Measures of Central Tendency and Dispersion - Verified Model Answer

STEP 1 - EXTRACT (Given Data)

Grouped frequency distribution, $n = 50$:

$\sum f = 8+10+15+9+6+2 = 50$ ✓

Required: mean, standard deviation, variance, coefficient of variation, plus conceptual explanation.

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STEP 2 - SOLVE

Part 1: Concepts

Measures of central tendency describe the center or typical value of a dataset: mean, median, mode.

Measures of dispersion describe the spread or variability of data about the central value.

Different measures of dispersion:

• Range (max − min)
• Quartile deviation / semi-interquartile range = $(Q_3 - Q_1)/2$
• Mean deviation (average absolute deviation)
• Standard deviation and variance
• Coefficient of variation (relative measure)

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Part 2: Computation

Midpoints and products:

$$\sum f = 50,\quad \sum fx = 176.0,\quad \sum fx^2 = 712.50$$

Mean: $$\bar{x} = \frac{\sum fx}{\sum f} = \frac{176.0}{50} = 3.52 \text{ lbs}$$

Variance and Standard Deviation (population form, standard for grouped data):

$$\sigma^2 = \frac{\sum fx^2}{n} - \bar{x}^2 = \frac{712.50}{50} - (3.52)^2$$ $$= 14.25 - 12.3904 = 1.8596 \approx 1.86 \text{ lbs}^2$$

$$\sigma = \sqrt{1.8596} = 1.3637 \approx 1.36 \text{ lbs}$$

(Note: the sample form with $n-1$ gives $s^2 = 92.98/49 = 1.8975 \approx 1.90$ and $s = 1.38$. Both are acceptable; the population form is the conventional choice for descriptive grouped-data problems and is the one reported above.)

Coefficient of Variation: $$CV = \frac{\sigma}{\bar{x}} \times 100 = \frac{1.3637}{3.52} \times 100 = 38.74%$$

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Summary Table

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Part 4: Interpretation

• Mean = 3.52 lbs: The average mass is about 3.52 lbs, centered in the modal 3-4 class (highest frequency = 15).
• Standard Deviation ≈ 1.36 lbs: Individual masses deviate from the mean by about 1.36 lbs on average, indicating a moderate spread.
• Variance ≈ 1.86 lbs²: The squared spread measure; SD is more interpretable being in original units.
• CV ≈ 38.7%: A relatively high CV, showing the data are fairly variable relative to the mean; since CV > 30% the distribution is not highly homogeneous.

The only judgement call here is the use of the sample ($n-1$) rather than the population ($n$) divisor, which yields slightly different SD, variance and CV values; both are defensible.

Correlation and Regression Analysis

Given Data

$n = 8$

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Meaning of Correlation Types

i. Positive Correlation: Two variables move in the same direction. When one increases, the other also increases (and vice versa). Example: height and weight.

ii. Negative Correlation: Two variables move in opposite directions. When one increases, the other decreases. Example: price and demand.

iii. Perfect Correlation: A change in one variable produces an exactly proportional change in the other, so all points lie on a straight line. Perfect positive gives $r = +1$; perfect negative gives $r = -1$.

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Calculation Table

Verification of sums: $\Sigma x = 47$, $\Sigma y = 1027$, $\Sigma x^2 = 321$, $\Sigma xy = 7146$, and $\Sigma y^2 = 160021$. All confirmed.

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Part (a): Correlation Coefficient

$$r = \frac{n\Sigma xy - \Sigma x \Sigma y}{\sqrt{[n\Sigma x^2 - (\Sigma x)^2][n\Sigma y^2 - (\Sigma y)^2]}}$$

Numerator: $$8(7146) - 47(1027) = 57168 - 48269 = 8899$$

First bracket: $$8(321) - 47^2 = 2568 - 2209 = 359$$

Second bracket: $$8(160021) - 1027^2 = 1280168 - 1054729 = 225439$$

$$r = \frac{8899}{\sqrt{359 \times 225439}} = \frac{8899}{\sqrt{80932601}} = \frac{8899}{8996.26}$$

$$\boxed{r \approx 0.989}$$

Interpretation: Since $r = 0.989$ is very close to $+1$, there is a very high positive correlation between weeks in the program and speed gain. As weeks increase, speed gain increases strongly.

