2080.1

STA154 · TU past paper

Basic Statistics 2080.1 question paper

The complete TU 2080.1 exam paper for Basic Statistics (STA154), all 12 questions with solved model answers written to the mark scheme.

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  1. 110 marksNumericalComparison of consistency between datasetsAnswer

    Measures of Central Tendency and Computer Consistency Analysis

    Measure Suitable Situation ----------------------------- Mean Data is fairly symmetric with no extreme outliers; all values must be used; further algebraic/statistical treatment (like SD, CV) is needed; quantitative data. Median Distribu...

  2. 210 marksNumericalAssumptions of linear regressionAnswer

    Simple Linear Regression Analysis

    1. Linearity: The mean of the response $Y$ is a linear function of $X$. 2. Independence: The error terms (observations) are independent. 3. Homoscedasticity: Constant variance of errors across all values of $X$. 4. Normality: The error t...
  3. 310 marksNumericalNormal distribution properties and applicaAnswer

    Discuss the measure properties of normal distribution. The burning time of an experimental rocket is a random variable having the normal distribution with mean 4.76 seconds and standard deviation 0.04 second respectively. What is probability that this kind of rocket will burn (i) Less than 4.68 seconds (ii) More than 4.80 seconds (iii) Anywhere from 4.70 to 4.82 seconds?μ=4.76,σ=0.04\mu = 4.76, \quad \sigma = 0.04μ=4.76,σ=0.04P(X<4.68), P(X>4.80), P(4.70<X<4.82)P(X < 4.68), ; P(X > 4.80), ; P(4.70 < X < 4.82)P(X<4.68),P(X>4.80),P(4.70<X<4.82)[10]

    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 ...
  4. 45 marksDescriptive versus inferential statisticsAnswer

    Highlight the difference between descriptive and inferential statistics. Describe the role of Statistics in information technology. [5]

    Aspect Descriptive Statistics Inferential Statistics -------------------------------------------------------- Purpose Summarizes and describes characteristics of a dataset Draws conclusions about a population based on sample data Scope W...

  5. 55 marksNumericalBar diagrams and Pareto diagramsAnswer

    Construct bar diagram and Pareto diagram of the following information.

    BrandDellAcerLenovoHPTotal
    No. of Sales501004060250

    [5]

    Model Answer: Bar Diagram and Pareto Diagram

    STEP 1 - Given Data

    BrandDellAcerLenovoHPTotal
    No. of Sales501004060250

    STEP 2 - Solution

    1. Bar Diagram

    In a bar diagram, each brand is shown as a separate rectangular bar whose height equals its number of sales. Bars are of equal width with equal gaps.

    • X-axis: Brand (Dell, Acer, Lenovo, HP)
    • Y-axis: Number of Sales

    Heights: Dell = 50, Acer = 100, Lenovo = 40, HP = 60

    No. of Sales
    110|
    100|          ████
     90|          ████
     80|          ████
     70|          ████
     60|          ████             ████
     50| ████     ████             ████
     40| ████     ████    ████     ████
     30| ████     ████    ████     ████
     20| ████     ████    ████     ████
     10| ████     ████    ████     ████
      0|__████_____████____████_____████__
         Dell     Acer   Lenovo    HP
                    Brand
    

    2. Pareto Diagram

    A Pareto diagram arranges categories in descending order of frequency (as bars) with a cumulative percentage line overlaid.

    Step 1: Arrange in descending order and compute cumulative percentage

    $$\text{Cumulative %} = \frac{\text{Cumulative Sales}}{250} \times 100$$

    BrandSalesCumulative SalesCumulative %
    Acer100100$\frac{100}{250}\times100 = 40%$
    HP60160$\frac{160}{250}\times100 = 64%$
    Dell50210$\frac{210}{250}\times100 = 84%$
    Lenovo40250$\frac{250}{250}\times100 = 100%$

