Revision: statistics and probability

Statistics and probability is certain to appear in the graduation exam. Part I usually has two or three questions that ask for a quick measure from a grouped frequency table or a probability. Part II or Part III usually has one real-life question on conditional probability or Bayes' theorem. Altogether this topic is usually worth about $1$ to $1.5$ marks.

The statistics part of this lesson uses one data set throughout. The daily phone use (minutes) of $40$ students is grouped as follows:

Time (minutes)	$[0, 30)$	$[30, 60)$	$[60, 90)$	$[90, 120)$	$[120, 150)$
Number of students	$4$	$10$	$14$	$8$	$4$
Midpoint	$15$	$45$	$75$	$105$	$135$
Cumulative frequency	$4$	$14$	$28$	$36$	$40$

Measures of central tendency

Theorem Formulas for grouped data

The data has classes $[u_1, u_2)$ , …, $[u_k, u_{k+1})$ , frequencies $n_1$ , …, $n_k$ , midpoints $x_1$ , …, $x_k$ and size $n$ .

Mean: $\bar x=\dfrac{n_1x_1+n_2x_2+\cdots+n_kx_k}{n}$ .
Median: if $[u_m, u_{m+1})$ is the first class with cumulative frequency $\ge\dfrac n2$ and $C$ is the cumulative frequency of the classes before it, then $M_e=u_m+\dfrac{\frac n2-C}{n_m}(u_{m+1}-u_m)$ . Quartiles $Q_1$ , $Q_3$ : replace $\dfrac n2$ by $\dfrac n4$ , $\dfrac{3n}{4}$ .
Mode: if $[u_m, u_{m+1})$ is the class with the highest frequency, then $M_o=u_m+\dfrac{n_m-n_{m-1}}{(n_m-n_{m-1})+(n_m-n_{m+1})}(u_{m+1}-u_m)$ .

For the table above, $\dfrac n2=20$ lies in the class $[60, 90)$ (because $14<20\le28$ ), so $M_e=60+\dfrac{20-14}{14}\cdot30\approx72.86$ . The class with the highest frequency is also $[60, 90)$ , so $M_o=60+\dfrac{4}{4+6}\cdot30=72$ .

Measures of spread

Example True/false: measures of spread

For the phone-use data above:

a) The range of the data is $150$ minutes.

b) The lower quartile is $Q_1=48$ .

c) The upper quartile lies in the class $[60, 90)$ .

d) The standard deviation of the data is approximately $33.51$ minutes.

Solution. We use the cumulative frequencies $4$ , $14$ , $28$ , $36$ , $40$ .

a) True. $R=150-0=150$ .

b) True. $\dfrac n4=10$ and $4<10\le14$ , so $Q_1$ lies in $[30, 60)$ :

$Q_1=30+\frac{10-4}{10}\cdot30=48.$

c) False. $\dfrac{3n}{4}=30$ and $28<30\le36$ , so $Q_3$ lies in the class $[90, 120)$ ; in fact $Q_3=90+\dfrac{30-28}{8}\cdot30=97.5$ .

d) True. We find the mean of the squares and subtract $\bar x^2=73.5^2$ :

$s^2=\frac{4\cdot15^2+10\cdot45^2+14\cdot75^2+8\cdot105^2+4\cdot135^2}{40}-73.5^2=6525-5402.25=1122.75.$

So $s=\sqrt{1122.75}\approx33.51$ minutes.

So the answers are a) True, b) True, c) False, d) True.

