Lesson 12: Normal Distribution

Calendar

Block I calendar with a red box around Thursday 17 September, Lesson 12, Normal Distribution.

What We Did: Lessons 1 through 11

  • Population vs sample, parameter (\(\mu\), \(\sigma\), \(p\)) vs statistic (\(\bar{x}\), \(s\), \(\hat{p}\)).
  • Random sampling buys generalization, random assignment buys causation.
  • Center: mean, median, trimmed mean. Spread: \(s^2\), \(s\), and the fourth spread \(f_s\).
  • Experiment, sample space \(\mathcal{S}\), event as a subset of \(\mathcal{S}\).
  • Union is “or”, intersection is “and”, complement is “not”.
  • Three axioms, the complement rule \(P(A') = 1 - P(A)\), and the addition rule \(P(A \cup B) = P(A) + P(B) - P(A \cap B)\).
  • Equally likely outcomes: \(P(A) = N(A)/N\).
  • Product rule: \(n_1 n_2 \cdots n_k\).
  • Permutation (order matters): \(P_{k,n} = \dfrac{n!}{(n-k)!}\).
  • Combination (order does not): \(\dbinom{n}{k} = \dfrac{n!}{k!\,(n-k)!}\).
  • Conditional probability: \(P(A \mid B) = \dfrac{P(A \cap B)}{P(B)}\).
  • Multiplication rule, Law of Total Probability, and Bayes’ Theorem.
  • Independent: \(P(A \mid B) = P(A)\), tested with \(P(A \cap B) = P(A)\,P(B)\).
  • \(P(\text{at least one}) = 1 - P(\text{none})\).
  • pmf: \(p_X(x) = P(X = x)\). cdf: \(F_X(x) = P(X \le x)\), a step function.
  • Binomial: BINS, \(p(x) = \dbinom{n}{x} p^x (1-p)^{n-x}\), \(E(X) = np\), \(V(X) = np(1-p)\).
  • Poisson: a rate and a window, \(p(x; \mu) = \dfrac{e^{-\mu}\mu^x}{x!}\), \(E(X) = V(X) = \mu\).
  • \(E(X) = \sum_x x\,p_X(x)\) and \(V(X) = E(X^2) - [E(X)]^2\).
  • Probability is area under the pdf: \(P(a \le X \le b) = \int_a^b f(x)\,dx\).
  • \(P(X = c) = 0\), so \(\le\) and \(<\) give the same answer.
  • cdf: \(F(x) = P(X \le x)\), and \(P(a \le X \le b) = F(b) - F(a)\).
  • Percentile: solve \(F(c) = P(X \le c) = p\) for \(c\).
  • \(E(X) = \int x\,f(x)\,dx\) and \(V(X) = E(X^2) - [E(X)]^2\).

Warm-Up

\[f_X(x) = \begin{cases} kx^3 & 0 \le x \le 2 \\ 0 & \text{otherwise} \end{cases}\]

a) Find \(k\).

The total area is \(1\).

\[\begin{aligned} \int_0^2 kx^3 \, dx &= 1 \\ \frac{kx^4}{4} \bigg|_0^2 &= 1 \\ 4k &= 1 \\ k &= \frac{1}{4} \end{aligned}\]

\[f_X(x) = \begin{cases} \dfrac{x^3}{4} & 0 \le x \le 2 \\ 0 & \text{otherwise} \end{cases}\]

The pdf x cubed over 4 on 0 to 2 with the whole region under it shaded.

b) Fully define the cdf.

\[F_X(x) = P(X \le x) = \int_0^x \frac{y^3}{4} \, dy = \frac{y^4}{16} \bigg|_0^x = \frac{x^4}{16}\]

\[F_X(x) = P(X \le x) = \begin{cases} 0 & x < 0 \\ \dfrac{x^4}{16} & 0 \le x \le 2 \\ 1 & x > 2 \end{cases}\]

c) What is the probability \(X\) lands more than one standard deviation above the mean or more than one standard deviation below it?