    Step 2: Draw bars (descending) with cumulative curve

    Sales                          Cumulative %
    110|                                    _____ 100%
    100| ████                          ___/       84%
     90| ████                      ___/           
     80| ████            ______●--/               64%
     70| ████        ___/    
     60| ████    ●--/         ████
     50| ████  /              ████     ████
     40| ●---/           ████ ████     ████ ████
     30| ████            ████ ████     ████ ████
     20| ████            ████ ████     ████ ████
     10| ████            ████ ████     ████ ████
      0|__████____________████_________████_████__
        Acer(40%)      HP(64%)     Dell(84%) Lenovo(100%)
    

    The line rises steeply for Acer and HP (together $64%$ of all sales) then flattens, illustrating the Pareto principle that a few categories account for the majority of the total.


    Result: Bars: Dell 50, Acer 100, Lenovo 40, HP 60. Pareto order Acer, HP, Dell, Lenovo with cumulative percentages $40%, 64%, 84%, 100%$.

  6. 65 marksNumericalKarl Pearson correlation coefficientAnswer

    Karl Pearson's Coefficient of Correlation

    From the following data on marks of 10 students in the two subjects, calculate the Karl Pearson's coefficient of correlation $r$ and interpret the result.

    $$\begin{array}{|c|cccccccccc|}\hline \text{Maths} & 55 & 70 & 40 & 30 & 90 & 80 & 60 & 80 & 90 & 80 \ \hline \text{Basic Statistics} & 65 & 40 & 30 & 50 & 60 & 70 & 50 & 50 & 60 & 70 \ \hline \end{array}$$

    [5]

    Marks of 10 students: Student Maths (x) Basic Statistics (y) --------- 1 55 65 2 70 40 3 40 30 4 30 50 5 90 60 6 80 70 7 60 50 8 80 50 9 90 60 10 80 70 $n = 10$ $$r = \frac{n\sum xy - \sum x \sum y}{\sqrt{[n\sum x^2 - (\sum x)^2][n\sum y...

  7. 75 marksNumericalExpected value and decision makingAnswer

    A jewelry dealer is interested in purchasing gold necklace for which probabilities are 0.18, 0.22, 0.33 and 0.27 respectively that it will be sold for a profit of Rs. 5000, Rs. 8000, 6000 and sell for a loss of Rs. 3000. Find expected profit and variance of profit. [5]

    • Outcomes (profit/loss): Rs. 5000, Rs. 8000, Rs. 6000, and a loss of Rs. 3000 (i.e. $-3000$) - Probabilities: 0.18, 0.22, 0.33, 0.27 Let $X$ = profit in Rs. $X$ 5000 8000 6000 -3000 ------------------------------ $P(X)$ 0.18 0.22 0.33 0...
  8. 85 marksNumericalPoisson distributionAnswer

    Fit a Poisson distribution to the following data.

    Defects (X = x)012345
    Number of pages142156692751

    [5]

    Defects (x) 0 1 2 3 4 5 --------------------- No. of pages (f) 142 156 69 27 5 1 $$\lambda = \bar{x} = \frac{\sum f x}{\sum f}$$ x f f·x ----------- 0 142 0 1 156 156 2 69 138 3 27 81 4 5 20 5 1 5 Total N = 400 Σfx = 400 $$\lambda = \fra...

  9. 95 marksNumericalBasic probability concepts and rulesAnswer

    The odds in against of A solving a problem as 8 to 6 and the odds in favor of B solving the same problem are 14 to 10. What is the probability that (a) both A and B will solve it? (b) A solve it but B fails to solve it? [5]

    • Odds against A solving = 8 : 6 - Odds in favor of B solving = 14 : 10 For A (odds against = 8 : 6): Odds against means unfavorable : favorable = 8 : 6, so: $$P(A) = \frac{6}{8+6} = \frac{6}{14} = \frac{3}{7}$$ $$P(A') = 1 - \frac{3}{7}...
  10. 105 marksNumericalFive number summaryAnswer