Conditional probability and the multiplication rule

Example Multiple choice: conditional probability from a table

Class 12A has $40$ students, of whom $24$ are girls. In the class, $15$ girls and $6$ boys belong to the maths club. A student of the class is chosen at random. Given that the student belongs to the maths club, the probability that the student is a girl is

A. $\dfrac58$ .

B. $\dfrac57$ .

C. $\dfrac38$ .

D. $\dfrac{21}{40}$ .

Solution. Let $B$ be the event "belongs to the club" and $A$ the event "is a girl". The event $B$ contains $15+6=21$ students, $15$ of them girls:

$P(A\mid B)=\frac{P(A\cap B)}{P(B)}=\frac{15/40}{21/40}=\frac{15}{21}=\frac57.$

The answer is B. Option A is $P(B\mid A)=\dfrac{15}{24}$ (the events swapped), option C is $P(A\cap B)$ and option D is $P(B)$ .

The law of total probability and Bayes' theorem

Tree diagram for a screening test

Example True/false: a screening test

A disease affects $2\%$ of a population. A test gives a positive result for $95\%$ of people with the disease and for $10\%$ of people without it. A person is chosen at random and tested. Let $B$ be the event "the person has the disease" and $A$ the event "the result is positive".

a) $P(\overline B)=0.98$ .

b) $P(A)=0.117$ .

c) If the result is positive, the chance that the person has the disease is more than $50\%$ .

d) $P(B\mid A)\approx0.16$ (to 2 decimal places).

Solution. We are given $P(B)=0.02$ , $P(A\mid B)=0.95$ , $P(A\mid\overline B)=0.1$ ; we read the branches of the tree diagram.

a) True. $P(\overline B)=1-0.02=0.98$ .

b) True. By the law of total probability:

$P(A)=0.02\cdot0.95+0.98\cdot0.1=0.019+0.098=0.117.$

c) False. By Bayes' theorem, $P(B\mid A)=\dfrac{0.019}{0.117}\approx0.162<0.5$ .

d) True. Rounding $0.162$ to 2 decimal places gives $0.16$ .

So the answers are a) True, b) True, c) False, d) True.

Key terms

English	Tiếng Việt
grouped data	mẫu số liệu ghép nhóm
class midpoint	giá trị đại diện
median	trung vị
quartile	tứ phân vị
mode	mốt
variance	phương sai
standard deviation	độ lệch chuẩn
conditional probability	xác suất có điều kiện
independent events	biến cố độc lập
law of total probability	công thức xác suất toàn phần
Bayes' theorem	công thức Bayes
tree diagram	sơ đồ cây

Exercises

The exercises go from easy to hard, and each one states its exam format.

Exercise 1 Multiple choice

The waiting times (minutes) of $10$ customers are grouped: the class $[0, 4)$ has $3$ people, $[4, 8)$ has $5$ people and $[8, 12)$ has $2$ people. The mean of the data is

A. $5.6$ . B. $6$ . C. $18.67$ . D. $3.6$ .

What you need and how to start

What you need: $\bar x=\dfrac{\sum n_ix_i}{n}$ , where $x_i$ are the class midpoints.

Steps:

The midpoints are $2$ , $6$ , $10$ .
Find the sum of $n_ix_i$ and divide by $10$ .

Show solution

$\bar x=\dfrac{3\cdot2+5\cdot6+2\cdot10}{10}=\dfrac{56}{10}=5.6$ . Option B is the mean of the three midpoints, option C divides by $3$ instead of $10$ , and option D uses the left endpoints.

Answer. A.

Exercise 2 Multiple choice

The test marks of $50$ students are given in the table. The median of the data (to 2 decimal places) is

Mark	$[5, 6)$	$[6, 7)$	$[7, 8)$	$[8, 9)$	$[9, 10)$
Number of students	$4$	$8$	$16$	$14$	$8$

A. $7.81$ . B. $7.5$ . C. $6.81$ . D. $7.93$ .

What you need and how to start

What you need: the median class is the first class with cumulative frequency $\ge\dfrac n2$ .

Steps:

Cumulative frequencies: $4$ , $12$ , $28$ , $42$ , $50$ ; $\dfrac n2=25$ .
Substitute into the formula with $C=12$ , $n_m=16$ .