\[\begin{aligned} E(X) &= \int_0^2 x \cdot \frac{x^3}{4} \, dx = \int_0^2 \frac{x^4}{4} \, dx = \frac{x^5}{20} \bigg|_0^2 = \frac{32}{20} = 1.6 \\ E(X^2) &= \int_0^2 x^2 \cdot \frac{x^3}{4} \, dx = \int_0^2 \frac{x^5}{4} \, dx = \frac{x^6}{24} \bigg|_0^2 = \frac{64}{24} = 2.6667 \\ V(X) &= E(X^2) - [E(X)]^2 = 2.6667 - 1.6^2 = 0.1067 \\ \sigma &= \sqrt{0.1067} = 0.3266 \end{aligned}\]

\[\mu \pm \sigma = 1.6 \pm 0.3266 = (1.2734,\ 1.9266)\]

\[P(X < 1.2734) + P(X > 1.9266) = F(1.2734) + 1 - F(1.9266)\]

\[= \frac{1.2734^4}{16} + 1 - \frac{1.9266^4}{16} = 0.1643 + 1 - 0.8611 = \mathbf{0.3032}\]

The pdf x cubed over 4 with both tails beyond one standard deviation shaded and a triangle at the mean, next to the cdf evaluated at 1.2734 and 1.9266.

The endpoints carry zero probability.

d) Find the 20th percentile.

Find \(c\) so that \(F(c) = P(X \le c) = 0.2\).

\[\begin{aligned} F(c) = P(X \le c) &= 0.2 \\ \frac{c^4}{16} &= 0.2 \\ c^4 &= 3.2 \\ c &= \sqrt[4]{3.2} = 1.3375 \end{aligned}\]

The pdf x cubed over 4 with the area from 0 to 1.3375 shaded, next to the cdf evaluated at 1.3375 with height 0.2.

What We’re Doing: Lesson 12

Objectives

  • Identify the normal and standard normal (\(z\)) distributions. (SLO 7)
  • Standardize values with \(z = (x-\mu)/\sigma\) and compute probabilities using the \(z\) table. (SLO 7)
  • Determine percentiles and \(z\) critical values for the normal distribution. (SLO 7)

Required Reading

Devore 4.3

Break!

Reese

DMath Frisbee!!

Math vs DPE

1-0

14-5

The Takeaway for Today

NoteKey Concepts from Lesson 12
  • The normal is a continuous distribution fixed by two parameters: \(X \sim N(\mu, \sigma^2)\), where \(E(X) = \mu\) and \(V(X) = \sigma^2\)
  • Probability is still area under the pdf, so \(P(X = a) = 0\) and \(\le\) matches \(<\)
  • pnorm(x, mean, sd) is the cdf \(F(x) = P(X \le x)\). The third slot is the standard deviation, not the variance
  • Standardize with \(z = \dfrac{x - \mu}{\sigma}\) to land on \(Z \sim N(0, 1)\). The area never changes, so pnorm(x, mu, sd) and pnorm(z) agree
  • The empirical rule: about 68%, 95%, and 99.7% of the area sits within 1, 2, and 3 standard deviations of \(\mu\)
  • qnorm(p, mean, sd) runs backwards, from a probability to the value that cuts it off
  • A critical value \(z_\alpha\) has area \(\alpha\) to its right

The Normal Distribution

The normal distribution is the most important one in all of probability and statistics.

Devore, Section 4.3

ImportantThe Normal pdf

\(X\) is normal with parameters \(\mu\) and \(\sigma^2\), written \(X \sim N(\mu, \sigma^2)\), if

\[f(x; \mu, \sigma) = \frac{1}{\sqrt{2\pi}\,\sigma}\, e^{-(x-\mu)^2 / (2\sigma^2)}, \qquad -\infty < x < \infty\]

  • \(\mu\) locates the center. \(\sigma\) sets the spread.
  • The support is all real numbers, so the curve never actually touches zero.
  • The curve is symmetric about \(\mu\), which makes \(\mu\) the mean and the median.

Three normal curves on shared axes. Two share a mean of 16 with standard deviations 2 and 4, and a third centered at 22 with standard deviation 2.

Mean, Variance, and Standard Deviation

For the normal, there is nothing to compute. The parameters are the mean and the variance.

