The Normal Distribution and Midterm 2 Review
Statistics I · Lecture 12
Today
- Normal distribution: worked examples
- Additivity and its corollaries
- Practice, and a mock test for Midterm 2
Normal Distribution (continued)
Example 3
If \(X\sim N(6,25)\), find \(P(6<X\leq 12)\)
\[ P(6<X\leq 12)=F(12)-F(6)= \] \[ P\left(\frac{6-6}{5}<\frac{X-6}{5}\leq \frac{12-6}{5}\right) \]
Or
\[ P(0<Z\leq 1.2)=\Phi(1.2)-\Phi(0)\\0.8849-0.5=0.3849 \]
Example 3
Example 4
Let \(X\sim N(6,25)\). Find \(P(X\leq -4)\) and \(P(X\geq 16)\)
\[ P(X\leq -4)=P\left(\frac{X-6}{5}\leq \frac{-4-6}{5}\right)=\\ P(Z\leq -2)=\Phi(-2)=1-\Phi(2)=0.0228 \]
Example 4
Example 5
Check that \(P(X\leq -4)=P(X\geq 16)\)
Because of symmetry: \(P(X\leq \mu-k)=P(X\geq \mu+k)\) \(\forall k\in\mathbb{R}\)
\(P(X\leq -4)=P(X\leq 6-10)=\\=P(X\geq 6+10)=P(X\geq 16)\)
TipCorollary
\(\Phi(-k)=P(Z\leq -k)=P(Z\geq k)=1-P(Z\leq k)=1-\Phi(k)\)
Example 6
Consider the distribution \(X\sim N(6,25)\)
Find \(P(0\leq X\leq 8)\)
\(P(0\leq X \leq 8)=P(0<X\leq 8)\) which is equivalent to
\[ P(-1.2<Z\leq 0.4)=\Phi(0.4)-\Phi(-1.2)=\\ \Phi(0.4)-\left[1-\Phi(1.2)\right]=\\ 0.6554-[1-0.8849]=0.5403 \]
Example 7
Consider the r.v. \(X\sim N(6, 25)\). Find \(P(|X-6|>10)\)
\[P(|X-6|>10)=1-P(|X-6|\leq 10)\]
\[1-P(-10\leq X-6\leq 10)=1-P(-2<Z\leq 2)\]
\[1-[\Phi(2)-\Phi(-2)]=1-[\Phi(2)-1+\Phi(2)]\]
\[2-2\Phi(2)=2-2\times 0.9772=0.0456\]
Example 8
Consider the r.v. \(X\sim N(6,25)\), find \(k\) such that \(P(X>k)=0.9\)
\(P(X>k)=0.9\Leftrightarrow 1-P(X\leq k)=0.9\)
Or
\[P(X\leq k)=0.10 \Leftrightarrow P\left(Z\leq \frac{k-6}{5}\right)=0.10\]
\[P\left(Z\leq \frac{k-6}{5}\right)=0.10\Leftrightarrow \Phi\left(\frac{k-6}{5}\right)=0.10\]
Because of symmetry we have \(P(Z\leq -z)=P(Z\geq z)\)
Example 8
Using the table, \(P(Z\geq z)=0.1\) means \(\Phi(z)=0.9\), so \(z=\Phi^{-1}(0.9)\approx 1.282\), and by symmetry \(\Phi^{-1}(0.1)\approx -1.282\). Substituting we get:
\[\Phi(-1.282)=0.1\Leftrightarrow \frac{k-6}{5}=-1.282\] \[k=6-1.282\times 5 = -0.41\]
Example 8
Normal Distribution
ImportantTheorem: Normal additivity
If \(X_1\sim N(\mu_1,\sigma_1^2)\) and \(X_2\sim N(\mu_2,\sigma_2^2)\) are independent, then for any \(a,b\in\mathbb{R}\) we have that \(T=aX_1+bX_2\), where \[T\sim N(\mu_T,\sigma_T^2)\]
To find \(\mu_T\) and \(\sigma_T^2\) remember the properties of the mean and variance.
