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LemonmathSelf-study sheet
Grade 12 · Lesson 7.8

Revision: statistics and probability

Name: Date: Time spent: min

Goals

Exercises

Exercise 1 Multiple choice

The waiting times (minutes) of 1010 customers are grouped: the class [0,4)[0, 4) has 33 people, [4,8)[4, 8) has 55 people and [8,12)[8, 12) has 22 people. The mean of the data is

A. 5.65.6. B. 66. C. 18.6718.67. D. 3.63.6.

Exercise 2 Multiple choice

The test marks of 5050 students are given in the table. The median of the data (to 2 decimal places) is

Mark [5,6)[5, 6) [6,7)[6, 7) [7,8)[7, 8) [8,9)[8, 9) [9,10)[9, 10)
Number of students 44 88 1616 1414 88

A. 7.817.81. B. 7.57.5. C. 6.816.81. D. 7.937.93.

Exercise 3 True/false

The masses (kg) of 2020 students: the class [45,50)[45, 50) has 55 students, [50,55)[50, 55) has 1010 and [55,60)[55, 60) has 55.

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

b) The lower quartile is Q1=50Q_1=50.

c) The variance of the data is 12.512.5.

d) The standard deviation of the data is 12.512.5 kg.

Exercise 4 Short answer

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

Class [0,10)[0, 10) [10,20)[10, 20) [20,30)[20, 30) [30,40)[30, 40)
Frequency 55 1212 1515 88

Exercise 5 Multiple choice

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

A. 0.90.9. B. 0.20.2. C. 0.70.7. D. 0.30.3.

Exercise 6 Short answer

A box holds 1010 cards, 44 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).

Exercise 7 Short answer

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

Exercise 8 Written answer

In a city, the probability of rain on a given day is 0.30.3. If it rains, the probability that Minh is late for school is 0.40.4; if it does not rain, that probability is 0.10.1. 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.

Exercise 9 Short answer

Find the standard deviation of the following grouped data.

Class [2,4)[2, 4) [4,6)[4, 6) [6,8)[6, 8) [8,10)[8, 10)
Frequency 44 88 66 22

Exercise 10 Short answer (challenge)

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

Mini test

15 minutes, without looking back at the lesson. Part I: circle one letter; Part II: write T or F for each statement; Part III: write the answer in the box.

Part I. Multiple choice

Question 1. A set of grouped data has the class [10,20)[10, 20) with frequency 22, [20,30)[20, 30) with frequency 55 and [30,40)[30, 40) with frequency 33. The mean of the data is
A. 2525.B. 2626.C. 86.6786.67.D. 2121.
Question 2. For the data in the previous question, the class containing the upper quartile Q3Q_3 is
A. [10,20)[10, 20).B. [20,30)[20, 30).C. [30,40)[30, 40).D. It cannot be determined.
Question 3. Let AA, BB be events with P(B)>0P(B)>0. The probability of AA given that BB has happened is
A. P(A∣B)=P(A∩B)P(A)P(A\mid B)=\dfrac{P(A\cap B)}{P(A)}.B. P(A∣B)=P(A)⋅P(B)P(A\mid B)=P(A)\cdot P(B).C. P(A∣B)=P(A∩B)P(B)P(A\mid B)=\dfrac{P(A\cap B)}{P(B)}.D. P(A∣B)=P(A)P(B)P(A\mid B)=\dfrac{P(A)}{P(B)}.
Question 4. Given P(A)=0.3P(A)=0.3 and P(B∣A)=0.5P(B\mid A)=0.5, the value of P(A∩B)P(A\cap B) is
A. 0.150.15.B. 0.80.8.C. 0.60.6.D. 0.20.2.

Part II. True or false

Question 1. A bag holds 55 red and 33 white marbles. Two marbles are taken one after the other, without replacement.
a) The probability that the first marble is red is 58\dfrac58.T / F
b) The probability that the second marble is red, given that the first is red, is 47\dfrac47.T / F
c) The probability that both marbles are red is 514\dfrac{5}{14}.T / F
d) The probability that the second marble is red is 47\dfrac47. T / F

Part III. Short answer

Question 1. A set of grouped data has the class [0,2)[0, 2) with frequency 33, [2,4)[2, 4) with frequency 44 and [4,6)[4, 6) with frequency 33. Find its standard deviation (to 2 decimal places).
Question 2. Given P(B)=0.4P(B)=0.4, P(A∣B)=0.5P(A\mid B)=0.5 and P(A∣B‾)=0.25P(A\mid\overline B)=0.25, find P(B∣A)P(B\mid A) (to 2 decimal places).

Score: ........ / 10

Hints (only when stuck)

Exercise 1.

What you need: xˉ=∑nixin\bar x=\dfrac{\sum n_ix_i}{n}, where xix_i are the class midpoints.

Steps:

  1. The midpoints are 22, 66, 1010.
  2. Find the sum of nixin_ix_i and divide by 1010.
Exercise 2.

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

Steps:

  1. Cumulative frequencies: 44, 1212, 2828, 4242, 5050; n2=25\dfrac n2=25.
  2. Substitute into the formula with C=12C=12, nm=16n_m=16.
Exercise 3.

What you need: the mode lies in the class with the highest frequency; s2=∑ni(xi−xˉ)2ns^2=\dfrac{\sum n_i(x_i-\bar x)^2}{n} and s=s2s=\sqrt{s^2}.

Steps:

  1. The midpoints are 47.547.5, 52.552.5, 57.557.5 and the mean is xˉ=52.5\bar x=52.5.
  2. Cumulative frequencies 55, 1515, 2020; n4=5\dfrac n4=5.
Exercise 4.

What you need: ΔQ=Q3−Q1\Delta_Q=Q_3-Q_1; the positions of Q1Q_1, Q3Q_3 are n4\dfrac n4, 3n4\dfrac{3n}{4}.

Steps:

  1. n=40n=40; cumulative frequencies 55, 1717, 3232, 4040.
  2. Position 1010 is in [10,20)[10, 20) and position 3030 is in [20,30)[20, 30).
Exercise 5.

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

Steps:

  1. Find P(A∩B)P(A\cap B).
  2. Substitute into the addition rule.
Exercise 6.

What you need: P(A∩B)=P(A)⋅P(B∣A)P(A\cap B)=P(A)\cdot P(B\mid A).

Steps:

  1. P(first red)=410P(\text{first red})=\dfrac{4}{10}.
  2. After one red card is taken, 33 of the 99 remaining cards are red.
Exercise 7.

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

Steps:

  1. Find P(marked)=0.3⋅0.9+0.7⋅0.05P(\text{marked})=0.3\cdot0.9+0.7\cdot0.05.
  2. Divide the branch "spam and marked" by that result.
Exercise 8.

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

Steps:

  1. Use the law of total probability for the event "late".
  2. Use Bayes' theorem for "rain given late".
Exercise 9.

What you need: s2=∑nixi2n−xˉ2s^2=\dfrac{\sum n_ix_i^2}{n}-\bar x^2.

Steps:

  1. The midpoints are 33, 55, 77, 99; find xˉ\bar x.
  2. Find the mean of the squares, subtract xˉ2\bar x^2, then take the square root.
Exercise 10.

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

Steps:

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

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