STA154 · TU past paper

Basic Statistics 2081 question paper

The complete TU 2081 exam paper for Basic Statistics (STA154), all 12 questions with solved model answers written to the mark scheme.

Past Papers2082 2081 2080.1 2080 2079 2078

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110 marksNumericalComparison of consistency between datasetsHideAnswer
Response Time Consistency Comparison

A cloud service provider is evaluating the response times of two different server clusters over 10 requests. The company wants to compare the variability in response times between the two clusters. Here are the response times (in milliseconds) for the two clusters. Which cluster has more consistent response times?

$\begin{array}{|c|cccccccccc|}\hline \text{Cluster A} & 88 & 92 & 94 & 85 & 90 & 95 & 89 & 87 & 91 & 86 \\ \hline \text{Cluster B} & 130 & 135 & 132 & 140 & 128 & 133 & 137 & 131 & 129 & 138 \\ \hline \end{array}$

[10]
Response Time Consistency Comparison

A cloud service provider is evaluating the response times of two different server clusters over 10 requests. The company wants to compare the variability in response times between the two clusters. Here are the response times (in milliseconds) for the two clusters. Which cluster has more consistent response times?

$$\begin{array}{|c|cccccccccc|}\hline \text{Cluster A} & 88 & 92 & 94 & 85 & 90 & 95 & 89 & 87 & 91 & 86 \ \hline \text{Cluster B} & 130 & 135 & 132 & 140 & 128 & 133 & 137 & 131 & 129 & 138 \ \hline \end{array}$$

[10]
Cluster A (ms): 88, 92, 94, 85, 90, 95, 89, 87, 91, 86 (n = 10) Cluster B (ms): 130, 135, 132, 140, 128, 133, 137, 131, 129, 138 (n = 10) Since the two clusters have very different means, the correct measure to compare consistency is the...
210 marksNumericalCentral moments and raw momentsHideAnswer
Compute four central moments and hence find the mean, standard deviation, the measure of skewness, and the measure of kurtosis from central moments. Also, comment on the nature of the data. 120, 135, 150, 125, 140, 160, 170, 155, 165, 150 [10]
Data values (n = 10): $120,\ 135,\ 150,\ 125,\ 140,\ 160,\ 170,\ 155,\ 165,\ 150$ $\bar{x} = \frac{\sum xi}{n} = \frac{120+135+150+125+140+160+170+155+165+150}{10} = \frac{1470}{10} = 147$ $xi$ $di$ $di^2$ $di^3$ $di^4$ -------------...
Data values (n = 10): $$120,\ 135,\ 150,\ 125,\ 140,\ 160,\ 170,\ 155,\ 165,\ 150$$ $$\bar{x} = \frac{\sum xi}{n} = \frac{120+135+150+125+140+160+170+155+165+150}{10} = \frac{1470}{10} = 147$$ $xi$ $di$ $di^2$ $di^3$ $di^4$ -------------...
310 marksNumericalKarl Pearson correlation coefficientHideAnswer
Question

A software company tracks the number of hours its employees spend on coding (X) and the corresponding number of bugs they encounter during testing (Y).

$\begin{array}{|c|cccccccccc|}\hline \text{Hours (X)} & 20 & 25 & 30 & 35 & 40 & 45 & 50 & 55 & 60 & 65 \\ \hline \text{Bugs (Y)} & 5 & 7 & 8 & 12 & 15 & 13 & 17 & 18 & 20 & 25 \\ \hline \end{array}$

(a) Calculate the Pearson correlation coefficient.

(b) Find the regression equation of Y on X.

(c) Predict the number of bugs encountered if an employee spends 48 hours coding.

[10+0]
Question

A software company tracks the number of hours its employees spend on coding (X) and the corresponding number of bugs they encounter during testing (Y).

$$\begin{array}{|c|cccccccccc|}\hline \text{Hours (X)} & 20 & 25 & 30 & 35 & 40 & 45 & 50 & 55 & 60 & 65 \ \hline \text{Bugs (Y)} & 5 & 7 & 8 & 12 & 15 & 13 & 17 & 18 & 20 & 25 \ \hline \end{array}$$

(a) Calculate the Pearson correlation coefficient.

(b) Find the regression equation of Y on X.

(c) Predict the number of bugs encountered if an employee spends 48 hours coding.

[10+0]
Model Answer: Correlation and Regression Analysis

Given Data

X 20 25 30 35 40 45 50 55 60 65
Y 5 7 8 12 15 13 17 18 20 25

$n = 10$ , predict Y at $X = 48$ .

