Statistics for
Data Science 1

1.1CDF and interval probability

• The CDF never decreases, starts from 0 and ends at 1.

1.2Worked questions

Question 1: daily rainfall

The CDF of the daily rainfall $X$ (in mm) is $F(x) = 1 - e^{-0.5x}$ for $x \ge 0$. Find the probability that the rainfall exceeds 10 mm.

Question 2: a piecewise CDF with an unknown k

The CDF of a random variable $X$ is below. If $P(X \le 4) = \frac{3}{4}$, find $P(1 \le X \le 5)$.

1. Pick the right piece. $x = 4$ lies in $2 \le x < 6$, so $F(4) = k + \frac{4 - 2}{8} = k + \frac{1}{4}$.
2. Use the given value. $k + \frac{1}{4} = \frac{3}{4}$, so $k = \frac{3}{4} - \frac{1}{4} = \frac{1}{2}$.
3. Interval probability. $P(1 \le X \le 5) = F(5) - F(1)$. Here $F(5) = k + \frac{5 - 2}{8} = \frac{1}{2} + \frac{3}{8} = \frac{7}{8}$ and $F(1) = \frac{k}{4} = \frac{1}{8}$.
4. So $P(1 \le X \le 5) = \frac{7}{8} - \frac{1}{8} = \frac{6}{8} = \frac{3}{4}$.

1.3Going deeper

1. Check the pieces join up. At $x = 2$: $\frac{kx^2}{4} = k$ and $k + \frac{2-2}{8} = k$. At $x = 6$: $k + \frac{4}{8} = 1$ also gives $k = \frac{1}{2}$. Two different conditions give the same $k$.
2. Which piece? Always find which interval $x$ falls in before substituting.
3. Less than and at most. $P(X > a) = 1 - F(a)$ because the whole probability is 1 and $F(a)$ is the part up to $a$.

• In a discrete variable, the CDF is a staircase; each step sits at a value $X$ can take.
• The height of the step is the probability of exactly that value.

2.1Worked question: reading probabilities off a staircase

A random variable has the CDF below. Find $P(X = 0)$, $P(X = 1)$ and $P(X = 2)$.

2.2Going deeper

1. Closed dot at the top of the step. The CDF includes the value itself ($\le$), so at $x = 0$ it is already $\frac{1}{3}$.
2. From CDF to PMF. Every jump height is a PMF value, so the staircase carries the whole PMF: 0 $\to$ $\frac{1}{3}$, 1 $\to$ $\frac{1}{6}$, 2 $\to$ $\frac{1}{2}$.
3. A mixed variable has both: jumps at some values and a smooth rise between them (the right picture of Figure 1).

So end points can be included or left out:

3.1Worked question: finding the exponent k

A random variable has the CDF below. Find $P(1 < X < 2)$.

1. Find k from the join at x = 3. The CDF cannot jump, so the middle piece must reach 1 at $x = 3$: $\frac{(3-1)^k}{8} = 1$, so $\frac{2^k}{8} = 1$ and $2^k = 8$, hence $k = 3$.
2. Interval probability. $P(1 < X < 2) = F(2) - F(1) = \frac{(2-1)^3}{8} - 0 = \frac{1}{8}$.

3.2Going deeper

1. Why a single value has probability 0. The probability of an interval of width $w$ shrinks to 0 as $w \to 0$, so the single point $x_0$ gets $F(x_0) - F(x_0) = 0$.
2. Probability 0 is not impossible. The event $X = x_0$ can be a possible outcome yet have probability 0 (extra remark).
3. Contrast with discrete. For a discrete variable $P(1 < X < 2)$ and $P(1 \le X \le 2)$ can differ; for a continuous one they never do.

For a continuous random variable $P(X = x)$ is 0, so we use a density function $f(x)$. Probability is the area under $f$ over an interval, and a high density means $X$ takes values around there more often.

The PDF is easier for probability calculations and shows the shape of the distribution.

4.1A valid PDF and finding k

The density must never be negative and must enclose total area 1. Using this we find $k$.

Worked question: find k

A random variable has PDF $f_X(x) = ke^{-x}$ for $x \ge 0$ and 0 otherwise. Find $k$.

The upper limit gives $e^{-\infty} = 0$, and the lower limit gives $\frac{e^{0}}{-1} = -1$, so the bracket is $0 - (-1) = 1$.

4.2Going deeper

1. The PDF can be bigger than 1. It is a density, not a probability; only areas are probabilities. A PDF equal to 5 on a very narrow interval is fine (extra remark).
2. CDF and PDF are two views of one thing. The CDF accumulates the PDF; the PDF is the slope of the CDF. For $F(x) = 1 - e^{-0.5x}$ the PDF is $f(x) = 0.5e^{-0.5x}$.
3. Always integrate over where the PDF is non-zero. Here that is $x \ge 0$; outside it $f = 0$.

