It enables estimation or prediction of the dependent variable's value from known values ...

Model Answer: Build System Consistency Analysis

Given Data

Build System A (MB): 88, 92, 94, 85, 90, 95, 89, 87, 91, 86 Build System B (MB): 130, 135, 132, 140, 128, 133, 137, 131, 129, 138 $n = 10$ for each

Method

Since the two data sets have different means (different units/scale of build sizes), the correct measure of relative consistency is the Coefficient of Variation (CV). The lower the CV, the more consistent.

Step 1: Means

System A: $$\bar{x}_A = \frac{88+92+94+85+90+95+89+87+91+86}{10} = \frac{897}{10} = 89.7 \text{ MB}$$

System B: $$\bar{x}_B = \frac{130+135+132+140+128+133+137+131+129+138}{10} = \frac{1333}{10} = 133.3 \text{ MB}$$

Step 2: Variance and Standard Deviation

System A - squared deviations from 89.7:

$$\sum (x-\bar{x})^2 = 100.10$$ $$\sigma_A^2 = \frac{100.10}{10} = 10.01, \qquad \sigma_A = \sqrt{10.01} \approx 3.164 \text{ MB}$$

System B - squared deviations from 133.3:

$$\sum (x-\bar{x})^2 = 148.10$$ $$\sigma_B^2 = \frac{148.10}{10} = 14.81, \qquad \sigma_B = \sqrt{14.81} \approx 3.849 \text{ MB}$$

Step 3: Coefficient of Variation

$$CV_A = \frac{\sigma_A}{\bar{x}_A}\times 100 = \frac{3.164}{89.7}\times 100 \approx 3.53%$$

$$CV_B = \frac{\sigma_B}{\bar{x}_B}\times 100 = \frac{3.849}{133.3}\times 100 \approx 2.89%$$

Conclusion

Consistency between two data sets with different means must be judged by the coefficient of variation, not the raw standard deviation.

• $CV_B \approx 2.89% < CV_A \approx 3.53%$

Therefore Build System B is more consistent, because relative to its own average build size its build sizes vary less. Although System A has a smaller absolute standard deviation, its builds are much smaller in magnitude, so a fair (relative) comparison shows System B fluctuates proportionally less around its mean.

Compare with the coefficient of variation, not the raw standard deviation: when the means differ substantially (89.7 against 133.3), the CV is the statistically correct basis, and it points to System B as the more consistent system.

1. Symmetry: The curve is perfectly symmetric about the mean $\mu$; the two halves are mirror images. 2. Bell-shaped: The probability density function is bell-shaped with its peak at $x = \mu$. 3. Mean = Median = Mode = $\mu$: All three ...

Model Answer: Equipment Reliability

Step 1 - EXTRACT: Given Data

• Equipment functions only when all of A, B, C are working.
• $P(A \text{ fails}) = 0.15$
• $P(B \text{ fails}) = 0.05$
• $P(C \text{ fails}) = 0.10$
• Component failures assumed independent.
• Find: $P(\text{equipment does not fail in one year})$

Step 2 - SOLVE

Probability each component does NOT fail:

$$P(A \text{ works}) = 1 - 0.15 = 0.85$$ $$P(B \text{ works}) = 1 - 0.05 = 0.95$$ $$P(C \text{ works}) = 1 - 0.10 = 0.90$$

Condition: Equipment does not fail only when all three components work.

Using the multiplication rule for independent events:

$$P(\text{no failure}) = P(A) \cdot P(B) \cdot P(C) = 0.85 \times 0.95 \times 0.90$$

$$= 0.8075 \times 0.90 = 0.72675$$

Final Answer

$$\boxed{P(\text{equipment does not fail}) = 0.72675 \approx 0.7268 ;(72.68%)}$$

The value is 0.7268 (exactly 0.72675).

• Distribution: Poisson - Mean rate: $\lambda = 5$ requests per minute - Interval: 1 minute - (a) Find $P(X = 3)$ - (b) Find $P(X 2)$ Poisson formula: $$P(X = k) = \frac{e^{-\lambda}\lambda^k}{k!}, \quad \lambda = 5$$ Note: $e^{-5} \appr...

Model Answer: Build System Consistency Analysis

Given Data

Build System A (MB): 88, 92, 94, 85, 90, 95, 89, 87, 91, 86 Build System B (MB): 130, 135, 132, 140, 128, 133, 137, 131, 129, 138 $n = 10$ for each

Method

Since the two data sets have different means (different units/scale of build sizes), the correct measure of relative consistency is the Coefficient of Variation (CV). The lower the CV, the more consistent.