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Part (b): Regression Equation of Speed Gain (y) on Weeks (x)

$$\hat{y} = a + bx$$

$$b = \frac{n\Sigma xy - \Sigma x \Sigma y}{n\Sigma x^2 - (\Sigma x)^2} = \frac{8899}{359} = 24.79$$

$$\bar{x} = \frac{47}{8} = 5.875, \quad \bar{y} = \frac{1027}{8} = 128.375$$

$$a = \bar{y} - b\bar{x} = 128.375 - 24.79(5.875) = 128.375 - 145.64 = -17.26$$

$$\boxed{\hat{y} = -17.26 + 24.79x}$$

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Part (c): Estimate for 6 Weeks and Slope Interpretation

$$\hat{y} = -17.26 + 24.79(6) = -17.26 + 148.74 = 131.48$$

$$\boxed{\hat{y} \approx 131.48 \text{ units of speed gain}}$$

Slope interpretation: $b = 24.79$ means that for each additional week in the program, the reading speed gain increases by about 24.79 units on average.

Joint Probability Density Function - Verified Solution

STEP 1 - Given Data

$$f(x, y) = kxy, \quad 0 < x < 4, \quad 1 < y < 5$$ $$f(x, y) = 0, \quad \text{otherwise}$$

Required:

• (a) constant $k$
• (b) $P(X > 3, Y < 2)$
• (c) $P(1 < X < 2, 2 < Y < 3)$

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STEP 2 - Solve

Part (a): Value of k

Normalization condition:

$$\int_{1}^{5} \int_{0}^{4} kxy , dx , dy = 1$$

Inner integral (over $x$):

$$\int_{0}^{4} x , dx = \left[\frac{x^2}{2}\right]_0^4 = \frac{16}{2} = 8$$

So:

$$k \int_{1}^{5} 8y , dy = 8k \left[\frac{y^2}{2}\right]_1^5 = 8k \cdot \frac{25 - 1}{2} = 8k \cdot 12 = 96k$$

Setting equal to 1:

$$96k = 1 \implies \boxed{k = \frac{1}{96}}$$

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Part (b): P(X > 3, Y < 2)

The region intersects the support: $3 < x < 4$ and $1 < y < 2$.

$$P = \frac{1}{96} \int_{1}^{2} \int_{3}^{4} xy , dx , dy$$

Inner integral:

$$\int_{3}^{4} x , dx = \left[\frac{x^2}{2}\right]_3^4 = \frac{16 - 9}{2} = \frac{7}{2}$$

Then:

$$P = \frac{1}{96} \cdot \frac{7}{2} \int_{1}^{2} y , dy = \frac{7}{192} \left[\frac{y^2}{2}\right]_1^2 = \frac{7}{192} \cdot \frac{4-1}{2} = \frac{7}{192} \cdot \frac{3}{2} = \frac{21}{384}$$

$$\boxed{P(X > 3, Y < 2) = \frac{7}{128} \approx 0.0547}$$

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Part (c): P(1 < X < 2, 2 < Y < 3)

Both limits lie within the support.

$$P = \frac{1}{96} \int_{2}^{3} \int_{1}^{2} xy , dx , dy$$

Inner integral:

$$\int_{1}^{2} x , dx = \left[\frac{x^2}{2}\right]_1^2 = \frac{4 - 1}{2} = \frac{3}{2}$$

Then:

$$P = \frac{1}{96} \cdot \frac{3}{2} \int_{2}^{3} y , dy = \frac{3}{192} \left[\frac{y^2}{2}\right]_2^3 = \frac{1}{64} \cdot \frac{9-4}{2} = \frac{1}{64} \cdot \frac{5}{2}$$

$$\boxed{P(1 < X < 2, 2 < Y < 3) = \frac{5}{128} \approx 0.0391}$$

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Summary

Sampling, Simple Random Sampling, and Stratified Random Sampling

Sampling

When one-by-one study of all units of a population is not possible due to factors like time, cost, manpower, resources, and destructive nature of study, we take a small representative part from the population for study. This small representative part selected for study from the population is called a sample, and the process of selecting a sample from a population is called sampling.

Example: A pathologist takes a syringe of blood as a sample to find out a disease.

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Simple Random Sampling

Simple random sampling is the most common and simplest method of sampling in which each sample unit is selected from a population with equal probability. Every unit in the population has a fixed and equal chance of being selected in the sample.

Example: Suppose a teacher wants to select 5 students from a class of 50. Each student is assigned a number from 1 to 50, and 5 numbers are drawn randomly (using a lottery or random number table). Every student has an equal probability (5/50 = 1/10) of being selected.

Limitations of Simple Random Sampling

• Requires an up-to-date sampling frame.
• When sampling units are widely spread geographically, the cost of collecting data may be high in terms of time and money.
• For a given precision, it usually requires a larger sample size compared to stratified random sampling.

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Stratified Random Sampling

When units in the population are not similar in nature, the population is first divided into subgroups called strata before the sample is drawn. Then a simple random sample is drawn from each stratum in proportion to its size. This method is called stratified random sampling.

Rules for Stratification

• The strata should be non-overlapping and should together comprise the whole population.
• Strata should be as homogeneous within groups and heterogeneous between groups as possible.