Computation Table

X Y XY X² Y²
20 5 100 400 25
25 7 175 625 49
30 8 240 900 64
35 12 420 1225 144
40 15 600 1600 225
45 13 585 2025 169
50 17 850 2500 289
55 18 990 3025 324
60 20 1200 3600 400
65 25 1625 4225 625

Totals:
- $\sum X = 425$
- $\sum Y = 140$
- $\sum XY = 6785$
- $\sum X^2 = 20125$
- $\sum Y^2 = 2314$
Verification of totals: $\sum Y^2 = 25+49+64+144+225+169+289+324+400+625 = 2314$ ✓

(a) Pearson Correlation Coefficient

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

Numerator: $10(6785) - 425(140) = 67850 - 59500 = 8350$

Denominator part 1: $10(20125) - 425^2 = 201250 - 180625 = 20625$

Denominator part 2: $10(2314) - 140^2 = 23140 - 19600 = 3540$

$r = \frac{8350}{\sqrt{20625 \times 3540}} = \frac{8350}{\sqrt{73012500}} = \frac{8350}{8544.74} = 0.9772$

$\boxed{r \approx 0.977}$

Watch the $X^2$ total: add the $X^2$ column carefully, since $\sum X^2 = 20125$ carries straight into $n\sum X^2 - (\sum X)^2 = 20625$ and into the product under the root, and a slip there changes $r$ , $b$ and the prediction together.

Interpretation: Strong positive correlation; as coding hours increase, bugs tend to increase.

(b) Regression Equation of Y on X

$b = \frac{n\sum XY - \sum X \sum Y}{n\sum X^2 - (\sum X)^2} = \frac{8350}{20625} = 0.40485$

Means: $\bar{X} = \frac{425}{10} = 42.5, \qquad \bar{Y} = \frac{140}{10} = 14$

Intercept: $a = \bar{Y} - b\bar{X} = 14 - 0.40485(42.5) = 14 - 17.2061 = -3.2061$

$\boxed{\hat{Y} = -3.2061 + 0.4048X}$

(c) Prediction for X = 48

$\hat{Y} = -3.2061 + 0.40485(48) = -3.2061 + 19.4328 = 16.2267$

$\boxed{\hat{Y} \approx 16.23 \approx 16 \text{ bugs}}$

Conclusion: About 16 bugs are expected for 48 coding hours.
Model Answer: Correlation and Regression Analysis

Given Data

X 20 25 30 35 40 45 50 55 60 65
Y 5 7 8 12 15 13 17 18 20 25

$n = 10$, predict Y at $X = 48$.

Computation Table

X Y XY X² Y²
20 5 100 400 25
25 7 175 625 49
30 8 240 900 64
35 12 420 1225 144
40 15 600 1600 225
45 13 585 2025 169
50 17 850 2500 289
55 18 990 3025 324
60 20 1200 3600 400
65 25 1625 4225 625

Totals:
- $\sum X = 425$
- $\sum Y = 140$
- $\sum XY = 6785$
- $\sum X^2 = 20125$
- $\sum Y^2 = 2314$
Verification of totals: $\sum Y^2 = 25+49+64+144+225+169+289+324+400+625 = 2314$ ✓

(a) Pearson Correlation Coefficient

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

Numerator: $$10(6785) - 425(140) = 67850 - 59500 = 8350$$

Denominator part 1: $$10(20125) - 425^2 = 201250 - 180625 = 20625$$

Denominator part 2: $$10(2314) - 140^2 = 23140 - 19600 = 3540$$

$$r = \frac{8350}{\sqrt{20625 \times 3540}} = \frac{8350}{\sqrt{73012500}} = \frac{8350}{8544.74} = 0.9772$$

$$\boxed{r \approx 0.977}$$

Watch the $X^2$ total: add the $X^2$ column carefully, since $\sum X^2 = 20125$ carries straight into $n\sum X^2 - (\sum X)^2 = 20625$ and into the product under the root, and a slip there changes $r$, $b$ and the prediction together.

Interpretation: Strong positive correlation; as coding hours increase, bugs tend to increase.