Step 1: Means

System A: $$\bar{x}_A = \frac{88+92+94+85+90+95+89+87+91+86}{10} = \frac{897}{10} = 89.7 \text{ MB}$$

System B: $$\bar{x}_B = \frac{130+135+132+140+128+133+137+131+129+138}{10} = \frac{1333}{10} = 133.3 \text{ MB}$$

Step 2: Variance and Standard Deviation

System A - squared deviations from 89.7:

$$\sum (x-\bar{x})^2 = 100.10$$ $$\sigma_A^2 = \frac{100.10}{10} = 10.01, \qquad \sigma_A = \sqrt{10.01} \approx 3.164 \text{ MB}$$

System B - squared deviations from 133.3:

$$\sum (x-\bar{x})^2 = 148.10$$ $$\sigma_B^2 = \frac{148.10}{10} = 14.81, \qquad \sigma_B = \sqrt{14.81} \approx 3.849 \text{ MB}$$

Step 3: Coefficient of Variation

$$CV_A = \frac{\sigma_A}{\bar{x}_A}\times 100 = \frac{3.164}{89.7}\times 100 \approx 3.53%$$

$$CV_B = \frac{\sigma_B}{\bar{x}_B}\times 100 = \frac{3.849}{133.3}\times 100 \approx 2.89%$$

Conclusion

Consistency between two data sets with different means must be judged by the coefficient of variation, not the raw standard deviation.

• $CV_B \approx 2.89% < CV_A \approx 3.53%$

Therefore Build System B is more consistent, because relative to its own average build size its build sizes vary less. Although System A has a smaller absolute standard deviation, its builds are much smaller in magnitude, so a fair (relative) comparison shows System B fluctuates proportionally less around its mean.

Compare with the coefficient of variation, not the raw standard deviation: when the means differ substantially (89.7 against 133.3), the CV is the statistically correct basis, and it points to System B as the more consistent system.

Model Answer: Equipment Reliability

Step 1 - EXTRACT: Given Data

• Equipment functions only when all of A, B, C are working.
• $P(A \text{ fails}) = 0.15$
• $P(B \text{ fails}) = 0.05$
• $P(C \text{ fails}) = 0.10$
• Component failures assumed independent.
• Find: $P(\text{equipment does not fail in one year})$

Step 2 - SOLVE

Probability each component does NOT fail:

$$P(A \text{ works}) = 1 - 0.15 = 0.85$$ $$P(B \text{ works}) = 1 - 0.05 = 0.95$$ $$P(C \text{ works}) = 1 - 0.10 = 0.90$$

Condition: Equipment does not fail only when all three components work.

Using the multiplication rule for independent events:

$$P(\text{no failure}) = P(A) \cdot P(B) \cdot P(C) = 0.85 \times 0.95 \times 0.90$$

$$= 0.8075 \times 0.90 = 0.72675$$

Final Answer

$$\boxed{P(\text{equipment does not fail}) = 0.72675 \approx 0.7268 ;(72.68%)}$$

The value is 0.7268 (exactly 0.72675).

Regression is a statistical technique used to model and analyze the relationship between a dependent variable and one or more independent variables. It enables estimation or prediction of the dependent variable's value from known values ...

1. Symmetry: The curve is perfectly symmetric about the mean $\mu$; the two halves are mirror images. 2. Bell-shaped: The probability density function is bell-shaped with its peak at $x = \mu$. 3. Mean = Median = Mode = $\mu$: All three ...

Model Answer: Correlation and Regression Analysis

Given Data

$n = 10$

STEP 1: Compute the sums

$\sum X = 375,\ \sum Y = 132,\ \sum XY = 5820,\ \sum X^2 = 16125,\ \sum Y^2 = 2112$

a) Pearson Correlation Coefficient

$$r = \frac{n\sum XY - \sum X \sum Y}{\sqrt{[n\sum X^2 - (\sum X)^2][n\sum Y^2 - (\sum Y)^2]}}$$

Compute components:

• $n\sum XY = 10 \times 5820 = 58200$
• $\sum X \sum Y = 375 \times 132 = 49500$
• $n\sum X^2 = 161250,\quad (\sum X)^2 = 140625 \Rightarrow 20625$
• $n\sum Y^2 = 21120,\quad (\sum Y)^2 = 17424 \Rightarrow 3696$

$$r = \frac{58200 - 49500}{\sqrt{20625 \times 3696}} = \frac{8700}{\sqrt{76245000}}$$

$$\sqrt{76245000} = 8731.95$$

$$r = \frac{8700}{8731.95} = 0.9963$$

$r \approx 0.996$: very strong positive correlation.