Example 9
Let the r.v.s \(X\sim N(6,4)\) and \(Y\sim N(6,4)\), with \(T=0.5 X-Y\) Find \(\mu_T\) and \(\sigma_T^2\)
\[\mu_T=E[0.5X-Y]=0.5E[X]-E[Y]=\\ 0.5 \times 6 - 6 = -3\]
\[\sigma_T^2=V[0.5 X- Y]=V[0.5 X]+V[-Y]=\\ 0.5^2 V[X]+V[Y]= 0.25 \times 4 + 4 = 5\]
Example 9
Find \(P(T>0)\)
\(T\sim N(-3, 5)\)
\[P(T>0)=P\left(Z>\frac{0+3}{\sqrt{5}}\right)=1-P(Z\leq 1.34)=\\ 1-\Phi(1.34)=0.0901\]
Example 9
Normal Distribution
ImportantCorollaries
- \(X_i\sim N(\mu,\sigma^2)\) for \(i=1,\dots,n\), i.i.d.
- \(T=X_1+\dots+X_n\) with \(\bar{X}=\frac{T}{n}\)
Then:
- \(T\sim N(n\times \mu, n\sigma^2)\)
- \(\bar{X}\sim N\left(\mu, \frac{\sigma^2}{n}\right)\)
Example 10
Let \(X_i\sim N(120, 64)\) be r.v.s representing the number of bank deposits made in a specific day. Then \(T=X_1+\dots+X_5\) are the weekly deposits.
Find the probability of the weekly deposits exceed 620.
\(T\sim N(600, 320)\) because of the Normal additivity property.
Example 10
\[P(T>620)=P\left(Z>\frac{620-5\times 120}{\sqrt{5}\times 8}\right)=\\1-P(Z\leq 1.12)=1-\Phi(1.12)=0.1314\]
❓ Normal · Question 1
If \(X\sim N(\mu,\sigma^2)\), then \(Z=\frac{X-\mu}{\sigma}\) follows:
A. \(N(\mu,\sigma^2)\)
B. an Exponential
C. a Uniform
D. \(N(0,1)\)
✅ D. Standardizing centres the variable at 0 and rescales it to unit variance. That is why one table is enough for every Normal.
❓ Normal · Question 2
For the standard Normal, \(\Phi(-z)\) equals:
A. \(1-\Phi(z)\)
B. \(\Phi(z)\)
C. \(-\Phi(z)\)
D. \(2\Phi(z)\)
✅ A. The standard Normal is symmetric about 0, so the area to the left of \(-z\) equals the area to the right of \(z\).
✏️ Normal · Question 3
Let \(X\sim N(6,25)\).
a) Compute \(P(X\leq 14)\) and \(P(X>2)\).
b) Find \(k\) such that \(P(X\leq k)=0.90\).
c) Let \(Y\sim N(4,9)\) be independent of \(X\). Compute \(P(X+Y>15)\).
✅ Normal · Solution
Here \(\mu=6\) and \(\sigma=\sqrt{25}=5\).
a) \(P(X\leq 14)=\Phi\left(\frac{14-6}{5}\right)=\Phi(1.6)=0.9452\)
\(P(X>2)=1-\Phi\left(\frac{2-6}{5}\right)=1-\Phi(-0.8)=\Phi(0.8)=0.7881\)
b) \(\Phi(z)=0.90\) gives \(z\approx 1.28\) from the table (\(1.282\) interpolated), so \(k=6+1.282\times 5\approx 12.41\)
c) By additivity, \(X+Y\sim N(6+4,\ 25+9)=N(10,34)\), so \[P(X+Y>15)=1-\Phi\left(\frac{15-10}{\sqrt{34}}\right)=1-\Phi(0.86)=1-0.8051=0.1949\]
Practice
Practice
The exercises go from easy to hard:
- 🟢 warm-up: one formula, one step
- 🟡 standard: two or three steps
- 🟠 exam level: this is what you will find in the midterm
- 🔴 stretch: you need to combine several things
After these, a mock test with Topic 3 problems, like in the midterm.
🟢 Practice · Exercise 1
Let \(X\sim N(12,4)\), and consider \(P(a\leq X\leq 15)=0.7332\). Find \(a\).
\(P(X\leq 15)=\Phi\left(\frac{15-12}{2}\right)=\Phi(1.5)=0.9332\)
So \(\Phi\left(\frac{a-12}{2}\right)=0.9332-0.7332=0.2\), and from the table \(\frac{a-12}{2}\approx -0.84\)
\[a\approx 12-2\times 0.84=10.32\]
🟡 Practice · Exercise 2
The weight of a package 📦, in grams, is \(X\sim N(250,100)\).
