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CHANCE & DATA

There are five different types of graph you need to be able to read:1. Scatter graph

Look for trend lines and outliers2. Times series

Look for a trend over time3. Box and whisker plot

Look for median and spread4. Dot Plot

Look for median, spread & outliers5. Histogram/Bar graph

Look to read values

This is when more than one event occurs• Addition Law Probability of E OR F = +P(E) P(F)• Multiplication Law Probability of E AND F = ×P(E) P(F)• Probability Trees

▶ A tool for working out probabilities ▶ Multiply along branches ▶ Probabilities of branches from the

same point add to 1

MULTI LEVELPROBABILITY p 139

GRAPHING INTRODUCTION p 126 Each time you should follow the same

pattern in your explanation:1. Problem - Define or list the problems/

issues in what you have read2. Explain yourself, using your

understanding of statistics (this could be mean, median, mode etc.). Give definitions as well as pros and cons

3. Improve - Discuss how the situation could be improved, which assumptions to change, or under which conditions their assumptions would be correct

1. Probability is a measure of the likelihood or chance of an event occurring.

2. Events have a probability between 0 and 1 (0 is impossible, 1 is certain)

3. Probability of an event E is written P(E)4. There are two ways of measuring

probability: ▶ Theoretical Calculations ▶ Experimentally, using the proportion

of times the event occurs. The experi-mental probability of an event E is:

PROBABILITY p 136

GRAPHING ANALYSIS p 129

CHANCE & DATA

1. Problem - Make a statement about whether the PROBABILITY is correct

2. Explain how you worked out the correct answer by showing all working and giving reasons

3. Improve - Give any improvements and list possible circumstances where the situation could have been correct

PROBABILITY ANALYSIS p 144

THE SKILLS YOU NEED TO KNOW:

DEMONSTRATE UNDERSTANDING OF CHANCE AND DATA4 CREDITS (91037)

P(E) Number of times E occursTotal number of events

=

Note: Problems may rely on knowledge from earlier chance and data sections.

There are five different types of graph you need to be able to read:1. Scatter graph (look for trend line

and outliers)2. Times series (look for a trend over

time)

3. Box and whisker plot (look for median and spread)

4. Dot Plot (look out for median, spread and outliers)

5. Histogram/Bar graph (look to read values)

For a complete tutorial on this topic visit www.learncoach.co.nz

GRAPHING INTRODUCTION

NCEA QUESTIONS1. Tuahu’s grandfather told him that a person’s arm-

span is often the same as their height (your arm-span is the distance from the fingertips of your left hand to the fingertips of your right hand, when your arms are stretched out). Tuahu wondered if this was true. He collected measurements from 100 randomly selected year 10 boys and girls. He drew a scattergraph of the results. He added a line of best fit to the graph. The results are shown on the graph below and some statistics are listed in the table to the right.a. What is the height of the tallest person on the

graph? b. What is the height of the person with the

smallest arm-span? c. How many people have an arm-span between

120 and 135 cm?

Statistics Height Arm-spanmean 166 162minimum 105 60lower quartile 160 157median 165 165upper quartile 173 171maximum 201 208range 96 148inter-quartile range 13 14

SUMMARY

CHANCE & DATA

2. The graphs below show the monthly average minimum and maximum temperatures for both Rome and Nairobi for three years from 2007-9.

a. Which city has the smallest yearly fluctuations and what is the difference (for the maximum temperature).

b. For each year in Rome state the hottest month.c. Which city has the largest difference between

its highest maximum and lowest minimum and what is it.

3. A group of year 9 students and a group of year 13 students were shown a page with 72 dots on it for three seconds. They were then asked to guess how many dots they saw. The results are shown below:

a. For year 9 there were 3 outliers, give their approximate values.

b. In this situation, which measure of average best represents the data and give a reason.

c. Give the interquartile range for both year groups.

4. Pete wanted to set up a tomato growing business. He had 2 possible locations with different soil types and slightly different climates. He decided to do a 1 year trial to see which site produced the higher yield. He set up 24 rows of plants at each site and over the year measured the weight of the tomatoes he harvested. Give the approximate range of weights for each site.