ImportantMean and variance of a normal

\[E(X) = \mu, \qquad V(X) = \sigma^2, \qquad \sigma_X = \sigma\]

Substitute \(x = \mu + \sigma z\), where \(\phi(z)\) is the standard normal pdf. Then \(\int \phi(z)\,dz = 1\), and \(z\,\phi(z)\) is odd, so it integrates to \(0\).

\[\begin{aligned} E(X) &= \int_{-\infty}^{\infty} x\,f(x)\,dx \\ &= \int_{-\infty}^{\infty} (\mu + \sigma z)\,\phi(z)\,dz \\ &= \mu(1) + \sigma(0) \\ &= \mu \\[1em] V(X) &= E\left[(X - \mu)^2\right] \\ &= \sigma^2 \int_{-\infty}^{\infty} z^2\,\phi(z)\,dz \\ &= \sigma^2 \left( \Big[-z\,\phi(z)\Big]_{-\infty}^{\infty} + \int_{-\infty}^{\infty} \phi(z)\,dz \right) \\ &= \sigma^2(0 + 1) \\ &= \sigma^2 \end{aligned}\]

The last step is integration by parts with \(u = z\) and \(dv = z\,\phi(z)\,dz\), so \(v = -\phi(z)\).

Fully Specify the Distribution

Cadet 2 mile run times average 16 minutes with a standard deviation of 2 minutes.

NoteFully specify

\[X \sim N(\mu = 16,\ \sigma^2 = 2^2)\]

  • Designate the random variable. \(X\) is a cadet’s 2 mile run time in minutes.
  • Name the distribution. Normal.
  • Specify the parameters. \(\mu = 16\), \(\sigma = 2\).

So \(E(X) = 16\) minutes, \(V(X) = 4\), and \(\sigma = 2\) minutes.

The Cumulative Distribution Function

ImportantThe Normal cdf

\[F(x) = P(X \le x) = \int_{-\infty}^{x} \frac{1}{\sqrt{2\pi}\,\sigma}\, e^{-(t-\mu)^2 / (2\sigma^2)} \, dt\]

Same idea as Lesson 11: \(F(x)\) is the area under the pdf to the left of \(x\). This integral has no closed form, so R evaluates it.

Find \(P(X < 13)\).

\[P(X < 13) = F(13) = \int_{-\infty}^{13} \frac{1}{2\sqrt{2\pi}}\, e^{-(x-16)^2/8} \, dx\]

Left, the run time normal pdf with the area left of 13 shaded. Right, the run time normal cdf with a dashed orange line up from 13 to the curve and across to 0.0668 on the y axis.

Computing Normal Probabilities in R

Importantpnorm is the cdf

\[\texttt{pnorm(x, mean, sd)} \;=\; F(x) \;=\; P(X \le x)\]

  • The third argument is the standard deviation \(\sigma\), not the variance.
  • Leave the arguments off and R assumes \(\mu = 0\) and \(\sigma = 1\).
  • dnorm is the height of the pdf. It is not a probability.

Same questions as Lesson 11, now for the run times, \(X \sim N(16, 2^2)\).

a) \(P(X < 13)\)

\[P(X < 13) = \int_{-\infty}^{13} \frac{1}{2\sqrt{2\pi}}\, e^{-(x-16)^2/8} \, dx = F(13)\]

The run time normal curve with the area to the left of 13 shaded.

pnorm(13, 16, 2) = 0.0668

b) \(P(X \le 13)\)

\[P(X \le 13) = \int_{-\infty}^{13} \frac{1}{2\sqrt{2\pi}}\, e^{-(x-16)^2/8} \, dx = F(13)\]

The run time normal curve with the area to the left of 13 shaded.

pnorm(13, 16, 2) = 0.0668

c) \(P(X > 19)\)

\[P(X > 19) = \int_{19}^{\infty} \frac{1}{2\sqrt{2\pi}}\, e^{-(x-16)^2/8} \, dx = 1 - F(19)\]

The run time normal curve with the area to the right of 19 shaded.

1 - pnorm(19, 16, 2) = 0.0668

d) \(P(X \ge 19)\)

\[P(X \ge 19) = \int_{19}^{\infty} \frac{1}{2\sqrt{2\pi}}\, e^{-(x-16)^2/8} \, dx = 1 - F(19)\]

The run time normal curve with the area to the right of 19 shaded.