(a) Find the probability that a package weighs between 230.4 and 269.6 grams.
(b) Packages farther away than 15.3 grams from the mean are rejected. Find the probability that a package is rejected.
(c) Find \(k\) such that \(P(X<k)=0.9\), and interpret it.
(d) Pick 4 packages. Find the probability that the total weight is above its mean.
✅ Practice · Exercise 2 · Solution
\(\sigma=10\). (a) \(P\left(|Z|<\frac{19.6}{10}\right)=P(|Z|<1.96)=0.95\)
(b) \(P(|Z|>1.53)=2(1-\Phi(1.53))=2(1-0.9370)=0.126\)
(c) \(\Phi(z)=0.9\Rightarrow z\approx 1.28\), so \(k=250+1.28\times 10=262.8\) grams. 90% of the packages weigh less than 262.8 grams.
(d) The total is Normal, and a Normal has half of its probability above its mean. So \(0.5\), no computation needed 😎
🟠 Practice · Exercise 3
Let \(X\sim N(1,1)\) and \(Z\sim N(0,1)\) be independent.
(a) Find \(P(X<Z+1)\).
(b) Find \(a\) such that \(P(X>Z+a)=0.1\).
Put everything on one side: \(D=X-Z\sim N(1-0,\ 1+1)=N(1,2)\). The variances add, even if we subtract.
(a) \(P(D<1)=0.5\), since 1 is the mean of \(D\).
(b) \(P(D>a)=0.1\Rightarrow\frac{a-1}{\sqrt{2}}\approx 1.282\Rightarrow a\approx 1+1.282\times 1.414\approx 2.813\)
🔴 Practice · Exercise 4
True or false?
(a) If \(X\sim N(0,1)\) and \(Y\sim N(1,2)\) are independent, then \(2X+Y\sim N(1,6)\).
(b) \(\Phi(-x)=\Phi(x)\) for all \(x\in\mathbb{R}\).
(c) \(\phi(-x)=1-\phi(x)\) for all \(x\in\mathbb{R}\).
(a) ✅ \(E=2(0)+1=1\) and \(V=4(1)+2=6\)
(b) ❌ \(\Phi(-x)=1-\Phi(x)\). The areas in the tails are symmetric, not the cumulative probabilities.
(c) ❌ The density is symmetric, \(\phi(-x)=\phi(x)\). You mixed up (b) and (c) 😉
Mock Test
Mock Test
Three problems, with the style and the length of the midterm. Try them in 40 minutes, with the tables and a calculator, and no slides.
Mock Test · Problem 1
A call center ☎️ receives, on average, 3 calls every 10 minutes, following a Poisson process.
(a) Find the probability of no calls in 10 minutes.
(b) Find the probability of at least 2 calls in 5 minutes.
(c) Find the expected waiting time until the next call, and the probability of waiting more than 5 minutes.
✅ Mock Test · Problem 1 · Solution
(a) \(X\sim Poi(3)\): \(P(X=0)=F(0)=0.0498\)
(b) In 5 minutes, \(Y\sim Poi(1.5)\): \(P(Y\geq 2)=1-F(1)=1-0.5578=0.4422\)
(c) The time between calls, in minutes, is \(T\sim Exp(0.3)\), so \(E[T]=\frac{10}{3}\approx 3.33\) minutes.
\(P(T>5)=e^{-1.5}\approx 0.2231\), which is exactly \(P(Y=0)\) from (b): waiting more than 5 minutes is the same as no calls in 5 minutes.
Mock Test · Problem 2
The weight of a bag of coffee ☕, in grams, is \(X\sim N(500,16)\). Bags weighing less than 494.88 grams are rejected.
(a) Find the probability that a bag is rejected.
(b) A box has 10 bags, independent. Find the probability that at most one bag in the box is rejected.
(c) Find the probability that a box of 10 bags weighs less than 4980 grams.
(d) Find \(k\) such that 95% of the bags weigh more than \(k\).