5. A business owner wants to compare the Saturday traffic with the Sunday traffic to her website.

a. Give the min and max values for Saturday and for Sunday.

b. Use the trend lines to determine which day has the greatest increase in traffic?

c. Give details about any outliers that can be seen in the graph above (week and values).

6. A survey of attendance of students at a movie theatre is carried out on Mondays and Tuesdays over several months.

a. Which day of the week did the movie theatre have their maximum number of students attend over the survey period?

b. Which day of the week had the fewest students ever attend over the survey period?

c. What is the approximate median number of patrons on Day 1

d. Tuesday is the night with the greatest variability in the number of patrons. Is Tuesday Day 1 or Day 2 (give reasons)?

PRACTICE QUESTIONS

Year 13 Year 9Mean 51.8 57.8

Median 51 50

Mode 45 48

Range 55 131

Minimum 28 28

Lower quartile 45 48

Upper quartile 60.5 55

Maximum 83 159

Number Surveyed 20 27

ANSWERSNCEA1. a. Height of the tallest person on the graph

= 201 cm (Achieved)b. Height of the person with the smallest arm

span = 160 cm (Achieved)c. Number of people with arm-span between 120

and 135 cm = 2 (Achieved)

PRACTICE2.

a. Nairobi: 26 - 21 = 5° (Achieved)b. 07: August

08: July 09: July/August (same) (Achieved)

c. Rome = 27- 4 = 23° (±1) (Achieved)

3. a. 100, 110 and 160. (Achieved)b. Median - the mode never is and in this situation

the outliers pull up the mean. (Achieved)c. Year 9: 55 - 48 = 7

Year 13: 60.5 - 45 = 15.5 (Achieved)

4. Site A: 6 - 4 = 2 kg Site B: 9 - 2 = 6 -7 kg (Achieved)

5. a. Saturday: Min 0, Max 7000. Sunday : Min 2250, Max 9300. (All within 500) (Achieved)

b. Sunday (Achieved)c. Week 6: Saturday had 0 visitors

Week 9: Saturday had 7000 and Sunday 9300 visitors. (Achieved)

6. a. Day 2 (Achieved)b. Day 2 (Achieved)c. 325 (Achieved)d. Day 2 - This is because it has the largest

inter-quartile range and the longest whiskers. (Achieved)

Study Tip:

Study GroupsCan help with exchange of ideas and with motivation.More isn’t always merrier.The larger the study group, the more likely you are to get dis-tracted.Small study groups (up to 4) work better.

CHANCE & DATA

GRAPHING ANALYSISSUMMARY

Each time you should follow the same pattern in your explanation:1. Problem - Define or list the problems/issues in what you have read (because there will be

problems!)2. Explain yourself, using your understanding of statistics (this could be mean, median, mode,

probability calculations, range, upper/lower quartile, inter-quartile range etc.). Give definitions as well as pros and cons.

3. Improve - Discuss how you could improve the situation, which assumptions need to change, or under which conditions their assumptions would be correct.For a complete tutorial on this topic visit www.learncoach.co.nz

1. Richard wants to move overseas to a warmer city. He would like to move to either Rome or Nairobi. The graphs below show the monthly average, minimum and maximum temperatures in each city for three years from 2007 to 2009. Richard decides to move to Rome because he thinks:

• The temperature in Rome is higher than the temperature in Nairobi, so Rome is warmer.

• The maximum temperature in Rome peaks every year, which is more pleasant to live in.

• The temperature in Rome is less variable over a year, so this is more pleasant.

• The temperature appears to be rising in Rome, so it will get warmer in future.

• There is less difference between the maximum and minimum temperature in Rome, so it will be more comfortable.