1 - pnorm(19, 16, 2) = 0.0668

e) \(P(15 < X < 18)\)

\[P(15 < X < 18) = \int_{15}^{18} \frac{1}{2\sqrt{2\pi}}\, e^{-(x-16)^2/8} \, dx = F(18) - F(15)\]

The run time normal curve with the area between 15 and 18 shaded.

pnorm(18, 16, 2) - pnorm(15, 16, 2) = 0.5328

f) \(P(14 \le X \le 18)\)

\[P(14 \le X \le 18) = \int_{14}^{18} \frac{1}{2\sqrt{2\pi}}\, e^{-(x-16)^2/8} \, dx = F(18) - F(14)\]

The run time normal curve with the area between 14 and 18 shaded, one standard deviation either side of 16.

pnorm(18, 16, 2) - pnorm(14, 16, 2) = 0.6827

\(14\) and \(18\) are one standard deviation either side of \(\mu = 16\).

g) \(P(12 < X < 20)\)

\[P(12 < X < 20) = \int_{12}^{20} \frac{1}{2\sqrt{2\pi}}\, e^{-(x-16)^2/8} \, dx = F(20) - F(12)\]

The run time normal curve with the area between 12 and 20 shaded, two standard deviations either side of 16.

pnorm(20, 16, 2) - pnorm(12, 16, 2) = 0.9545

\(12\) and \(20\) are two standard deviations either side of \(\mu = 16\).

h) \(P(X = 16)\)

\[P(X = 16) = \int_{16}^{16} \frac{1}{2\sqrt{2\pi}}\, e^{-(x-16)^2/8} \, dx = F(16) - F(16) = 0\]

pnorm(16, 16, 2) - pnorm(16, 16, 2) = 0.0000

Warningdnorm is not a probability

dnorm(16, 16, 2) = 0.1995

That is the height of the pdf at \(16\), not \(P(X = 16)\). A line has no width, so the probability is \(0\), and every \(\le\) can be read as \(<\).

The Standard Normal Distribution

ImportantStandardizing

If \(X \sim N(\mu, \sigma^2)\), then

\[Z = \frac{X - \mu}{\sigma} \sim N(0, 1)\]

Subtracting \(\mu\) slides the center to \(0\). Dividing by \(\sigma\) squeezes the spread to \(1\). \(Z\) is the standard normal, and a value of \(z\) says how many standard deviations \(x\) sits from the mean.

For a 13 minute run time,

\[z = \frac{x - \mu}{\sigma} = \frac{13 - 16}{2} = -1.5\]

which reads “13 minutes is 1.5 standard deviations below the mean.”

NoteThe area does not move

Standardizing relabels the axis. It does not change the picture.

Two normal curves side by side. On the left, X normal with mean 16 and standard deviation 2 with the area left of 13 shaded. On the right, the standard normal with the area left of negative 1.5 shaded. The two shaded regions have the same shape.

Animation of the run time normal being standardized. The curve slides left by 16 so it centers at 0, then squeezes by a factor of 2 into the standard normal. The shaded area left of 13 follows along and ends as the area left of negative 1.5, labeled 0.0668 throughout.

Finding Probabilities with \(X\) and \(Z\)

Problem Not standardized Standardized Answer
\(P(X < 13)\) pnorm(13, 16, 2) pnorm(-1.5) 0.0668
\(P(X > 19)\) 1 - pnorm(19, 16, 2) 1 - pnorm(1.5) 0.0668
\(P(15 < X < 18)\) pnorm(18, 16, 2) - pnorm(15, 16, 2) pnorm(1) - pnorm(-0.5) 0.5328
NoteWhy do this?

A \(z\) score is a common scale. A cadet who runs 13 minutes (\(z = -1.5\)) and a cadet who deadlifts 240 pounds in a population with \(\mu = 200\) and \(\sigma = 30\) (\(z = 1.33\)) can now be ranked against their own distributions, even though minutes and pounds have nothing to do with each other.

It is also what every inference procedure in Block II runs on.

The Empirical Rule

Standardizing puts every normal on the same \(Z\) scale, so the area within \(k\) standard deviations of the mean is the same for every normal.

ImportantThe 68, 95, 99.7 rule
  • \(P(\mu - \sigma \le X \le \mu + \sigma) = P(-1 \le Z \le 1) = 0.6827\)
  • \(P(\mu - 2\sigma \le X \le \mu + 2\sigma) = P(-2 \le Z \le 2) = 0.9545\)
  • \(P(\mu - 3\sigma \le X \le \mu + 3\sigma) = P(-3 \le Z \le 3) = 0.9973\)

Three copies of the standard normal curve with the areas within one, two, and three of 0 shaded, labeled 0.6827, 0.9545, and 0.9973.

Problem Not standardized Standardized Answer
Within 1 sd pnorm(18, 16, 2) - pnorm(14, 16, 2) pnorm(1) - pnorm(-1) 0.6827
Within 2 sd pnorm(20, 16, 2) - pnorm(12, 16, 2) pnorm(2) - pnorm(-2) 0.9545
Within 3 sd pnorm(22, 16, 2) - pnorm(10, 16, 2) pnorm(3) - pnorm(-3) 0.9973

Parts f) and g) were exactly this: \(P(14 \le X \le 18) = P(-1 \le Z \le 1)\) and \(P(12 < X < 20) = P(-2 < Z < 2)\).