(b) Regression Equation of Y on X

$$b = \frac{n\sum XY - \sum X \sum Y}{n\sum X^2 - (\sum X)^2} = \frac{8350}{20625} = 0.40485$$

Means: $$\bar{X} = \frac{425}{10} = 42.5, \qquad \bar{Y} = \frac{140}{10} = 14$$

Intercept: $$a = \bar{Y} - b\bar{X} = 14 - 0.40485(42.5) = 14 - 17.2061 = -3.2061$$

$$\boxed{\hat{Y} = -3.2061 + 0.4048X}$$

(c) Prediction for X = 48

$$\hat{Y} = -3.2061 + 0.40485(48) = -3.2061 + 19.4328 = 16.2267$$

$$\boxed{\hat{Y} \approx 16.23 \approx 16 \text{ bugs}}$$

Conclusion: About 16 bugs are expected for 48 coding hours.
45 marksNumericalPie chartsHideAnswer
In a survey of 50 IT students, the following programming languages were preferred: Python (20), Java (15), C++ (10), and JavaScript (5). Create a pie chart representing the preferences. [5]
Total students surveyed = 50 Language Count ----------------- Python 20 Java 15 C++ 10 JavaScript 5 Check: $20 + 15 + 10 + 5 = 50$ ✓ The angle for each sector is computed as: $$\text{Angle} = \frac{\text{Count}}{\text{Total}} \times 360°...
Total students surveyed = 50 Language Count ----------------- Python 20 Java 15 C++ 10 JavaScript 5 Check: $20 + 15 + 10 + 5 = 50$ ✓ The angle for each sector is computed as: $$\text{Angle} = \frac{\text{Count}}{\text{Total}} \times 360°...
55 marksNumericalPoisson distributionHideAnswer
A server records the number of requests per minute, which follows a Poisson distribution with a mean of 5 requests per minute. (a) What is the probability that exactly 3 requests occur in a given minute? (b) What is the probability that more than 2 requests will occur? λ=5\lambda = 5λ=5[5]
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:

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:
65 marksNumericalConfidence interval for population proportHideAnswer
A sample survey of 80 customers shows that 56 are satisfied with an IT service. Estimate the proportion of satisfied customers in the population with a 95% confidence interval. $n = 80, \quad x = 56$ [5]
A sample survey of 80 customers shows that 56 are satisfied with an IT service. Estimate the proportion of satisfied customers in the population with a 95% confidence interval. $n = 80, \quad x = 56$ [5]
Sample size: $n = 80$ - Number satisfied: $x = 56$ - Confidence level: $95\%$ , so $z{\alpha/2} = 1.96$ $\hat{p} = \frac{x}{n} = \frac{56}{80} = 0.70$ $$SE = \sqrt{\frac{\hat{p}(1-\hat{p})}{n}} = \sqrt{\frac{0.70 \times 0.30}{80}} = ...

Sample size: $n = 80$ - Number satisfied: $x = 56$ - Confidence level: $95%$, so $z{\alpha/2} = 1.96$ $$\hat{p} = \frac{x}{n} = \frac{56}{80} = 0.70$$ $$SE = \sqrt{\frac{\hat{p}(1-\hat{p})}{n}} = \sqrt{\frac{0.70 \times 0.30}{80}} = ...
75 marksNumericalConditional probability and Bayes theoremHideAnswer
A database server has a 95% reliability rate. Even if the server does not fail, there is a 5% chance that it will still report an error. If the server reports an error, what is the probability that the server has actually failed? $P(\text{Failure}) = 0.05, \quad P(\text{False positive}) = 0.05$ [5]
A database server has a 95% reliability rate. Even if the server does not fail, there is a 5% chance that it will still report an error. If the server reports an error, what is the probability that the server has actually failed? $P(\text{Failure}) = 0.05, \quad P(\text{False positive}) = 0.05$ [5]
Server reliability = 95%, hence probability of failure: $P(F) = 0.05$ - Probability of no failure: $P(\bar{F}) = 1 - 0.05 = 0.95$ - False positive rate (error reported even though no failure): $P(E \mid \bar{F}) = 0.05$ - Assumpt...

Server reliability = 95%, hence probability of failure: $$P(F) = 0.05$$ - Probability of no failure: $$P(\bar{F}) = 1 - 0.05 = 0.95$$ - False positive rate (error reported even though no failure): $$P(E \mid \bar{F}) = 0.05$$ - Assumpt...
85 marksNumericalStandard normal distribution and Z-scoresHideAnswer
The resolution time for customer support queries follows a normal distribution with mean 30 minutes and standard deviation 5 minutes. (a) What percentage of queries are resolved in less than 25 minutes? (b) What percentage are resolved between 25 and 35 minutes? μ=30,σ=5\mu = 30, \quad \sigma = 5μ=30,σ=5[5]
Mean: $\mu = 30$ minutes - Standard deviation: $\sigma = 5$ minutes - Distribution: Normal, $X \sim N(30, 5^2)$ --- Standardize using $Z = \dfrac{X - \mu}{\sigma}$ : $Z = \frac{25 - 30}{5} = \frac{-5}{5} = -1$ Look up the cumulative p...