✅ Mock Test · Problem 2 · Solution
(a) \(P(X<494.88)=\Phi\left(\frac{494.88-500}{4}\right)=\Phi(-1.28)=1-0.8997\approx 0.10\)
(b) \(N\sim Bin(10,0.1)\) counts the rejected bags: \(P(N\leq 1)=0.7361\)
(c) \(T=X_1+\dots+X_{10}\sim N(5000,160)\): \(P(T<4980)=\Phi\left(\frac{-20}{\sqrt{160}}\right)=\Phi(-1.58)=1-0.9429=0.0571\)
(d) \(P(X>k)=0.95\Rightarrow\frac{k-500}{4}=-1.645\Rightarrow k\approx 493.42\) grams.
Mock Test · Problem 3
In a factory 🏭, 4% of the parts are defective. Parts are inspected one by one, independently.
(a) How many parts are expected to be inspected until the first defective one?
(b) Find the probability that the first defective part appears after the 10th inspection.
(c) A batch of 100 parts contains exactly 4 defective ones. You take 5 without replacement. Find the probability that none is defective, exactly and with the Binomial approximation.
✅ Mock Test · Problem 3 · Solution
(a) \(X\sim Geo(0.04)\), \(E[X]=\frac{1}{0.04}=25\) parts.
(b) \(P(X>10)=0.96^{10}\approx 0.6648\)
(c) \(Y\sim Hypergeometric(100,4,5)\):
\[P(Y=0)=\frac{\binom{4}{0}\binom{96}{5}}{\binom{100}{5}}\approx 0.8119\]
\(\frac{n}{N}=0.05<0.1\), so \(Y\approx Bin(5,0.04)\) and \(P(Y=0)\approx 0.96^5\approx 0.8154\). Very close.
📝 Homework
Problem set 3.2, Questions 8 to 13.
Good luck in the midterm 🍀
References
- Murteira, B.; Silva Ribeiro, C.; Andrade e Silva, J. & Pimenta, C., Introdução à Estatística, Escolar Editora, McGraw-Hill, 2010
- Paulino, C. D. & Branco, J. A. (2005). Exercícios de Probabilidade e Estatística. Escolar Editora
- Pimenta, F., Andrade e Silva, J.; Silva Ribeiro, C. & Murteira, B., Introdução à Estatística, 3ª Edição, Escolar Editora, 2015
- Ferreira, T., Custódio, S.G., Modelos Probabilísticos, Síntese Teórica e Exercícios Resolvidos, Edições Sílabo (1ª Edição), 2023
Appendix
Standard Normal Table
| z | .00 | .01 | .02 | .03 | .04 | .05 | .06 | .07 | .08 | .09 |
|---|---|---|---|---|---|---|---|---|---|---|
| 0.0 | 0.5000 | 0.5040 | 0.5080 | 0.5120 | 0.5160 | 0.5199 | 0.5239 | 0.5279 | 0.5319 | 0.5359 |
| 0.1 | 0.5398 | 0.5438 | 0.5478 | 0.5517 | 0.5557 | 0.5596 | 0.5636 | 0.5675 | 0.5714 | 0.5753 |
| 0.2 | 0.5793 | 0.5832 | 0.5871 | 0.5910 | 0.5948 | 0.5987 | 0.6026 | 0.6064 | 0.6103 | 0.6141 |
| 0.3 | 0.6179 | 0.6217 | 0.6255 | 0.6293 | 0.6331 | 0.6368 | 0.6406 | 0.6443 | 0.6480 | 0.6517 |
| 0.4 | 0.6554 | 0.6591 | 0.6628 | 0.6664 | 0.6700 | 0.6736 | 0.6772 | 0.6808 | 0.6844 | 0.6879 |
| 0.5 | 0.6915 | 0.6950 | 0.6985 | 0.7019 | 0.7054 | 0.7088 | 0.7123 | 0.7157 | 0.7190 | 0.7224 |
