2078

STA154 · TU past paper

Basic Statistics 2078 question paper

The complete TU 2078 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

    Measuring Dispersion and Student Performance Analysis

    Dispersion measures the scatter or spread of data about a central value. Methods are classified as absolute (same units as data) and relative (unit-free ratios/percentages). Absolute measures: 1. Range = Largest value − Smallest value (

  2. 210 marksNumericalSimple linear regression modelAnswer

    Regression Analysis: Experience and Performance

    The following data gives the experience of Computer Operators in years and their performance as given by the number of good parts turned out per 100 pieces.

    Experience (X)1612184310512
    Performance (Y)8788896878807583

    i) Fit the regression equation of performance rating on experience.

    ii) Estimate the probable performance of an operator with 8 years of experience and interpret the regression coefficient.

    [10]

    X (Experience, yrs) 16 12 18 4 3 10 5 12 --------------------------- Y (Performance) 87 88 89 68 78 80 75 83 n = 8, estimate required at X = 8. X Y XY X² -------------- 16 87 1392 256 12 88 1056 144 18 89 1602 324 4 68 272 16 3 78 234 9 ...

  3. 310 marksNumericalStandard normal distribution and Z-scoresAnswer

    Define standard normal distribution. For a certain type of computers, the length of time between charges of the battery is normally distribbuted with a mean of 50 hours and a standard deviation of 15 hours. He owns one of these computers and wants to know the probability that the length of time will be (i) between 50 and 70 hours, (ii) more than 60 hours, and (iii) less than 45 hours?[10]

    • Mean $\mu = 50$ hours - Standard deviation $\sigma = 15$ hours - $X \sim N(50, 15^2)$, normally distributed - Required: - (i) $P(50 \le X \le 70)$ - (ii) $P(X 60)$ - (iii) $P(X < 45)$ The standard normal distribution is a normal distri...
  4. 45 marksNumericalBasic probability concepts and rulesAnswer

    Suppose that after 10 years of service, 40% of computers have problems with motherboards (MB), 30% have problems with hard drives (HD), and l5% have problems with both MB and HD. What is the probability that a 10-year old computer still has fully functioning MB and HD? [5]

    • $P(MB) = 0.40$ (probability of motherboard problem) - $P(HD) = 0.30$ (probability of hard drive problem) - $P(MB \cap HD) = 0.15$ (probability of both problems) - Find: $P(\text{no MB problem AND no HD problem})$ Step 1: Probability of...
  5. 55 marksSampling error and non-sampling errorAnswer

    Describe sampling error and non-sampling error. [5]

    Sampling error is the difference between a sample statistic and the corresponding population parameter, arising solely because a sample is used instead of the entire population. Characteristics: - Occurs due to random variation in the se...

  6. 65 marksMeasurement scales and typesAnswer

    What is measurement of scale? Describe different types of measurement scale. [5]

    Measurement of Scale

    Definition

    Measurement of scale refers to the system or framework used to categorize, quantify, and classify data based on the properties and relationships of the values assigned to variables. It defines the nature of information contained in data and determines which statistical operations and analyses are appropriate.

    In other words, a scale of measurement is a classification that describes the nature of information within the numbers assigned to variables.


    Types of Measurement Scales

    There are four main types of measurement scales:

    1. Nominal Scale

    • Definition: The most basic level of measurement where data is categorized into distinct, non-overlapping categories with no inherent order or ranking.
    • Characteristics:
      • Categories are mutually exclusive
      • No numerical relationship between categories
      • No ordering or hierarchy
    • Examples: Gender (Male/Female), Religion, Blood type (A, B, AB, O), Color
    • Allowed Operations: Counting, frequency, mode

    2. Ordinal Scale

    • Definition: Data is organized into categories that have a natural order or ranking, but the intervals between categories are not equal or meaningful.
    • Characteristics:
      • Categories can be ranked or ordered
      • Differences between ranks are not uniform
      • Relative position matters, not absolute value
    • Examples: Educational level (Primary, Secondary, Bachelor's, Master's), Customer satisfaction (Poor, Fair, Good, Excellent), Military rank
    • Allowed Operations: Median, mode, ranking, correlation

    3. Interval Scale

    • Definition: Numerical scale where the distance between any two adjacent units is equal, but there is no true zero point (zero is arbitrary).
    • Characteristics:
      • Equal intervals between values
      • No true/absolute zero
      • Allows meaningful differences
    • Examples: Temperature (Celsius/Fahrenheit), Calendar years, IQ scores
    • Allowed Operations: Mean, standard deviation, addition, subtraction

    4. Ratio Scale

    • Definition: The highest level of measurement with equal intervals between values AND a true/absolute zero point, allowing for meaningful ratios.
    • Characteristics:
      • Equal intervals between values
      • True zero point exists
      • Ratios are meaningful
      • Most flexible scale
    • Examples: Height, Weight, Age, Income, Distance, Time
    • Allowed Operations: All statistical operations including mean, median, mode, standard deviation, multiplication, division

    Summary Table

    ScaleOrderEqual IntervalsTrue ZeroExample
    NominalNoNoNoGender
    OrdinalYesNoNoSatisfaction rating
    IntervalYesYesNoTemperature (°C)
    RatioYesYesYesWeight (kg)
  7. 75 marksNumericalPie chartsAnswer

    The following table shows the numbers of hours spent by a child on different events on a working day. Represent the adjoining information on a pie chart.

