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Oddly enough most people think they are better than average.
Average Value Theorem. If f is a continuous function on [a, b], then its average or mean value on [a, b] is given by favg = $\frac{\int_{a}^{b} f(x)dx}{b-a}$
A weighted average is a method of computing a mean where some elements contribute more than others. The weights reflect the relative significance of each data point. In other words, some elements or data points have different weights or importance assigned to them. The weights reflect the relative significance or contribution of each element to the overall average.
For a discrete list of elements $x_1, \dots, x_n$ with corresponding weights $w_1, \dots, w_n$, the weighted average is calculated by multiplying each value by its weight, and dividing by the sum of all the weights:
$\boxed{Weighted\ Average = \frac{w_1x_1 + w_2x_2 + \dots + w_nx_n}{w_1 + \dots + w_n}}$
The continuous analogue of a weighted average requires integrals. Let $f(x)$ be a function defined on $[a, b]$, and let $w(x)$ be a non-negative weight function on the same interval.
Continuous Weighted Average, $\boxed{f_{avg, w} = \frac{\int_{a}^{b} f(x)w(x) \, dx}{\int_{a}^{b} w(x) \, dx}}$
A hot fluid flows through a central cylindrical metal pipe (radius = 5 cm, height = 20 cm) at a temperature of 90°C. It is surrounded by a larger cylindrical layer extending to a radius of 15 cm (height = 20 cm). The temperature decreases linearly from 90°C at the inner wall to 23°C at the outer wall. We need to find the weighted average temperature of this outer layer.
Solution (Figure 3):
:
Find the weight function:
Because the region is a cylinder, the temperature depends on the radial distance $x$. The “weight” function is the volume of an infinitesimally thin cylindrical shell: $w(x) = 2\pi xh$.
Next, we find the linear equation for the temperature $T(x)$ passing through the points $(5, 90)$ and $(15, 23)$:
$T - 90 = \frac{23 - 90}{15 - 5}(x - 5) = \frac{-67}{10}(x - 5)$
$10T - 900 = -67x + 335$
Sover for T, $T(x) = \frac{1235}{10} - \frac{67}{10}x$
Next, we calculate the weighted average using the formula:
$\text{Average} = \frac{\int_{a}^{b} T(x)w(x)dx}{\int_{a}^{b} w(x)dx} = \frac{\int_{5}^{15} T(x) \cdot 2\pi x h \, dx}{\int_{5}^{15} 2\pi x h \, dx}$
Since the constant $2\pi h$ appears in both the numerator and denominator, it cancels out: $\text{Average} = \frac{\int_{5}^{15} x\left(\frac{1235}{10} - \frac{67}{10}x\right)dx}{\int_{5}^{15} x \, dx}$
$\text{Average} = \frac{\int_{5}^{15} \left(\frac{1235}{10}x - \frac{67}{10}x^2\right)dx}{\int_{5}^{15} x \, dx}$
Now, we integrate and evaluate from 5 to 15:
$\text{Average} = \frac{\left[ \frac{1235}{20}x^2 - \frac{67}{30}x^3 \right]_{5}^{15}}{\left[ \frac{x^2}{2} \right]_{5}^{15}}$
$\text{Average} = \frac{(13893.75 - 7537.5) - (1543.75 - 279.17)}{112.5 - 12.5}$
$\text{Average} = \frac{6356.25 - 1264.58}{100} = \frac{5091.67}{100} = 50.91^\circ\text{C}$
Let’s select a point at random along a distribution defined by the curve $w(x) = 1 - x^2$ where $x \in [-1, 1]$. What are the probabilities that $x \ge \frac{1}{2}$ and that $-\frac{1}{2} \le x \le -\frac{1}{4}$?
Solution (Figure 1.d):
For a continuous probability distribution, the probability of finding $x$ in a given interval is the ratio of the area under the curve in that interval (the part) to the total area under the curve (the whole): $\boxed{P(x_1 < x < x_2) = \frac{\int_{x_1}^{x_2} w(x)dx}{\int_{a}^{b} w(x)dx}}$
First, let’s calculate the total area (the whole) from $-1$ to $1$:
$\int_{-1}^{1} (1 - x^2)dx = \left[ x - \frac{x^3}{3} \right]_{-1}^{1} = \left(1 - \frac{1}{3}\right) - \left(-1 + \frac{1}{3}\right) = \frac{2}{3} - \left(-\frac{2}{3}\right) = \frac{4}{3}$
Part A: Probability that $x > \frac{1}{2}$
$\int_{1/2}^{1} (1 - x^2)dx = \left[ x - \frac{x^3}{3} \right]_{1/2}^{1} = \left(1 - \frac{1}{3}\right) - \left(\frac{1}{2} - \frac{1}{24}\right) = \frac{2}{3} - \frac{11}{24} = \frac{16}{24} - \frac{11}{24} = \frac{5}{24}$
Therefore, $P\left(x > \frac{1}{2}\right) = \frac{5/24}{4/3} = \frac{15}{96} = \frac{5}{32}$
Part B: Probability that $-\frac{1}{2} \le x \le -\frac{1}{4}$
By the symmetry of the parabola $1 - x^2$, the area from $-\frac{1}{2}$ to $-\frac{1}{4}$ is identical to the area from $\frac{1}{4}$ to $\frac{1}{2}$.