Show solution

$M_e=7+\dfrac{25-12}{16}\cdot1=7.8125\approx7.81$ . Option C takes $C=28$ (including the median class), option D uses the frequency $14$ of the next class, and option B is just the midpoint of the class.

Answer. A.

Exercise 3 True/false

The masses (kg) of $20$ students: the class $[45, 50)$ has $5$ students, $[50, 55)$ has $10$ and $[55, 60)$ has $5$ .

a) The mode of the data lies in the class $[55, 60)$ .

b) The lower quartile is $Q_1=50$ .

c) The variance of the data is $12.5$ .

d) The standard deviation of the data is $12.5$ kg.

What you need and how to start

What you need: the mode lies in the class with the highest frequency; $s^2=\dfrac{\sum n_i(x_i-\bar x)^2}{n}$ and $s=\sqrt{s^2}$ .

Steps:

The midpoints are $47.5$ , $52.5$ , $57.5$ and the mean is $\bar x=52.5$ .
Cumulative frequencies $5$ , $15$ , $20$ ; $\dfrac n4=5$ .

Show solution

a) False. The class with the highest frequency is $[50, 55)$ .

b) True. The first class with cumulative frequency $\ge5$ is $[45, 50)$ : $Q_1=45+\dfrac{5-0}{5}\cdot5=50$ .

c) True. $s^2=\dfrac{5\cdot(-5)^2+10\cdot0^2+5\cdot5^2}{20}=\dfrac{250}{20}=12.5$ .

d) False. $s=\sqrt{12.5}\approx3.54$ kg.

Answer. a) False; b) True; c) True; d) False.

Exercise 4 Short answer

Find the interquartile range of the following grouped data (to 1 decimal place).

Class	$[0, 10)$	$[10, 20)$	$[20, 30)$	$[30, 40)$
Frequency	$5$	$12$	$15$	$8$

What you need and how to start

What you need: $\Delta_Q=Q_3-Q_1$ ; the positions of $Q_1$ , $Q_3$ are $\dfrac n4$ , $\dfrac{3n}{4}$ .

Steps:

$n=40$ ; cumulative frequencies $5$ , $17$ , $32$ , $40$ .
Position $10$ is in $[10, 20)$ and position $30$ is in $[20, 30)$ .

Show solution

$Q_1=10+\frac{10-5}{12}\cdot10=\frac{85}{6}\approx14.17,\qquad Q_3=20+\frac{30-17}{15}\cdot10=\frac{86}{3}\approx28.67.$

To avoid rounding errors, we subtract the fractions: $\Delta_Q=\dfrac{86}{3}-\dfrac{85}{6}=\dfrac{87}{6}=14.5$ .

Answer. $14.5$ .

Exercise 5 Multiple choice

$A$ and $B$ are independent events with $P(A)=0.4$ and $P(B)=0.5$ . The probability $P(A\cup B)$ is

A. $0.9$ . B. $0.2$ . C. $0.7$ . D. $0.3$ .

What you need and how to start

What you need: $P(A\cup B)=P(A)+P(B)-P(A\cap B)$ and $P(A\cap B)=P(A)P(B)$ .

Steps:

Find $P(A\cap B)$ .
Substitute into the addition rule.

Show solution

$P(A\cap B)=0.2$ , so $P(A\cup B)=0.4+0.5-0.2=0.7$ . Option A treats $A$ , $B$ as mutually exclusive; option B is $P(A\cap B)$ ; option D is $1-0.7$ .

Answer. C.

Exercise 6 Short answer

A box holds $10$ cards, $4$ of which are red. Two cards are taken one after the other, without replacement. Find the probability that both cards are red (to 2 decimal places).

What you need and how to start

What you need: $P(A\cap B)=P(A)\cdot P(B\mid A)$ .

Steps:

$P(\text{first red})=\dfrac{4}{10}$ .
After one red card is taken, $3$ of the $9$ remaining cards are red.