    Drive D of a computer having 25 folders contain the following number of files. (a) List the five number summaries (b) Construct a box-plot for the data (c) Are the data skewed? 4, 0, 5, 2, 3, 1, 4, 3, 2, 3, 4, 3, 1, 6, 3, 2, 3, 4, 2, 1, 0, 3, 2, 5, 4 [5]

    Number of files in 25 folders: $$4, 0, 5, 2, 3, 1, 4, 3, 2, 3, 4, 3, 1, 6, 3, 2, 3, 4, 2, 1, 0, 3, 2, 5, 4$$ $n = 25$ Step 1: Sort the data $$0, 0, 1, 1, 1, 2, 2, 2, 2, 2, 3, 3, 3, 3, 3, 3, 3, 4, 4, 4, 4, 4, 5, 5, 6$$ Let me verify the c...

  11. 115 marksNumericalInterval estimation and confidence intervaAnswer

    Define interval estimation. In an examination, the random sample of 10 students selected from examination and their marks obtained in an examination was 45, 55, 20, 54, 40, 55, 51, 35, 43 and 46. Find 99% confidence interval for population mean. Assuming that the population from which samples are drawn is normally distributed. [5]

    Model Answer: Interval Estimation and 99% Confidence Interval

    Definition of Interval Estimation

    Interval estimation is a method of statistical inference in which we estimate a population parameter by constructing a range (interval) of values, together with a stated confidence level, within which the true parameter value is expected to lie. It is expressed as:

    $$\text{Point Estimate} \pm \text{Margin of Error}$$

    The endpoints of the interval are called confidence limits, and the associated probability (e.g., 99%) is the confidence coefficient.


    STEP 1 - Given Data

    • Sample size: $n = 10$
    • Marks: $45, 55, 20, 54, 40, 55, 51, 35, 43, 46$
    • Confidence level: $99%$
    • Population is normally distributed; population variance unknown → use t-distribution

    STEP 2 - Solve

    Step 1: Sample Mean

    $$\sum x_i = 45+55+20+54+40+55+51+35+43+46 = 444$$

    $$\bar{x} = \frac{444}{10} = 44.4$$

    Step 2: Sample Standard Deviation

    $x$$x-\bar{x}$$(x-\bar{x})^2$
    450.60.36
    5510.6112.36
    20-24.4595.36
    549.692.16
    40-4.419.36
    5510.6112.36
    516.643.56
    35-9.488.36
    43-1.41.96
    461.62.56

    $$\sum(x-\bar{x})^2 = 1068.4$$

    $$s = \sqrt{\frac{1068.4}{n-1}} = \sqrt{\frac{1068.4}{9}} = \sqrt{118.711} = 10.895 \approx 10.90$$

    Step 3: t-value

    $$df = n-1 = 9, \quad \alpha = 0.01, \quad \alpha/2 = 0.005$$

    $$t_{0.005,,9} = 3.250$$

    Step 4: Margin of Error

    $$E = t_{\alpha/2}\cdot\frac{s}{\sqrt{n}} = 3.250 \times \frac{10.895}{\sqrt{10}}$$

    $$\frac{10.895}{3.1623} = 3.4453$$

    $$E = 3.250 \times 3.4453 = 11.197 \approx 11.20$$

    Step 5: Confidence Interval

    $$CI = \bar{x} \pm E = 44.4 \pm 11.20$$

    $$\text{Lower limit} = 44.4 - 11.20 = 33.20$$ $$\text{Upper limit} = 44.4 + 11.20 = 55.60$$


    Final Answer

    The 99% confidence interval for the population mean is $(33.20,; 55.60)$, i.e., $44.4 \pm 11.20$.

    We are 99% confident that the true population mean marks lie between approximately 33.2 and 55.6.

  12. 125 marksParameter and statistic distinctionAnswer

    Write short notes on any two: (a) Sampling and non-sampling error (b) Parameter and statistic. [5]

    Sampling Error: - Arises due to the use of a sample instead of the entire population - Occurs because a sample is only a part of the population and cannot perfectly represent all characteristics - Can be reduced by increasing sample size...