b) Regression Equation of Y on X

$$b = \frac{n\sum XY - \sum X \sum Y}{n\sum X^2 - (\sum X)^2} = \frac{8700}{20625} = 0.42182$$

Means:

$$\bar{X} = \frac{375}{10} = 37.5,\qquad \bar{Y} = \frac{132}{10} = 13.2$$

Intercept:

$$a = \bar{Y} - b\bar{X} = 13.2 - (0.42182)(37.5) = 13.2 - 15.818 = -2.618$$

$$\boxed{\hat{Y} = -2.618 + 0.4218,X}$$

c) Prediction for X = 38 hours

$$\hat{Y} = -2.618 + 0.4218(38) = -2.618 + 16.029 = 13.41$$

Approximately 13 issues resolved (13.4 if retaining decimals).

Since $r = 0.996$ and $X = 38$ lies inside the observed range $[15,60]$, the prediction is reliable.

Verified Model Answer: Expected Value and Variance

STEP 1 - Given Data

Probability distribution of $Y$ (bug reports closed per hour):

Check: $\sum P(Y) = 0.10 + 0.18 + 0.32 + 0.30 + 0.10 = 1.00$ ✓ (valid distribution)

STEP 2 - Solve

Part 1: Expected Value $E(Y)$

$$E(Y) = \sum Y \cdot P(Y)$$

$$E(Y) = 0 + 0.18 + 0.64 + 0.90 + 0.40 = 2.12 \text{ bug reports/hour}$$

Part 2: Variance $\text{Var}(Y)$

Compute $E(Y^2) = \sum Y^2 \cdot P(Y)$:

$$E(Y^2) = 5.76$$

Apply the variance formula:

$$\text{Var}(Y) = E(Y^2) - [E(Y)]^2 = 5.76 - (2.12)^2$$

$$(2.12)^2 = 4.4944$$

$$\text{Var}(Y) = 5.76 - 4.4944 = 1.2656 \approx 1.27$$

Final Results

• Expected number of bug reports closed per hour: $E(Y) = 2.12$
• Variance: $\text{Var}(Y) = 1.2656 \approx 1.27$
• (Standard deviation: $\sigma = \sqrt{1.2656} \approx 1.125$)

Model Answer: Probability of At Least 80% Watching Entertainment Channel

STEP 1 - Given Data

• Population proportion watching entertainment channel: $p = 0.80$
• Probability of not watching: $q = 1 - p = 0.20$
• Sample size: $n = 5$
• Required: $P(\text{at least } 80% \text{ of sample watch})$

STEP 2 - Solve

Distribution: Let $X$ = number of viewers (out of 5) who watch the entertainment channel. Since each viewer independently watches with probability $p = 0.80$, $X$ follows a Binomial distribution:

$$P(X = k) = \binom{5}{k}(0.80)^k(0.20)^{5-k}$$

Interpret "at least 80% of 5": $$0.80 \times 5 = 4 \text{ viewers}$$

So we need $P(X \geq 4) = P(X = 4) + P(X = 5)$.

For $X = 4$: $$P(X = 4) = \binom{5}{4}(0.80)^4(0.20)^1 = 5 \times 0.4096 \times 0.20$$ $$= 5 \times 0.08192 = 0.4096$$

For $X = 5$: $$P(X = 5) = \binom{5}{5}(0.80)^5(0.20)^0 = 1 \times 0.32768 \times 1 = 0.32768$$

Sum: $$P(X \geq 4) = 0.4096 + 0.32768 = 0.73728$$

Final Answer

$$\boxed{P(X \geq 4) \approx 0.7373 \text{ or } 73.73%}$$

The probability that at least 80% of the viewers in a random sample of five watch the entertainment channel is approximately 0.737 (73.7%).

• Distribution: Poisson - Mean rate: $\lambda = 5$ requests per minute - Interval: 1 minute - (a) Find $P(X = 3)$ - (b) Find $P(X 2)$ Poisson formula: $$P(X = k) = \frac{e^{-\lambda}\lambda^k}{k!}, \quad \lambda = 5$$ Note: $e^{-5} \appr...

• Sample size: $n = 50$ - $\Sigma x = 525$ - $\Sigma x^2 = 5625$ - Confidence level: 99% $$\bar{x} = \frac{\Sigma x}{n} = \frac{525}{50} = 10.5 \text{ km/liter}$$ Using the sample estimate of variance (with $n$ divisor, common at this le...