Purposes of Stratification

• To make the sample more representative.
• For greater accuracy.
• For administrative convenience.

Example: Suppose a researcher wants to study the income level of 1000 people in a city divided into three income groups:

A simple random sample is drawn from each stratum proportionally, ensuring all income groups are represented in the final sample of 100.

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Key Difference

Requisites for a Good Average & Distribution Analysis

Requisites for a Good Average

1. Rigidly defined - It should have a clear, unambiguous definition.
2. Based on all observations - It should use every value in the data.
3. Easy to understand and compute - Simple to calculate and interpret.
4. Capable of further algebraic treatment - Suitable for further mathematical work.
5. Least affected by sampling fluctuations - Stable across samples.
6. Not unduly affected by extreme values - Robust against abnormal observations.

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Given Data

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

Top 10% means the highest 10% of scores, i.e. the marks exceeded by 90% of students. This is the 90th percentile $P_{90}$.

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

Locate 180 in the cf column: cf = 169 (up to 60-70), cf = 189 (up to 70-80). So the 180th value lies in class 70-80.

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

$$P_{90} = L + \frac{\frac{90N}{100} - cf}{f} \times h = 70 + \frac{180 - 169}{20} \times 10$$

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

$$\boxed{P_{90} = 75.5 \text{ marks}}$$

The minimum marks obtained by the top 10% of students is 75.5.

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Part (ii): Modal Marks

Highest frequency = 60, so modal class is 50-60.

• $L = 50$, $f_1 = 60$, $f_0 = 50$, $f_2 = 45$, $h = 10$

$$\text{Mode} = L + \frac{f_1 - f_0}{2f_1 - f_0 - f_2} \times h$$

$$= 50 + \frac{60 - 50}{2(60) - 50 - 45} \times 10 = 50 + \frac{10}{120 - 95} \times 10$$

$$= 50 + \frac{10}{25} \times 10 = 50 + 4 = 54$$

$$\boxed{\text{Mode} = 54 \text{ marks}}$$

The modal marks of the distribution is 54.

Conditional Probability and Prisoner Escape Problem

Step 1 - Given Data

• Roads: A, B, C selected at random, so $P(A) = P(B) = P(C) = \dfrac{1}{3}$
• $P(E \mid A) = \dfrac{1}{10}$
• $P(E \mid B) = \dfrac{1}{8}$
• $P(E \mid C) = \dfrac{1}{5}$

where $E$ = event of escaping.

Definition: Conditional Probability

Conditional probability is the probability of an event occurring given that another event has already occurred. For events $A$ and $B$:

$$P(A \mid B) = \frac{P(A \cap B)}{P(B)}, \quad P(B) > 0$$

Step 2 - Solve

Part (i): Probability of Escape

By the Theorem of Total Probability:

$$P(E) = P(A)P(E \mid A) + P(B)P(E \mid B) + P(C)P(E \mid C)$$

$$P(E) = \frac{1}{3}\cdot\frac{1}{10} + \frac{1}{3}\cdot\frac{1}{8} + \frac{1}{3}\cdot\frac{1}{5}$$

$$P(E) = \frac{1}{3}\left(\frac{1}{10} + \frac{1}{8} + \frac{1}{5}\right)$$

LCM of $10, 8, 5 = 40$:

$$\frac{1}{10} + \frac{1}{8} + \frac{1}{5} = \frac{4 + 5 + 8}{40} = \frac{17}{40}$$

$$P(E) = \frac{1}{3}\cdot\frac{17}{40} = \frac{17}{120} \approx 0.1417$$

Part (ii): Probability Road A Chosen Given Escape

By Bayes' Theorem:

$$P(A \mid E) = \frac{P(A)P(E \mid A)}{P(E)} = \frac{\frac{1}{3}\cdot\frac{1}{10}}{\frac{17}{120}} = \frac{\frac{1}{30}}{\frac{17}{120}}$$

$$P(A \mid E) = \frac{1}{30}\times\frac{120}{17} = \frac{120}{510} = \frac{4}{17} \approx 0.2353$$

Summary

• $P(\text{escape}) = \dfrac{17}{120} \approx 0.142$
• $P(A \mid \text{escape}) = \dfrac{4}{17} \approx 0.235$

Binomial Distribution

STEP 1 - Given data

• Probability of success (hit) per shot: $p = 0.3$
• Probability of failure (miss): $q = 1 - 0.3 = 0.7$
• Number of independent trials (arrows): $n = 6$
• Required: $P(\text{at least one hit}) = P(X \ge 1)$

All required data present.

STEP 2 - Solution

Definition

A binomial distribution is a discrete probability distribution giving the probability of obtaining exactly $x$ successes in $n$ independent trials, where each trial results in only one of two outcomes (success or failure) with a constant success probability $p$.