$\int_{1/4}^{1/2} (1 - x^2)dx = \left[ x - \frac{x^3}{3} \right]_{1/4}^{1/2} = \left(\frac{1}{2} - \frac{1}{24}\right) - \left(\frac{1}{4} - \frac{1}{192}\right)= \frac{11}{24} - \frac{47}{192} = \frac{88}{192} - \frac{47}{192} = \frac{41}{192}$
Therefore, $P\left(-\frac{1}{2} \le x \le -\frac{1}{4}\right) = \frac{41/192}{4/3} = \frac{123}{768} = \frac{41}{256} \approx 0.16$
When a golfer aims at a flagstick, their hits are normally distributed around the target. The number of hits follows a 2D bell-shaped curve, meaning the density of hits at a radial distance $x$ from the center is proportional to $ce^{-x^2}$. What is the probability that the ball lands in a circular ring between radius $r_1$ and $r_2$?
Solution (Figure 4):
To calculate this, we integrate over concentric cylindrical shells.
The volume of each shell is its circumference ($2\pi x$) multiplied by its thickness ($dx$) and its height ($ce^{-x^2}$).
$P(r_1 < x < r_2) = \frac{\int_{r_1}^{r_2} 2\pi x \cdot ce^{-x^2}dx}{\int_{0}^{\infty} 2\pi x \cdot ce^{-x^2}dx} = \frac{\pi c \int_{r_1}^{r_2} 2xe^{-x^2}dx}{\pi c \int_{0}^{\infty} 2xe^{-x^2}dx}$
🧠 We’re told the hit density in the plane is proportional to $f(x,y) \propto e^{-x^2 - y^2}$.
In polar coordinates, $r = \sqrt{x^2 + y^2}$, so this becomes $f(r,\theta) \propto e^{-r^2}$.
The probability that the ball lands in a small ring between radii (r) and (r + dr) is:
(i) Area element in polar coordinates: $dA = r dr d\theta$.
(ii) Density: $ce^{-r^2}$ for some constant (c)
So the probability mass in that ring is $dP = \int_0^{2\pi} ce^{-r^2} r dr d\theta = 2\pi c r e^{-r^2}dr.$
This is exactly our “circumference $\times$ thickness $\times$ height” picture:
where circumference (= $2\pi r$), thickness (= dr), “height” (= $ce^{-r^2}$).
Calling them cylindrical shells is just a 3D metaphor; mathematically, it’s the area of a thin circular ring in 2D.
Calculate the Part (Numerator):
$\text{Part} = \pi c \left[ -e^{-x^2} \right]_{r_1}^{r_2} = \pi c \left( -e^{-r_2^2} - \left(-e^{-r_1^2}\right) \right) = \pi c \left( e^{-r_1^2} - e^{-r_2^2} \right)$
Key Calculus Note:
The integral relies on the fact that the derivative of $-e^{-x^2}$ perfectly matches the core of our integrand: $\frac{d}{dx}\left(-e^{-x^2}\right) = -e^{-x^2} \cdot (-2x) = 2xe^{-x^2}$
Calculate the Whole (Denominator):
$\text{Whole} = \pi c \left[ -e^{-x^2} \right]_{0}^{\infty} = \pi c \left( -e^{-\infty} - \left(-e^{0}\right) \right) = \pi c (0 + 1) = \pi c$
Final Probability: $P(r_1 < x < r_2) = \frac{\pi c \left( e^{-r_1^2} - e^{-r_2^2} \right)}{\pi c} = e^{-r_1^2} - e^{-r_2^2}$
When hitting a golf ball toward the flagstick, the probability distribution of the ball’s landing location follows a 2D bell-shaped curve (normal distribution). We need to calculate the probability that the ball lands in a pond, which occupies a $20^\circ$ angular sector between radial distances $r_1$ and $r_2$ from the hole.