Going Backwards: Percentiles and Critical Values

ImportantDefinition: Percentile

The \((100p)\)th percentile is the value \(c\) that satisfies \[F(c) = P(X \le c) = p\]

90th percentile of run times. Find \(c\) so that \(F(c) = P(X \le c) = 0.9\).

The run time normal cdf. A dashed orange line runs across from 0.9 on the y axis to the curve, then down to the x axis at 18.5631.

By hand, solve for \(c\):

\[F(c) = P(X \le c) = \int_{-\infty}^{c} \frac{1}{2\sqrt{2\pi}}\, e^{-(x-16)^2/8} \, dx = 0.9\]

There is no closed form to solve for \(c\), so R runs the cdf backwards.

Importantqnorm is the inverse cdf

\[\texttt{qnorm(p, mean, sd)} \;=\; c \text{ where } F(c) = P(X \le c) = p\]

qnorm(0.90, 16, 2) = 18.5631

The Same Percentile in \(z\) Units

The 90th percentile of \(Z\) is qnorm(0.9) = 1.2816. Unstandardize it with \(x = \mu + z\sigma\):

\[c = 16 + 2(1.2816) = 18.5631\]

The 90th percentile run time is \(1.2816\) standard deviations above the mean.

Two normal curves side by side. On the left, run times with the area left of 18.5631 shaded. On the right, the standard normal with the area left of 1.2816 shaded. Both shaded areas are 0.90.

Critical Values

That \(1.2816\) has area \(0.90\) to its left, so it has area \(0.10\) to its right. Statisticians name it by the right tail: \(z_{0.10}\).

Important\(z_\alpha\)

\(z_\alpha\) has area \(\alpha\) to its right, so it is the \(100(1-\alpha)\)th percentile of \(Z\): qnorm(1 - alpha).

Critical value Area to the right Area to the left R-Lite Answer
\(z_{0.10}\) 0.10 0.90 qnorm(0.90) 1.2816
\(z_{0.05}\) 0.05 0.95 qnorm(0.95) 1.6449
\(z_{0.025}\) 0.025 0.975 qnorm(0.975) 1.9600

\(z_{0.10}\), the 90th percentile

The standard normal curve with area 0.10 shaded to the right of 1.2816.

\(z_{0.05}\), the 95th percentile

The standard normal curve with area 0.05 shaded to the right of 1.6449.

\(z_{0.025}\), the 97.5th percentile

The standard normal curve with area 0.025 shaded to the right of 1.9600.

Those two numbers, \(1.645\) and \(1.96\), come back in every confidence interval in Block II.

Distributions So Far

Three named distributions. Naming one is not an answer. Fully specifying it is, and it is the same three steps every time: designate the random variable in words with units, name the distribution, specify the parameters as numbers.

Binomial Poisson Normal
Type Discrete Discrete Continuous
Use it when \(x\) successes in \(n\) fixed trials \(x\) events in a window Symmetric, bell shaped measurement
Fully specified \(X \sim \text{Binom}(n,\ p)\) \(X \sim \text{Pois}(\mu)\) \(X \sim N(\mu,\ \sigma^2)\)
Parameters \(n\) trials, \(p\) success probability \(\mu = \lambda t\), expected count \(\mu\) center, \(\sigma\) spread
Possible values \(x = 0, 1, \dots, n\) \(x = 0, 1, 2, \dots\) \(-\infty < x < \infty\)
pmf or pdf \(\dbinom{n}{x} p^x (1-p)^{n-x}\) \(\dfrac{e^{-\mu}\mu^x}{x!}\) \(\dfrac{1}{\sqrt{2\pi}\,\sigma} e^{-(x-\mu)^2/(2\sigma^2)}\)
\(E(X)\) \(np\) \(\mu\) \(\mu\)
\(V(X)\) \(np(1-p)\) \(\mu\) \(\sigma^2\)
d, the pmf or pdf dbinom(x, n, p) dpois(x, mu) dnorm(x, mean, sd)
p, the cdf \(P(X \le x)\) pbinom(x, n, p) ppois(x, mu) pnorm(x, mean, sd)
q, the cdf backwards qbinom(area, n, p) qpois(area, mu) qnorm(area, mean, sd)
Our example \(X \sim \text{Binom}(n = 5,\ p = 0.75)\) \(X \sim \text{Pois}(\mu = 5)\) \(X \sim N(\mu = 16,\ \sigma^2 = 2^2)\)
\(P(X = 3)\) dbinom(3, 5, 0.75) dpois(3, 5) \(0\), use an interval
\(P(X \le 3)\) pbinom(3, 5, 0.75) ppois(3, 5) pnorm(3, 16, 2)
\(90\)th percentile qbinom(0.90, 5, 0.75) qpois(0.90, 5) qnorm(0.90, 16, 2)
WarningWhere people lose points
  • dnorm is a height, not a probability. For a continuous variable \(P(X = x) = 0\).
  • dbinom and dpois are probabilities. \(P(X = x)\) is real for a discrete variable.
  • The normal takes \(\sigma\), not \(\sigma^2\). The Poisson takes \(\mu\), not \(\lambda\).
  • Leave mean and sd off and R assumes the standard normal, \(Z \sim N(0, 1)\).
  • p takes an \(x\) and returns an area. q takes an area and returns an \(x\).