| 0.6 | 0.7257 | 0.7291 | 0.7324 | 0.7357 | 0.7389 | 0.7422 | 0.7454 | 0.7486 | 0.7517 | 0.7549 |
| 0.7 | 0.7580 | 0.7611 | 0.7642 | 0.7673 | 0.7704 | 0.7734 | 0.7764 | 0.7794 | 0.7823 | 0.7852 |
| 0.8 | 0.7881 | 0.7910 | 0.7939 | 0.7967 | 0.7995 | 0.8023 | 0.8051 | 0.8078 | 0.8106 | 0.8133 |
| 0.9 | 0.8159 | 0.8186 | 0.8212 | 0.8238 | 0.8264 | 0.8289 | 0.8315 | 0.8340 | 0.8365 | 0.8389 |
| 1.0 | 0.8413 | 0.8438 | 0.8461 | 0.8485 | 0.8508 | 0.8531 | 0.8554 | 0.8577 | 0.8599 | 0.8621 |
| 1.1 | 0.8643 | 0.8665 | 0.8686 | 0.8708 | 0.8729 | 0.8749 | 0.8770 | 0.8790 | 0.8810 | 0.8830 |
| 1.2 | 0.8849 | 0.8869 | 0.8888 | 0.8907 | 0.8925 | 0.8944 | 0.8962 | 0.8980 | 0.8997 | 0.9015 |
| 1.3 | 0.9032 | 0.9049 | 0.9066 | 0.9082 | 0.9099 | 0.9115 | 0.9131 | 0.9147 | 0.9162 | 0.9177 |
| 1.4 | 0.9192 | 0.9207 | 0.9222 | 0.9236 | 0.9251 | 0.9265 | 0.9279 | 0.9292 | 0.9306 | 0.9319 |
| 1.5 | 0.9332 | 0.9345 | 0.9357 | 0.9370 | 0.9382 | 0.9394 | 0.9406 | 0.9418 | 0.9429 | 0.9441 |
| 1.6 | 0.9452 | 0.9463 | 0.9474 | 0.9484 | 0.9495 | 0.9505 | 0.9515 | 0.9525 | 0.9535 | 0.9545 |
| 1.7 | 0.9554 | 0.9564 | 0.9573 | 0.9582 | 0.9591 | 0.9599 | 0.9608 | 0.9616 | 0.9625 | 0.9633 |
| 1.8 | 0.9641 | 0.9649 | 0.9656 | 0.9664 | 0.9671 | 0.9678 | 0.9686 | 0.9693 | 0.9699 | 0.9706 |
| 1.9 | 0.9713 | 0.9719 | 0.9726 | 0.9732 | 0.9738 | 0.9744 | 0.9750 | 0.9756 | 0.9761 | 0.9767 |
| 2.0 | 0.9772 | 0.9778 | 0.9783 | 0.9788 | 0.9793 | 0.9798 | 0.9803 | 0.9808 | 0.9812 | 0.9817 |
| 2.1 | 0.9821 | 0.9826 | 0.9830 | 0.9834 | 0.9838 | 0.9842 | 0.9846 | 0.9850 | 0.9854 | 0.9857 |
| 2.2 | 0.9861 | 0.9864 | 0.9868 | 0.9871 | 0.9875 | 0.9878 | 0.9881 | 0.9884 | 0.9887 | 0.9890 |
| 2.3 | 0.9893 | 0.9896 | 0.9898 | 0.9901 | 0.9904 | 0.9906 | 0.9909 | 0.9911 | 0.9913 | 0.9916 |
| 2.4 | 0.9918 | 0.9920 | 0.9922 | 0.9925 | 0.9927 | 0.9929 | 0.9931 | 0.9932 | 0.9934 | 0.9936 |
| 2.5 | 0.9938 | 0.9940 | 0.9941 | 0.9943 | 0.9945 | 0.9946 | 0.9948 | 0.9949 | 0.9951 | 0.9952 |
| 2.6 | 0.9953 | 0.9955 | 0.9956 | 0.9957 | 0.9959 | 0.9960 | 0.9961 | 0.9962 | 0.9963 | 0.9964 |
| 2.7 | 0.9965 | 0.9966 | 0.9967 | 0.9968 | 0.9969 | 0.9970 | 0.9971 | 0.9972 | 0.9973 | 0.9974 |
| 2.8 | 0.9974 | 0.9975 | 0.9976 | 0.9977 | 0.9977 | 0.9978 | 0.9979 | 0.9979 | 0.9980 | 0.9981 |
| 2.9 | 0.9981 | 0.9982 | 0.9982 | 0.9983 | 0.9984 | 0.9984 | 0.9985 | 0.9985 | 0.9986 | 0.9986 |
| 3.0 | 0.9987 | 0.9987 | 0.9987 | 0.9988 | 0.9988 | 0.9989 | 0.9989 | 0.9989 | 0.9990 | 0.9990 |