    ActivityNo. of hours
    School6
    Sleep8
    Playing2
    Study4
    T.V.1
    Others3

    [5]

    Activity No. of hours ------------------------ School 6 Sleep 8 Playing 2 Study 4 T.V. 1 Others 3 $$\text{Total} = 6 + 8 + 2 + 4 + 1 + 3 = 24 \text{ hours}$$ Formula: $$\text{Central Angle} = \frac{\text{Hours for activity}}{\text{Total ...

  8. 85 marksNumericalCorrelation coefficient definition and proAnswer

    Define correlation coefficient? Calculate the coefficient of correlation for the following ages (in years) of husbands and wives and interpret it.

    Husband's age X:23272828293031333536
    Wife's age Y:18202721292727292829

    [5]

    Correlation Coefficient: Definition and Calculation

    Definition

    The correlation coefficient is a statistical measure that expresses the degree and direction of the linear relationship between two variables. It ranges from $-1$ to $+1$:

    • $+1$ = perfect positive correlation
    • $-1$ = perfect negative correlation
    • $0$ = no linear correlation

    Pearson's correlation coefficient:

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

    Given Data (n = 10)

    XYXYX²Y²
    2318414529324
    2720540729400
    2827756784729
    2821588784441
    2929841841841
    3027810900729
    3127837961729
    33299571089841
    35289801225784
    362910441296841

    Summations (verified)

    • $\sum X = 23+27+28+28+29+30+31+33+35+36 = 300$
    • $\sum Y = 18+20+27+21+29+27+27+29+28+29 = 255$
    • $\sum XY = 414+540+756+588+841+810+837+957+980+1044 = 7767$
    • $\sum X^2 = 529+729+784+784+841+900+961+1089+1225+1296 = 9138$
    • $\sum Y^2 = 324+400+729+441+841+729+729+841+784+841 = 6659$

    Note: the correct sums are $\sum X = 300$ (not 310), $\sum X^2 = 9138$ (not 9738) and $\sum Y^2 = 6659$ (not 6859). Rechecking $\sum Y^2$: $324+400=724$, $+729=1453$, $+441=1894$, $+841=2735$, $+729=3464$, $+729=4193$, $+841=5034$, $+784=5818$, $+841=6659$. Confirmed $6659$.

    Substitution

    Numerator: $$n\sum XY - \sum X\sum Y = 10(7767) - (300)(255) = 77670 - 76500 = 1170$$

    X-term: $$n\sum X^2 - (\sum X)^2 = 10(9138) - (300)^2 = 91380 - 90000 = 1380$$

    Y-term: $$n\sum Y^2 - (\sum Y)^2 = 10(6659) - (255)^2 = 66590 - 65025 = 1565$$

    $$r = \frac{1170}{\sqrt{1380 \times 1565}} = \frac{1170}{\sqrt{2159700}} = \frac{1170}{1469.59}$$

    $$\boxed{r \approx 0.796}$$

    Interpretation

    $r \approx +0.796$ indicates a strong positive correlation between husbands' and wives' ages. As the husband's age increases, the wife's age also tends to increase. This is consistent with real-world marriage patterns. The coefficient of determination $r^2 \approx 0.634$, so about 63% of the variation in wife's age is explained by husband's age.

    A result of $r \approx -0.646$ comes from using $\sum X = 310$ instead of $300$, together with wrong squared sums, which flips the sign of the numerator.

  9. 95 marksNumericalContinuous probability distributionsAnswer

    Suppose a continuous random variable X has the density function. Find; (i) value of k, (ii) P (0 < x < 0.5), and, (iii) E(X) and (iv) E(2X + 4)

    $$f(x) = \begin{cases} k(1-x)^2; & \text{for } 0 < x < 1 \ 0; & \text{elsewhere} \end{cases}$$

    [5]

    • Density: $f(x) = k(1-x)^2$ for $0 < x < 1$; $f(x) = 0$ elsewhere - Required: (i) $k$, (ii) $P(0<X<0.5)$, (iii) $E(X)$, (iv) $E(2X+4)$ --- Total probability must equal 1: $$\int{0}^{1} k(1-x)^2,dx = 1$$ $$k\left[-\frac{(1-x)^3}{3}\righ...
  10. 105 marksNumericalBinomial distributionAnswer