Board Problems

Problem 1: Ruck March Times

Times for a 12 mile ruck are normal with a mean of 105 minutes and a standard deviation of 12 minutes.

  1. Fully specify the distribution, then find \(P(X < 90)\).

  2. Find \(P(X > 120)\).

  3. Find \(P(95 \le X \le 115)\).

  4. The fastest 10% earn a pass. What time do they have to beat?

  1. \(X\) is a cadet ruck time in minutes, \(X \sim N(\mu = 105,\ \sigma^2 = 12^2)\).

\[z = \frac{90 - 105}{12} = -1.25, \qquad P(X < 90) = \mathbf{0.1056}\]

The ruck time normal curve with the area left of 90 shaded, next to the standard normal with the area left of negative 1.25 shaded.

pnorm(90, 105, 12) = 0.1056

  1. \(z = (120-105)/12 = 1.25\), and by symmetry this matches part (a): \(\mathbf{0.1056}\).

The ruck time normal curve with the area right of 120 shaded.

1 - pnorm(120, 105, 12) = 0.1056

  1. \[P(95 \le X \le 115) = F(115) - F(95) = \mathbf{0.5953}\]

The ruck time normal curve with the area between 95 and 115 shaded.

pnorm(115, 105, 12) - pnorm(95, 105, 12) = 0.5953

  1. The fastest 10% are the left tail, so this is the 10th percentile: \(\mathbf{89.62}\) minutes.

The ruck time normal curve with the leftmost 10 percent of the area shaded, cut off at 89.62.

qnorm(0.10, 105, 12) = 89.6214

Problem 2: Mortar Round Weight

A production line fills 60mm mortar rounds to a weight that is normal with \(\mu = 4.2\) kg and \(\sigma = 0.05\) kg. A round is in spec if it weighs between 4.1 and 4.3 kg.

  1. Without using R, what fraction of rounds are in spec? Which rule gets you there?

  2. Confirm part (a) in R, then give the fraction rejected.

  3. A round weighs 4.32 kg. How many standard deviations from the mean is that, and how unusual is a round at least that heavy?

  1. The spec limits are exactly \(\mu \pm 2\sigma = 4.2 \pm 0.10\), so the empirical rule gives \(\mathbf{0.9545}\).

The mortar round weight normal curve with the area within two standard deviations of 4.2 shaded, from 4.1 to 4.3.

  1. Same number out of R, and the rejected fraction is the complement, \(1 - 0.9545 = \mathbf{0.0455}\), about 1 round in 22.

pnorm(4.3, 4.2, 0.05) - pnorm(4.1, 4.2, 0.05) = 0.9545

  1. \[z = \frac{4.32 - 4.2}{0.05} = \mathbf{2.4}, \qquad P(X \ge 4.32) = \mathbf{0.0082}\]

Fewer than 1 round in 100 is that heavy, so pull the line and check the fill.

The mortar round weight normal curve with the area right of 4.32 shaded, next to the standard normal with the area right of 2.4 shaded.

1 - pnorm(4.32, 4.2, 0.05) = 0.0082

Problem 3: Two Different Scales

Cadet A shoots 36 on a rifle qualification where scores are \(N(31, 3^2)\). Cadet B scores 285 on an indoor simulator where scores are \(N(268, 14^2)\).