Mean: $\mu = 30$ minutes - Standard deviation: $\sigma = 5$ minutes - Distribution: Normal, $X \sim N(30, 5^2)$ --- Standardize using $Z = \dfrac{X - \mu}{\sigma}$: $$Z = \frac{25 - 30}{5} = \frac{-5}{5} = -1$$ Look up the cumulative p...
95 marksNumericalFive number summaryHideAnswer
The time taken (in seconds) to complete a file transfer operation is recorded for 11 sessions. Calculate the five-number summary and construct a box plot. 100, 120, 110, 115, 130, 105, 125, 122, 112, 108, 135 [5]
Transfer times (seconds), $n = 11$ : $100, 120, 110, 115, 130, 105, 125, 122, 112, 108, 135$ $100,\ 105,\ 108,\ 110,\ 112,\ 115,\ 120,\ 122,\ 125,\ 130,\ 135$ - Min = $100$ - Max = $135$ With $n = 11$ , position $= \frac{n+1}{2} = 6$ th...
Transfer times (seconds), $n = 11$: $$100, 120, 110, 115, 130, 105, 125, 122, 112, 108, 135$$ $$100,\ 105,\ 108,\ 110,\ 112,\ 115,\ 120,\ 122,\ 125,\ 130,\ 135$$ - Min = $100$ - Max = $135$ With $n = 11$, position $= \frac{n+1}{2} = 6$th...
105 marksNumericalExpected value and decision makingHideAnswer
A tech support team handles the following number of customer queries (X) per hour along with their probabilities. Find the expected number and variance of queries handled per hour.

$\begin{array}{|c|ccccc|}\hline X & 0 & 1 & 2 & 3 & 4 \\ \hline P(X) & 0.10 & 0.20 & 0.30 & 0.25 & 0.15 \\ \hline \end{array}$

[5]
A tech support team handles the following number of customer queries (X) per hour along with their probabilities. Find the expected number and variance of queries handled per hour.

$$\begin{array}{|c|ccccc|}\hline X & 0 & 1 & 2 & 3 & 4 \ \hline P(X) & 0.10 & 0.20 & 0.30 & 0.25 & 0.15 \ \hline \end{array}$$

[5]
Model Answer: Expected Value and Variance of Customer Queries

Given Data

X 0 1 2 3 4
P(X) 0.10 0.20 0.30 0.25 0.15

Check: $\sum P(X) = 0.10 + 0.20 + 0.30 + 0.25 + 0.15 = 1.00$ ✓ (valid distribution)

Part 1: Expected Value $E(X)$

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

X P(X) X·P(X)
0 0.10 0.00
1 0.20 0.20
2 0.30 0.60
3 0.25 0.75
4 0.15 0.60

$E(X) = 0 + 0.20 + 0.60 + 0.75 + 0.60 = 2.15 \text{ queries/hour}$

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

$\text{Var}(X) = E(X^2) - [E(X)]^2$

Step 1: Compute $E(X^2)$

X X² P(X) X²·P(X)
0 0 0.10 0.00
1 1 0.20 0.20
2 4 0.30 1.20
3 9 0.25 2.25
4 16 0.15 2.40

$E(X^2) = 0 + 0.20 + 1.20 + 2.25 + 2.40 = 6.05$

Step 2: $[E(X)]^2 = (2.15)^2 = 4.6225$

Step 3:

$\text{Var}(X) = 6.05 - 4.6225 = 1.4275$

Standard deviation (optional): $\sigma = \sqrt{1.4275} \approx 1.195$

Final Answer
- Expected number of queries per hour: $E(X) = 2.15$
- Variance: $\text{Var}(X) = 1.4275 \approx 1.43$
Model Answer: Expected Value and Variance of Customer Queries

Given Data

X 0 1 2 3 4
P(X) 0.10 0.20 0.30 0.25 0.15

Check: $\sum P(X) = 0.10 + 0.20 + 0.30 + 0.25 + 0.15 = 1.00$ ✓ (valid distribution)