5.1Worked question: how much petrol to stock

At a petrol station the daily petrol demand $X$ (in thousands of litres) has PDF $f_X(x) = k(1 - x)^4$ for $0 \le x \le 1$, and 0 otherwise. What should be the minimum supply $s$ (in thousands of litres) so that the probability of the petrol running out on any given day is only 1%? (a) 0.4321 (b) 0.3981 (c) 0.6019 (d) 0.9903

1. Find k. $\int_0^1 k(1-x)^4\, dx = k\left[\frac{-(1-x)^5}{5}\right]_0^1 = \frac{k}{5} = 1$, so $k = 5$ and $f(x) = 5(1-x)^4$.
2. Running out means demand above supply. We need $P(X > s) = 0.01$, that is $\int_s^1 5(1-x)^4\, dx = 0.01$.
3. Integrate. $5\left[\frac{(1-x)^5}{-5}\right]_s^1 = (1-s)^5$, so $(1-s)^5 = 0.01$.
4. Solve. $1 - s = (0.01)^{1/5} \approx 0.3981$, so $s = 1 - (0.01)^{1/5} \approx 0.6019$.

5.2Percentiles

Worked question: the 75th percentile

A random variable has PDF $f_X(x) = kx$ for $0 \le x < 2$, and 0 otherwise. Find the 75th percentile of $X$. (Hint: the value $k$ such that $P(X \le k) = 0.75$.) Note that the letter $k$ is used twice: first as the constant in the PDF, then as the percentile.

1. Find the constant. $\int_0^2 kx\, dx = 2k = 1$, so the constant is $\frac{1}{2}$ and $f(x) = \frac{x}{2}$.
2. Set up the percentile. $P(X \le k) = \int_0^k \frac{x}{2}\, dx = \left[\frac{x^2}{4}\right]_0^k = \frac{k^2}{4} = 0.75$.
3. Solve. $k^2 = 3$, so $k = \sqrt{3} \approx 1.732$ (the positive root, since $0 \le X < 2$).

5.3Going deeper

1. Tail probabilities are what safety margins ask for. “Running out only 1% of days” means choosing $s$ so the upper tail has area 0.01, i.e. $s$ is the 99th percentile of the demand.
2. The median is the 50th percentile. For $f(x) = \frac{x}{2}$ it solves $\frac{k^2}{4} = 0.5$, so $k = \sqrt{2}$ (extra remark).
3. Check the numbers. $(0.3981)^5 \approx 0.0100$, and $\frac{(\sqrt{3})^2}{4} = 0.75$.

For example $P(X < s \mid X > t)$ with $t < s$ becomes $\dfrac{P(t < X < s)}{P(X > t)}$.

6.1Practice on conditional probability

Question 1

A random variable has PDF $f_X(x) = 5x^4$ for $0 < x \le 1$, and 0 otherwise. Find $P\left(X \le \frac{8}{14} \mid X > \frac{1}{14}\right)$.

Question 2

Same PDF. Find $P\left(X \le \frac{6}{7} \mid X > \frac{1}{7}\right)$.

6.2The uniform and exponential distributions

Question 3: package delivery (Statistics 2 Quiz 1, 13 Jul 2025)

A package delivery time $T$ (in hours) is uniformly distributed over the interval $[a, b]$. The probability that the delivery takes less than 5 hours is equal to the probability that it takes more than 11 hours. If the value of $b$ is 14, find $P(5 < T < 11 \mid T > 7)$.

1. Find a. $P(T < 5) = \frac{5 - a}{b - a}$ and $P(T > 11) = \frac{b - 11}{b - a}$. Equal, so $5 - a = b - 11$, hence $a + b = 16$ and, with $b = 14$, $a = 2$.
2. Conditional probability. $P(5 < T < 11 \mid T > 7) = \frac{P(7 < T < 11)}{P(T > 7)}$, because inside $T > 7$ the event $5 < T < 11$ means $7 < T < 11$.
3. Lengths. The numerator has length $11 - 7 = 4$ and the denominator $14 - 7 = 7$ (the common width $b - a = 12$ cancels).

Question 4: finding a (Statistics 2 Quiz 1, 23 Feb 2025)

The PDF of a random variable $X$ is $f_X(x) = ae^{-3x}$ for $x \ge 0$, and 0 otherwise. Find the value of $a$.