Use the graphs given below to answer each of questions (a) to (f). You do not need to explain why the climate features happen.

a. Is Richard correct to believe that the temperature in Rome is higher than the temperature in Nairobi, so Rome is warmer? Justify your answer using the graphs.

b. Comment on how the maximum and minimum temperatures in the two cities vary over a year.

c. Richard has been told that both the maximum and minimum temperatures vary less in Rome than they do in Nairobi. Do the graphs support this? Justify your answer.

d. Richard thinks that “the temperature appears to be rising in Rome, so it will get warmer in the future”. Do you agree? Justify your answer, using the graphs.

e. Is Richard correct to say that “the difference between the maximum and minimum temperature is less in Rome”? Justify your answer using the graphs.

f. Richard wants to live somewhere warm. Should he choose Rome or Nairobi? Justify your answer by referring to the graphs. Discuss any limitations in the data, or any other research you would need to do before you could make a valid decision.

OLD NCEA QUESTIONS

2. In an investigation Melanie found that year nine students can estimate the number of dots on a page more accurately than year 13 students.Melanie selected ONE year 9 class and ONE year 13 class from her school to take part in her experiment. She held up a page with 72 dots drawn on it and allowed the students 3 seconds to look at the page. Each student wrote down how many dots they thought were on the page.Melanie then collected the guesses and analysed the data. “Using the results from my experiment I can conclude that year 9 students are better at estimating the number of dots because the mean guess for year 9 students is closer to the actual number of dots”, Melanie says.The guesses the students made are illustrated by the graph below.

Evaluate the report about Melanie’s investigation. Make at least FOUR evaluative statements about what Melanie did and what she concluded from her investigation.In your answer, you could consider:

• The data accuracy, including the data collection method

• The data display and analysis• The interpretation of the data• If the claim is sensible.

Additional Resources:

Year 13 Year 9Mean 51.8 57.8Median 51 50Mode 45 48

Range 55 131Minimum 28 28Lower quartile 45 48Upper quartile 60.5 55Maximum 83 159Number Surveyed 20 27

3. Tuahu’s grandfather told him that a person’s arm-span is often the same as their height (your arm-span is the distance from the fingertips of your left hand to the fingertips of your right hand, when your arms are stretched out). Tuahu wondered if this was true. He collected measurements from 100 randomly selected year 10 boys and girls. He drew a scattergraph of the results. He added a line of best fit to the graph. The results are shown on the graph below and some statistics are listed in the table below.

Statistics Height Arm-spanmean 166 162minimum 105 60lower quartile 160 157median 165 165upper quartile 173 171maximum 201 208range 96 148inter-quartile range 13 14

a. Why was a scattergraph appropriate to show the data Tuahu had collected?

b. There are some points on the graph that seem to be unlikely measurements for a year 10 student. Give the height and arm-span for THREE points that seem unlikely. Explain why you think they are unlikely measurements for a year 10 student.

c. Tuahu concludes from his graph that the statement made by his grandfather is correct: on average, a person’s arm-span is the same as their height. Is Tuahu’s conclusion valid? You should give at least TWO reasons for your answer

CHANCE & DATA

4. A local cinema wants to increase the number of customers by offering a ‘Students Half Price’ night. This cinema will offer the deal on the night where the least number of people attend. Therefore, the cinema needs to decide whether to offer the deal on day 1 or day 2. So it carries out a survey of attendance of students on these two days over several months. The results are shown below:

From the box and whisker plot the cinema decided to have the ‘Students Half Price’ night on Day 1, because day 1 had a lower maximum and a lower upper-quartile. Evaluate their choice, commenting on:

• The accuracy of their conclusions• The appropriateness of the data display• If their claim is sensible

5. Pete wanted to set up a tomato growing business. He had 2 possible locations with different soil types and different climates. He decided to do a 1 year trial to see which site produced the higher yield. He set up 24 rows of plants at each site and over the season measured the weight of the tomatoes he harvested. Using the data and graph below, answer the following questions, justifying your answer using the data and graph.a. Which site should Pete chose If he wants a

consistent yield from his crop?b. If Pete wants the highest yield possible, which

site should he chose?c. Pete has concluded he should choose Site

A. Discuss this using the given data. Include limitations of the data in your discussion.

Average Yield for Site A = 5.0kg, Site B = 6.1kg

6. A business owner wants to compare the Saturday traffic with the Sunday traffic to her website . She is also hoping that traffic to her website is increasing.a. Give possible explanations for any outliersb. Using the graph, write a short report for her.