  1. Standardize both scores.

  2. Which cadet performed better relative to the cadets they were shooting against?

  3. What percentile is each cadet in?

\[\begin{aligned} z_A &= \frac{36 - 31}{3} = \mathbf{1.6667} \\ z_B &= \frac{285 - 268}{14} = \mathbf{1.2143} \end{aligned}\]

  1. Cadet A. The raw scores are not comparable, 36 against 285 means nothing, but A is 1.67 standard deviations above their mean while B is 1.21 above theirs.

Two standard normal curves. The left shades the area below 1.6667 for Cadet A, the right shades the area below 1.2143 for Cadet B.

  1. Cadet A is in the 95.2nd percentile, Cadet B in the 88.8th.

pnorm(1.6667) = 0.9522
pnorm(1.2143) = 0.8877

Problem 4: Body Armor Sizing

Chest circumference for incoming cadets is normal with \(\mu = 39\) inches and \(\sigma = 2.2\) inches. The supply sergeant stocks a medium plate carrier that fits from 37 to 41 inches.

  1. What fraction of cadets fit the medium?

  2. Supply wants a small for the bottom 15% and a large for the top 15%. What two chest measurements are the cut points?

  3. Out of a class of 1200, roughly how many need something other than a medium?

  1. \[P(37 \le X \le 41) = F(41) - F(37) = \mathbf{0.6367}\]

The chest circumference normal curve with the area between 37 and 41 inches shaded.

pnorm(41, 39, 2.2) - pnorm(37, 39, 2.2) = 0.6367

  1. The 15th and 85th percentiles: \(\mathbf{36.72}\) and \(\mathbf{41.28}\) inches.

The chest circumference normal curve with the middle 70 percent shaded between 36.72 and 41.28 inches.

qnorm(c(0.15, 0.85), 39, 2.2) = 36.7198, 41.2802

  1. \(1 - 0.6367 = 0.3633\), and \(1200(0.3633) \approx \mathbf{436}\) cadets.

Problem 5: Stretch, Solve for the Parameters

Artillery rounds land at a range that is normal. Nobody tells you \(\mu\) or \(\sigma\), but the firing tables say 5% of rounds fall short of 100 meters and 10% carry past 140 meters.

  1. Write the two standardized equations these facts give you.

  2. Solve for \(\mu\) and \(\sigma\).

  3. Find \(P(110 < X < 130)\).

  1. \(P(X < 100) = 0.05\) puts 100 at \(z = -1.6449\), and \(P(X > 140) = 0.10\) puts 140 at \(z = 1.2816\).

\[\begin{aligned} \frac{100 - \mu}{\sigma} &= -1.6449 \\ \frac{140 - \mu}{\sigma} &= 1.2816 \end{aligned}\]

  1. Subtract the first equation from the second. The \(\mu\) terms cancel.

\[\begin{aligned} \frac{40}{\sigma} &= 1.2816 - (-1.6449) = 2.9265 \\ \sigma &= \frac{40}{2.9265} = \mathbf{13.6686} \\ \mu &= 100 + 1.6449(13.6686) = \mathbf{122.4829} \end{aligned}\]

The solved range normal curve centered at 122.48 with the lower 5 percent below 100 meters and the upper 10 percent above 140 meters shaded.

  1. \[P(110 < X < 130) = \mathbf{0.5283}\]

The solved range normal curve with the area between 110 and 130 meters shaded.

pnorm(130, 122.4829, 13.6686) - pnorm(110, 122.4829, 13.6686) = 0.5283

Before You Leave

Today

  • \(X \sim N(\mu, \sigma^2)\) is fixed by two parameters, and those parameters are the mean and the variance
  • pnorm is the cdf, qnorm runs it backwards, and the third argument is \(\sigma\)
  • Standardizing with \(z = (x-\mu)/\sigma\) relabels the axis without moving any area
  • A \(z\) score puts two different distributions on one scale
  • A critical value \(z_\alpha\) has area \(\alpha\) to its right

Any questions?

Next Lesson

Lesson 13: Exponential Distribution

  • Identify the exponential distribution and relate its parameter \(\lambda\) to the mean and standard deviation (both \(1/\lambda\))
  • Compute exponential probabilities using the pdf and cdf
  • Apply the memoryless property of the exponential distribution

Reading: Devore 4.4

Upcoming Graded Events

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