Show solution

$P=\dfrac{4}{10}\cdot\dfrac39=\dfrac{2}{15}\approx0.133$ .

Answer. $0.13$ .

Exercise 7 Short answer

In a company's inbox, $30\%$ of the emails are spam. A filter marks $90\%$ of spam emails as "spam" and also $5\%$ of normal emails. An email is marked "spam". Find the probability that it really is spam (to 2 decimal places).

What you need and how to start

What you need: the law of total probability and Bayes' theorem.

Steps:

Find $P(\text{marked})=0.3\cdot0.9+0.7\cdot0.05$ .
Divide the branch "spam and marked" by that result.

Show solution

$P(\text{marked})=0.27+0.035=0.305$ , so $P(\text{spam}\mid\text{marked})=\dfrac{0.27}{0.305}\approx0.885$ .

Answer. $0.89$ .

Exercise 8 Written answer

In a city, the probability of rain on a given day is $0.3$ . If it rains, the probability that Minh is late for school is $0.4$ ; if it does not rain, that probability is $0.1$ . Find the probability that Minh is late on a randomly chosen day. Given that Minh is late today, find the probability that it is raining today.

What you need and how to start

What you need: draw a tree diagram with the branches "rain" and "no rain".

Steps:

Use the law of total probability for the event "late".
Use Bayes' theorem for "rain given late".

Show solution

$P(\text{late})=0.3\cdot0.4+0.7\cdot0.1=0.12+0.07=0.19$ . Then $P(\text{rain}\mid\text{late})=\dfrac{0.12}{0.19}=\dfrac{12}{19}\approx0.63$ .

Answer. $0.19$ and about $0.63$ .

Exercise 9 Short answer

Find the standard deviation of the following grouped data.

Class	$[2, 4)$	$[4, 6)$	$[6, 8)$	$[8, 10)$
Frequency	$4$	$8$	$6$	$2$

What you need and how to start

What you need: $s^2=\dfrac{\sum n_ix_i^2}{n}-\bar x^2$ .

Steps:

The midpoints are $3$ , $5$ , $7$ , $9$ ; find $\bar x$ .
Find the mean of the squares, subtract $\bar x^2$ , then take the square root.

Show solution

$\bar x=\dfrac{12+40+42+18}{20}=5.6$ . $\dfrac{4\cdot9+8\cdot25+6\cdot49+2\cdot81}{20}=\dfrac{692}{20}=34.6$ , so $s^2=34.6-31.36=3.24$ and $s=1.8$ .

Answer. $1.8$ .

Exercise 10 Short answer (challenge)

A shop buys components from three suppliers X, Y, Z in the proportions $50\%$ , $30\%$ , $20\%$ . The defect rates of X, Y, Z are $1\%$ , $2\%$ , $3\%$ respectively. A component chosen at random is found to be faulty. Find the probability that it came from Z (to 2 decimal places).

What you need and how to start

What you need: the law of total probability with the three cases X, Y, Z; Bayes' theorem.

Steps:

Find $P(\text{faulty})$ as the sum of three branches.
Divide the branch for Z by $P(\text{faulty})$ .

Show solution

$P(\text{faulty})=0.5\cdot0.01+0.3\cdot0.02+0.2\cdot0.03=0.005+0.006+0.006=0.017$ . Then $P(Z\mid\text{faulty})=\dfrac{0.006}{0.017}\approx0.353$ .

Answer. $0.35$ .

Measures of central tendency

Measures of spread

Conditional probability and the multiplication rule

The law of total probability and Bayes' theorem

Key terms

Exercises

Mini test

Part I. Multiple choice

Part II. True or false

Part III. Short answer

a) The probability that the first marble is red is $\dfrac58$ .	True	False
b) The probability that the second marble is red, given that the first is red, is $\dfrac47$ .	True	False
c) The probability that both marbles are red is $\dfrac{5}{14}$ .	True	False
d) The probability that the second marble is red is $\dfrac47$ .	True	False