Include:• Trends for both days• Spread for both days• Outliers

7. Riverflow for the past hundred years for a river in New Zealand was obtained and a box and whisker plot created for each month.

a. Why are the upper whiskers much longer than the lower whiskers?

b. The local council are unsure whether or not they need to do more flood protection work. Currently, river flows over 650 cumecs (cubic metres per second) cause minor flooding. At 750 cumecs stop banks are breached and hundreds of houses are flooded. Give reasons for your recommendations. Include in your report:

• The likelihood of either flood event• Any limitations of the data

c. The river is also used for the town water supply. If water flow ever gets below 20 cumecs the town will get heavily fined if they keep drawing water. Do they need to look at alternative water supplies? Justify your answer. You may like to consider both:

• River flow issues• Water use issues (population, water use per

person, etc.)• Any other areas the council should look into

PRACTICE QUESTIONS

NCEA1. a. Problem: It is not correct for Richard to believe

that the temperature in Rome is higher than the temperature in Nairobi. Explain: Looking at the graphs, Nairobi’s average maximum temperatures never go below 20°C while Rome’s average maximum temperatures fall below 20°C for approximately half the year. Similarly, Nairobi’s average minimum temperatures never go below 10°C, while Rome’s average minimum temperatures fall below 10°C for approximately half the year. So for approximately half of the year, Rome is colder than Nairobi. (Merit)

b. Problem: The average maximum temperatures in Rome vary from low teens in January each year to high 20s in July/August each year. The average maximum temperatures in Nairobi vary from low 20s in July/August each year to high 20s in February each year. So there is greater variation in the maximum temperatures in Rome than there is in Nairobi. Explain: The average minimum temperatures in Rome vary from single digits in January each year to high teens in July/August each year. The average minimum temperatures in Nairobi remain in the low teens all year. So there is greater variation in the minimum temperatures in Rome than there is in Nairobi. Improvements: Rome is not always warmer, only at particular times of the year. The smooth, pronounced waves formed by Rome’s line graphs indicate a steady rise and fall in average temperatures over a year with peaks around July and troughs around January. Nairobi’s line graphs show less variation in temperature over a year, but also more fluctuation over a year — there are small peaks around February and small troughs around July. There is also a smaller peak around October and a smaller trough around November/December. (Excellence)

c. The graphs do not support what Richard has been told. In fact, the bigger waves in Rome’s line graphs indicate that both the maximum and minimum temperatures vary more in Rome than they do in Nairobi.

(Achieved)d. There is no evidence to support rising

temperatures in Rome. The line graphs show regular waves over year-long periods and there is no upward trend to the graphs.

(Merit)

e. Richard is correct in his claim. Overall, the vertical distance between Rome’s line graphs (typically around 10°C) is smaller than the vertical distance between Nairobi’s line graphs (typically around 13°C). (Merit)If Richard wants to live somewhere warm, he should choose Nairobi. For approximately half the year, Nairobi’s average temperatures are warmer than Rome’s average temperatures. Also, Nairobi’s average temperatures vary much less so it would feel warmer all year round, whereas Rome would feel much colder during certain months of the year (December to February). However, the graphs show only monthly averages. It would be sensible for Richard to look at some daily temperature data to see if there are any significant deviations from the monthly averages. There are other factors that influence how a climate feels, such as humidity, rainfall and wind. It would be sensible for Richard to investigate some of these other factors before making his decision. (Excellence)

2. Problem: Melanie’s claim that Year 9s are better than Year 13s at guessing the number of dots is not a sensible one. Explain: She has made this claim because the Yr 9 mean is closer to the actual number of dots than the Yr 13. However because of Year 9s 3 outliers, the median is the best measure of average. The Year 13 median is 51 and Year 9 is 50, so they are very similar. The Year 9s’ guesses are more consistent, with the Inter-quartile range for Yr 9 being 7 (55 - 48). whereas for Year 13 it is 15.5 (60.5 - 45) but greater consistency in guessing does not mean greater accuracy.The sample size for each is very small and only carried out in one class, in one school. Improvements: Melanie needs to carry out her investigations in a range of classes and schools to make her study more valid.The data collection method needs to be free of biases such as students being able to discuss their guesses among themselves during answering and knowing other students’ guesses before answering. (Excellence)

3. a. A scattergraph is used to display bivariate data. It allows us to plot points that display both variables in the same graph and thereby determine whether or not there is a relationship (correlation) between the two variables. Since Tuahu has collected data with two variables (height and armspan), a scattergraph is an appropriate way to display his data.