6.3Going deeper

1. Conditioning shrinks the sample space. Given $T > 7$, only $[7, 14]$ is left, of length 7; the uniform density on it is $\frac{1}{7}$, and the answer $\frac{4}{7}$ is the length 4 over 7.
2. The exponent rule. $\int_0^{\infty} ce^{-\lambda x}\, dx = \frac{c}{\lambda}$; setting it to 1 gives $c = \lambda$, which is why the exponential density is $\lambda e^{-\lambda x}$. In Question 4, $\lambda = 3$ so $a = 3$.
3. Uniform probabilities are length ratios. $P(c < T < d) = \frac{d - c}{b - a}$ for $a \le c < d \le b$ (extra remark).

7.1Expected value of a function, mean and variance

Question 5: variance of a continuous variable (Statistics 2 Quiz 1, 13 Jul 2025)

Let $X$ be a continuous random variable with PDF $f_X(x) = \frac{x}{2}$ for $0 \le x \le 2$, and 0 otherwise. Calculate $\text{Var}(X)$.

1. Mean. $E[X] = \int_0^2 x \cdot \frac{x}{2}\, dx = \frac{1}{2}\left[\frac{x^3}{3}\right]_0^2 = \frac{1}{2} \cdot \frac{8}{3} = \frac{4}{3}$.
2. Second moment. $E[X^2] = \int_0^2 x^2 \cdot \frac{x}{2}\, dx = \frac{1}{2}\left[\frac{x^4}{4}\right]_0^2 = \frac{1}{2} \cdot \frac{16}{4} = 2$.
3. Variance. $\text{Var}(X) = E[X^2] - (E[X])^2 = 2 - \left(\frac{4}{3}\right)^2 = 2 - \frac{16}{9} = \frac{2}{9} \approx 0.222$.

Question 6: a linear change

For the same $X$, define $Y = 72X + 5$. Find $\text{Var}(Y)$.

7.2Uniform [a, b]

Question 7: the value of a + b (Statistics 2 Quiz 1, 13 Jul 2025)

A package delivery time $T$ (in hours) is uniformly distributed over $[a, b]$. The probability that the delivery takes less than 5 hours is equal to the probability that it takes more than 11 hours. Find $a + b$.

7.3Exponential (λ)

Question 8: Jadeja batting (Statistics 2 Quiz 1, 5 Jun 2022)

During an IPL cricket match, the amount of time (in minutes) Jadeja bats follows the exponential distribution with an expected time of 50 minutes. What is the probability that Jadeja will bat for more than 40 minutes?

1. Find λ. $E[X] = \frac{1}{\lambda} = 50$, so $\lambda = \frac{1}{50}$ and $f(x) = \frac{1}{50}e^{-x/50}$.
2. Tail probability. $P(X > 40) = \int_{40}^{\infty} \frac{1}{50}e^{-x/50}\, dx = \left[-e^{-x/50}\right]_{40}^{\infty} = e^{-40/50}$.

Question 9: the service center (Statistics 2 Quiz 1, 13 Jul 2025)

The time (in minutes) a customer spends inside a service center follows an exponential distribution with an average of 10 minutes. What is the probability that a customer has to wait at least 12 more minutes in the center, given that he has already spent 10 minutes in the center?

1. Set up. The notes write $P(X > 12 \mid X > 10)$, reading “12 more” loosely as the interval from the start; the question asks for 12 more minutes after 10, which is $P(X > 22 \mid X > 10)$.
2. Rate. $\lambda = \frac{1}{10}$, so $P(X > t) = e^{-t/10}$.
3. Conditional probability. $P(X > 22 \mid X > 10) = \frac{P(X > 22)}{P(X > 10)} = \frac{e^{-2.2}}{e^{-1}} = e^{-1.2} \approx 0.301$. For comparison the notes' set-up $P(X > 12 \mid X > 10) = e^{-0.2} \approx 0.819$.

7.4Normal (μ, σ²)

The notes end here with the mean and variance of the normal distribution, written in terms of its two parameters.

7.5Going deeper

1. Same formulas as discrete, with integrals. The sum $\sum x\,P(X = x)$ of Week 3 becomes $\int x\, f(x)\, dx$.
2. Linear change of variance. $\text{Var}(aX + b) = a^2\,\text{Var}(X)$: the shift $b$ does nothing, the scale $a$ is squared (as in Statistics 1, Week 3).
3. Check against the formulas. For $f(x) = \frac{x}{2}$ on $[0, 2]$ the mean $\frac{4}{3}$ lies to the right of the middle 1, as the density is larger on the right.
4. Uniform check. For $U[2, 14]$: mean $\frac{2 + 14}{2} = 8$ (the midpoint) and variance $\frac{12^2}{12} = 12$.

Everything in Week 4 on one sheet, for a last-minute revision.

CDF and PDF

Named distributions

Rules of thumb

Values computed in the notes