Part 1: Expected Value $E(X)$

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

X P(X) X·P(X)
0 0.10 0.00
1 0.20 0.20
2 0.30 0.60
3 0.25 0.75
4 0.15 0.60

$$E(X) = 0 + 0.20 + 0.60 + 0.75 + 0.60 = 2.15 \text{ queries/hour}$$

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

$$\text{Var}(X) = E(X^2) - [E(X)]^2$$

Step 1: Compute $E(X^2)$

X X² P(X) X²·P(X)
0 0 0.10 0.00
1 1 0.20 0.20
2 4 0.30 1.20
3 9 0.25 2.25
4 16 0.15 2.40

$$E(X^2) = 0 + 0.20 + 1.20 + 2.25 + 2.40 = 6.05$$

Step 2: $[E(X)]^2 = (2.15)^2 = 4.6225$

Step 3:

$$\text{Var}(X) = 6.05 - 4.6225 = 1.4275$$

Standard deviation (optional): $\sigma = \sqrt{1.4275} \approx 1.195$

Final Answer
- Expected number of queries per hour: $E(X) = 2.15$
- Variance: $\text{Var}(X) = 1.4275 \approx 1.43$
115 marksDefinition and role of statistics in ITHideAnswer
Discuss the importance of statistics in the field of information technology. [5]
Statistics plays a crucial role in the field of Information Technology by providing methods to collect, analyze, and interpret data. It enables IT professionals to make informed decisions and optimize system performance. - Statistics ena...
125 marksSampling error and non-sampling errorHideAnswer
Write a short note on: (a) Sampling Error (b) Stratified sampling. [5]
Definition: Sampling error is the difference between a statistic (calculated from a sample) and the corresponding parameter (true value from the population). It occurs because a sample does not perfectly represent the entire population. ...

X	20	25	30	35	40	45	50	55	60	65
Y	5	7	8	12	15	13	17	18	20	25

X	Y	XY	X²	Y²
20	5	100	400	25
25	7	175	625	49
30	8	240	900	64
35	12	420	1225	144
40	15	600	1600	225
45	13	585	2025	169
50	17	850	2500	289
55	18	990	3025	324
60	20	1200	3600	400
65	25	1625	4225	625

X	20	25	30	35	40	45	50	55	60	65
Y	5	7	8	12	15	13	17	18	20	25

X	Y	XY	X²	Y²
20	5	100	400	25
25	7	175	625	49
30	8	240	900	64
35	12	420	1225	144
40	15	600	1600	225
45	13	585	2025	169
50	17	850	2500	289
55	18	990	3025	324
60	20	1200	3600	400
65	25	1625	4225	625

X	0	1	2	3	4
P(X)	0.10	0.20	0.30	0.25	0.15

X	P(X)	X·P(X)
0	0.10	0.00
1	0.20	0.20
2	0.30	0.60
3	0.25	0.75
4	0.15	0.60

X	X²	P(X)	X²·P(X)
0	0	0.10	0.00
1	1	0.20	0.20
2	4	0.30	1.20
3	9	0.25	2.25
4	16	0.15	2.40

X	0	1	2	3	4
P(X)	0.10	0.20	0.30	0.25	0.15

X	P(X)	X·P(X)
0	0.10	0.00
1	0.20	0.20
2	0.30	0.60
3	0.25	0.75
4	0.15	0.60

X	X²	P(X)	X²·P(X)
0	0	0.10	0.00
1	1	0.20	0.20
2	4	0.30	1.20
3	9	0.25	2.25
4	16	0.15	2.40

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Basic Statistics 2081 question paper

Model Answer: Correlation and Regression Analysis

Given Data

Computation Table

(a) Pearson Correlation Coefficient

(b) Regression Equation of Y on X

(c) Prediction for X = 48

Model Answer: Correlation and Regression Analysis

Given Data

Computation Table

(a) Pearson Correlation Coefficient

(b) Regression Equation of Y on X

(c) Prediction for X = 48

Model Answer: Expected Value and Variance of Customer Queries

Given Data

Part 1: Expected Value E(X)E(X)E(X)

Part 2: Variance Var(X)\text{Var}(X)Var(X)

Final Answer

Model Answer: Expected Value and Variance of Customer Queries

Given Data

Part 1: Expected Value $E(X)$

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

Final Answer

Model Answer: Correlation and Regression Analysis

Given Data

Computation Table

(a) Pearson Correlation Coefficient

(b) Regression Equation of Y on X

(c) Prediction for X = 48

Model Answer: Expected Value and Variance of Customer Queries

Given Data

Part 1: Expected Value E(X)E(X)E(X)

Part 2: Variance Var(X)\text{Var}(X)Var(X)

Final Answer

Part 1: Expected Value $E(X)$

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

Part 1: Expected Value $E(X)$

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