Model Answer: Box and Whisker Plot, Parameter and Statistic

a) Use of Box and Whisker Plot

A box and whisker plot (or boxplot) is a graphical method used to display the distribution and spread of numerical data. Its main uses include:

• Visual Summary of Data: Displays five-number summary - minimum, Q1 (first quartile), median (Q2), Q3 (third quartile), and maximum values in a single diagram.

• Identifying Outliers: Points that fall beyond 1.5 × IQR (Interquartile Range) from Q1 or Q3 are plotted separately as outliers, making them easily identifiable.

• Comparing Distributions: Multiple boxplots can be drawn side-by-side to compare the central tendency, spread, and skewness of different datasets or groups.

• Assessing Skewness: The position of the median line within the box and the length of whiskers indicate whether data is symmetrically distributed or skewed.

• Understanding Data Spread: The box width (IQR) shows where the middle 50% of data lies, while whiskers extend to show the range of typical values.

Example: In quality control, boxplots help compare product measurements across different batches to identify inconsistencies.

b) Parameter and Statistic

Key Relationship: Statistics are used as estimators of population parameters. For instance, the sample mean (x̄) is used to estimate the population mean (μ).

Model Answer: Fourth Central Moment and Kurtosis

STEP 1 - Given Data

• Standard deviation: $\sigma = 9$
• Distribution is symmetric
• Required: fourth central moment $\mu_4$ for the distribution to be (i) mesokurtic, (ii) platykurtic, (iii) leptokurtic

STEP 2 - Solution

Key relation (coefficient of kurtosis):

$$\beta_2 = \frac{\mu_4}{\mu_2^2} = \frac{\mu_4}{\sigma^4}$$

so $\mu_4 = \beta_2 \cdot \sigma^4$.

Compute $\sigma^4$:

$$\sigma^2 = 9^2 = 81$$ $$\sigma^4 = 81^2 = 6561$$

Kurtosis criteria:

• Mesokurtic: $\beta_2 = 3$
• Platykurtic: $\beta_2 < 3$
• Leptokurtic: $\beta_2 > 3$

(i) Mesokurtic ($\beta_2 = 3$)

$$\mu_4 = 3 \times 6561 = 19683$$

(ii) Platykurtic ($\beta_2 < 3$)

$$\mu_4 < 3 \times 6561$$ $$\mu_4 < 19683$$ (e.g. $\mu_4 = 15000$)

(iii) Leptokurtic ($\beta_2 > 3$)

$$\mu_4 > 3 \times 6561$$ $$\mu_4 > 19683$$ (e.g. $\mu_4 = 25000$)

Summary

• Number of employees: $N = 5000$ - Distribution: Normal - Mean: $\mu = 2000$ - Standard deviation: $\sigma = 200$ Standardization: $Z = \dfrac{X - \mu}{\sigma}$, so $X = \mu + Z\sigma$. --- Middle 60% leaves 20% in each tail. - Lower cu...

Differentiate Pareto Chart and Bar Diagram

Definition and Purpose

Bar Diagram:

• A basic statistical graph that displays data using rectangular bars of equal width
• Bars are arranged side-by-side or in groups
• Used to compare quantities across different categories
• Primarily for simple data comparison and visualization

Pareto Chart:

• A specialized bar chart combined with a line graph (cumulative percentage curve)
• Based on the Pareto principle (80/20 rule)
• Bars are arranged in descending order of frequency/magnitude
• Used to identify the most significant factors contributing to a problem

Key Differences

Example Context

• Bar Diagram: Comparing sales of different products in a month
• Pareto Chart: Identifying which product defects account for 80% of quality issues

Conclusion

While a bar diagram is a general-purpose tool for data comparison, a Pareto chart is a specialized analytical tool designed to prioritize problems and identify the most impactful factors in quality management and process improvement.

Variable Measurement Scale Best Statistical Method ----------------------------------------------------- Blood group Nominal Mode, Chi-square test, Frequency distribution Students' satisfaction (5-point Likert scale) Ordinal Median, Mode...

• Selection proportion for A : B : C = 4 : 2 : 3 - $P(D \mid A) = 0.3$ - $P(D \mid B) = 0.5$ - $P(D \mid C) = 0.8$ where $D$ = "democratization is introduced". Total parts $= 4 + 2 + 3 = 9$ $$P(A) = \frac{4}{9}, \quad P(B) = \frac{2}{9},...