Its probability mass function is:

$$P(X = x) = \binom{n}{x} p^x q^{,n-x}, \quad x = 0, 1, 2, \ldots, n$$

where $q = 1 - p$. It is a biparametric distribution with parameters $n$ and $p$.

Mean $= np$, Variance $= npq$.

Conditions for applicability

1. The number of trials $n$ is fixed and finite.
2. Each trial has only two mutually exclusive outcomes (success/failure).
3. The trials are independent.
4. The probability of success $p$ is constant across all trials.

Numerical part

Here each shot is a trial with two outcomes (hit/miss), $n = 6$ fixed, independent shots, and constant $p = 0.3$: so the binomial model applies.

$$X \sim B(n = 6,\ p = 0.3)$$

Using the complement:

$$P(X \ge 1) = 1 - P(X = 0)$$

$$P(X = 0) = \binom{6}{0}(0.3)^0 (0.7)^6 = 1 \cdot 1 \cdot (0.7)^6$$

$$(0.7)^6 = 0.117649$$

Therefore:

$$P(X \ge 1) = 1 - 0.117649 = 0.882351$$

$$\boxed{P(X \ge 1) \approx 0.8824}$$

Interpretation: There is about an 88.24% chance the archer hits the target at least once in six shots.

Normal Distribution: Features and Probability

Features of Normal Distribution

1. Bell-shaped and symmetric about the mean $\mu$; the two halves are mirror images.
2. Mean = Median = Mode, all located at the center $\mu$.
3. Completely defined by two parameters: mean $\mu$ and standard deviation $\sigma$.
4. Total area under the curve equals $1$ (total probability = 1).
5. Asymptotic: the curve approaches but never touches the x-axis, ranging from $-\infty$ to $+\infty$.
6. It follows the empirical rule: about 68% of values lie within $\mu \pm \sigma$, 95% within $\mu \pm 2\sigma$, and 99.7% within $\mu \pm 3\sigma$.
7. PDF:

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

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Numerical Solution

Given data:

• Mean $\mu = 18.25$ s
• Standard deviation $\sigma = 0.34$ s
• Required: $P(X \geq 17.64)$

Step 1: Standardize

$$Z = \frac{X - \mu}{\sigma} = \frac{17.64 - 18.25}{0.34} = \frac{-0.61}{0.34} = -1.79$$

Step 2: Express probability

$$P(X \geq 17.64) = P(Z \geq -1.79)$$

Step 3: Apply symmetry

$$P(Z \geq -1.79) = P(Z \leq 1.79) = 0.5 + P(0 \leq Z \leq 1.79)$$

Step 4: Z-table value

$$P(0 \leq Z \leq 1.79) = 0.4633$$

Step 5: Final result

$$P(X \geq 17.64) = 0.5 + 0.4633 = \boxed{0.9633}$$

Interpretation: There is approximately a 96.33% probability that a print takes at least 17.64 seconds to develop.

Random Variable: Definition, E(X), Variance, and E(g(X))

Step 1 - Extract (Given Data)

• Experiment: rolling a balanced (fair) die
• Random variable $X$ = number of points shown
• Possible values: $x = 1, 2, 3, 4, 5, 6$
• $P(X = x) = \frac{1}{6}$ for each value (balanced die)
• Function required: $g(X) = 2X^2 + 1$

All data present and sufficient.

Step 2 - Solve

Definition of Random Variable

A random variable is a real-valued function that assigns a numerical value to each outcome in the sample space of a random experiment. It may be discrete (countable values) or continuous (values over an interval).

Probability Distribution

$$P(X = x) = \frac{1}{6}, \quad x = 1, 2, 3, 4, 5, 6$$

Calculation of E(X)

$$E(X) = \sum x \cdot P(X=x) = \frac{1+2+3+4+5+6}{6} = \frac{21}{6} = \frac{7}{2} = 3.5$$

Calculation of Variance

$$E(X^2) = \frac{1^2+2^2+3^2+4^2+5^2+6^2}{6} = \frac{1+4+9+16+25+36}{6} = \frac{91}{6}$$

$$\text{Var}(X) = E(X^2) - [E(X)]^2 = \frac{91}{6} - \left(\frac{7}{2}\right)^2 = \frac{91}{6} - \frac{49}{4}$$

$$= \frac{182}{12} - \frac{147}{12} = \frac{35}{12} \approx 2.917$$

Expected Value of g(X) = 2X² + 1

$$E(2X^2 + 1) = 2E(X^2) + 1 = 2 \cdot \frac{91}{6} + 1 = \frac{182}{6} + \frac{6}{6} = \frac{188}{6} = \frac{94}{3} \approx 31.33$$

Summary