Solution: (Figure 5)
As established in previous exercises, the probability density function for the distance $x$ from the hole is $ce^{-x^2}$. The total volume of this probability space (the “whole”) is $\pi c$.
First, we find the probability of the ball landing anywhere in the full circular ring between $r_1$ and $r_2$: $\text{Part (Full Ring)} = \int_{r_1}^{r_2} 2\pi x \cdot c e^{-x^2} dx = \pi c (e^{-r_1^2} - e^{-r_2^2})$ (Prefvious exercise).
However, the pond does not cover the entire ring. It only occupies a $20^\circ$ slice out of a full $360^\circ$ circle. We multiply the full ring’s probability by this fraction: $\text{Part (Pond)} = \frac{20}{360} \cdot \pi c (e^{-r_1^2} - e^{-r_2^2}) = \frac{1}{18} \pi c (e^{-r_1^2} - e^{-r_2^2})$
To find the final probability, we divide this specific part by the whole: $P(\text{Pond}) = \frac{\frac{1}{18} \pi c (e^{-r_1^2} - e^{-r_2^2})}{\pi c} = \frac{1}{18}(e^{-r_1^2} - e^{-r_2^2})$
Suppose a youngster is throwing darts at a dartboard while his little brother stands nearby. The locations of the youngster’s hits are normally distributed around the center of the board, heavily concentrated near the bullseye but with a small chance of landing far away. What is the probability that the brother gets hit?
Solution:
We start with the general rule of probability: finding the ratio of the “part” (where the brother is) to the “whole” (all possible places the dart could land).
Let us consider a ring of the dart board. The inside of the ring will have radius $r_1$ and the outside will have radius $r_2$. If we plot $y = e^{-r^2}$, we get a “side view” of our probability distribution where the heigh indicate that the darts are most likely to land near the center of the dartboard and less likely to land far out from the center -Figure 1.a-.
To calculate the probability of a dart landing in the ring, we observe the area between $r_1$ and $r_2$ below $y = e^{-r^2}$ and above the r-axis, and compute the volume of revolution of this area revolving about the center of the dart board (y-axis).
Taking into consideration that the probability of a hit is constant at any given radio, the probability of hitting inside the ring is $P(r_1 \le r \le r_2) = \int_{r_1}^{r_2} (2\pi r)e^{-r^{2}}dr = -\pi e^{-r^{2}}\bigg|_{r_1}^{r_2} = -\pi e^{-r_2^{2}}-(-\pi e^{-r_1^{2}}) = \pi (e^{-r_1^{2}}-e^{-r_2^{2}}).$
Determine the youngster’s accuracy.
Let’s assume the youngster hits the actual dartboard (radius $a$) exactly half the time: $P(0 \le r \le a) = \frac{1}{2}$
Then, $e^{-0^2} - e^{-a^2} = \frac{1}{2} \implies 1 - e^{-a^2} = \frac{1}{2} \implies e^{-a^2} = \frac{1}{2}$.
Define the brother’s location.
We approximate the little brother’s location using two parameters:
$\star$ Distance: He is standing at twice the board’s radius away, so he occupies the space between $2a$ and $3a$.
$\star$ Angle: He is standing to the right of the board, roughly between 3 o’clock and 5 o’clock. On a 12-hour clock face, this spans 2 hours, which is $\frac{2}{12} = \frac{1}{6}$ of the full circle.
Calculate the probability of hitting the brother.
We take $\frac{1}{6}$ of the probability of landing in the ring between $2a$ and $3a$: $P(\text{Brother}) = \frac{1}{6} \cdot P(2a \le r \le 3a)$.
Therefore, $P(\text{Brother}) = \frac{1}{6} \left( e^{-(2a)^2} - e^{-(3a)^2} \right) = \frac{1}{6} \left( e^{-4a^2} - e^{-9a^2} \right)$
Using exponent rules, we rewrite this so we can substitute our known value of $e^{-a^2} = \frac{1}{2}$: $P(\text{Brother}) = \frac{1}{6} \left[ \left(e^{-a^2}\right)^4 - \left(e^{-a^2}\right)^9 \right] = \frac{1}{6} \left[ \left(\frac{1}{2}\right)^4 - \left(\frac{1}{2}\right)^9 \right]$
$P(\text{Brother}) = \frac{1}{6} \left( \frac{1}{16} - \frac{1}{512} \right)$
Because $\frac{1}{512}$ is negligibly small compared to $\frac{1}{16}$, we can approximate the result as: $P(\text{Brother}) \approx \frac{1}{6} \times \frac{1}{16} \approx \frac{1}{96} \approx 1 \%$ (this ignores the subtraction entirely).
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