    During one stage in the manufacture of integrated circuit chips, a coating must be applied. If 70% of chips received a thick enough Coating. find the probability that among 15 chips (1) at least 12 will have thick enough coatings, and (2) exactly 10 will have thick enough coatings. [5]

    Model Answer: Binomial Probability

    STEP 1 - Given Data

    • Probability of thick enough coating: $p = 0.70$
    • Probability of not thick enough: $q = 1 - p = 0.30$
    • Sample size: $n = 15$

    Binomial distribution: $P(X = r) = \binom{n}{r} p^r q^{n-r}$


    STEP 2 - Solve

    (1) P(at least 12) = P(X ≥ 12)

    $$P(X \geq 12) = P(12) + P(13) + P(14) + P(15)$$

    P(X = 12): $$\binom{15}{12}(0.7)^{12}(0.3)^{3} = 455 \times 0.0138413 \times 0.027$$

    $(0.7)^{12} = 0.01384128$

    $= 455 \times 0.01384128 \times 0.027 = 0.17004$

    P(X = 13): $$\binom{15}{13}(0.7)^{13}(0.3)^{2} = 105 \times 0.00968889 \times 0.09$$

    $(0.7)^{13} = 0.00968889$

    $= 105 \times 0.00968889 \times 0.09 = 0.09156$

    P(X = 14): $$\binom{15}{14}(0.7)^{14}(0.3)^{1} = 15 \times 0.00678223 \times 0.3$$

    $(0.7)^{14} = 0.00678223$

    $= 15 \times 0.00678223 \times 0.3 = 0.03052$

    P(X = 15): $$\binom{15}{15}(0.7)^{15}(0.3)^{0} = 1 \times 0.00474756 \times 1 = 0.00475$$

    Total: $$P(X \geq 12) = 0.17004 + 0.09156 + 0.03052 + 0.00475 = 0.2969$$

    $$\boxed{P(X \geq 12) \approx 0.2969}$$

    (2) P(exactly 10) = P(X = 10)

    $$P(X = 10) = \binom{15}{10}(0.7)^{10}(0.3)^{5}$$

    $\binom{15}{10} = 3003$

    $(0.7)^{10} = 0.0282475$

    $(0.3)^{5} = 0.00243$

    $$= 3003 \times 0.0282475 \times 0.00243$$

    $$= 3003 \times 0.00006864 = 0.2061$$

    $$\boxed{P(X = 10) \approx 0.2061}$$


    Summary of Answers

    1. P(at least 12) ≈ 0.2969 or 29.69%
    2. P(exactly 10) ≈ 0.2061 or 20.61%
  11. 115 marksNumericalCoefficient of kurtosisAnswer

    The standard deviation of a symmetric distribution is 7. Compute the possible value of fourth central moment for the distribution to be (i) mesokurtic (ii) platykurtic, and (iii) leptokurtic. [5]

    Model Answer: Fourth Central Moment and Kurtosis

    Step 1 - Given Data

    • Standard deviation: $\sigma = 7$
    • Distribution is symmetric
    • Required: value of the fourth central moment $\mu_4$ for the distribution to be (i) mesokurtic, (ii) platykurtic, (iii) leptokurtic

    Step 2 - Solve

    Key Relationship

    The coefficient of kurtosis is: $$\beta_2 = \frac{\mu_4}{\mu_2^2} = \frac{\mu_4}{\sigma^4}$$

    since $\mu_2 = \sigma^2$.

    Rearranging: $$\mu_4 = \beta_2 \times \sigma^4$$

    Compute $\sigma^4$

    $$\sigma^4 = 7^4 = 2401$$

    (i) Mesokurtic

    Same peakedness as normal distribution: $\beta_2 = 3$. $$\mu_4 = 3 \times 2401 = \boxed{7203}$$

    (ii) Platykurtic

    Flatter than normal: $\beta_2 < 3$. $$\mu_4 < 3 \times 2401 = \boxed{7203} \quad (\mu_4 < 7203)$$

    (iii) Leptokurtic

    More peaked than normal: $\beta_2 > 3$. $$\mu_4 > 3 \times 2401 = \boxed{7203} \quad (\mu_4 > 7203)$$

    Summary Table

    TypeCondition$\mu_4$
    Mesokurtic$\beta_2 = 3$$\mu_4 = 7203$
    Platykurtic$\beta_2 < 3$$\mu_4 < 7203$
    Leptokurtic$\beta_2 > 3$$\mu_4 > 7203$

    The symmetry condition confirms odd central moments are zero, but it does not affect the kurtosis computation, which depends only on $\mu_4$ and $\sigma$.

  12. 125 marksParameter and statistic distinctionAnswer

    Write short notes on the following: a.) Choice of appropriate measure of central tendency b.)Parameter and statistic [5]

    The selection of an appropriate measure of central tendency depends on several factors: 1. Nature of Data - Mean: Best for continuous, normally distributed data without extreme outliers - Median: Preferred for skewed distributions or whe...