(Excellence)

ANSWERS

CHANCE & DATAb. There are three points on the scattergraph that

appear to be unlikely measurements: (105,160) — it is unlikely that someone with a height of 105 cm would have an arm-span as long as 160 cm; (160,60) — it is unlikely that someone with a height of 160 cm would have an arm-span as short as 60 cm; (173,80) — it is unlikely that someone with a height of 173 cm would have an arm-span as short as 80 cm; (Merit - 2 correct)

c. Tuahu’s conclusion is valid because • There is a line of best fit, which suggests a

trend of a person’s arm-span being the same as their height.

• Most of the points are close to the line of best fit.

However, there are a number of outliers with unlikely measurements for a Year 10 student, so further investigation into the validity of Tuahu’s data collection would be sensible. (Excellence)

PRACTICE4. Problem: Their choice of Day 1 is not sensible.

Explain: Day 1 does have the lower maximum and upper quartile, but also has higher median, higher lower quartile and higher minimum. Day 2’s lower minimum, lower quartile and median suggest it would be beneficial to run the promotion on Day 2.Day 2 has much greater variation in number of customers attending as shown by higher maximum, lower minimum and larger lower and upper quartiles suggesting students may be more likely to be persuaded to come out on Day 2 by half price tickets.Improvements: However, we do not know which months the survey was carried out over. It may have included the 3 month student holidays when either the students may leave the town or may be working and have more money to attend the movies. From the box and whisker plot neither trends nor outliers can be seen.Presenting the survey data in another form (such as a time series or dot plot) may give more useful information. (Excellence)

5. a. Pete should use Site A if he wants consistent yield. From the graph it can be seen that for Site A the difference between the highest and lowest yield is about 2kg, whereas for Site B the yield difference between highest and lowest is about 6.5kg. (Merit)

b. Since the average yield for site A is 5.0kg and for site B is 6.1kg, Pete should choose Site B for highest yield. (Achieved)

c. If Pete wants consistent yield, then Site A appears to be the better option. Although the overall yield from Site B is higher, it gives such varying yields from each plant (6.5kg) that Pete may be concerned about what will happen under different weather patterns other years.His results were only from 1 summer so that may have been a particularly wet or dry, sunny or cloudy, windy or still summer. The plant quality (even in the same variety) may vary from year to year and different tomato plant varieties may give different results. Testing over a few summers with different plant varieties would give a more accurate picture. (Excellence)

6. a. The weekend with Saturday maximum of 7000 and the Sunday maximum of 9300 are both high outliers. Reasons for that could be:

• The website has advertised special deals

• It is just before Christmas• The weather is bad meaning people are

more likely to be inside and going on the internet

A Saturday low outlier of just above zero could be caused by:

• A special Saturday event or unexpectedly good weather meaning people are outside

• The website being down or internet problems meaning people are unable to access the site. (Merit)

b. Sunday traffic is generally heavier than Saturday as can be seen by a higher Sunday trendline.Saturday traffic has a wider spread than Sunday as can be seen by the distance of each point from the trendline.Saturday traffic is increasing slightly as can be seen by the trendline rising from an average of 2000 hits per Saturday to just over 3500. Sunday traffic is increasing more quickly than Saturday’s. This is shown by a steeper trendline. Average traffic has increased from just over 3000 hits to about 6800. (Excellence)

7. a. With river flow there is a minimum flow rate (0) but no maximum rate. Over 100 years there would have been many floods of various sizes which would account for the large whiskers on the upper part of the box and whisker plot.

(Merit)

b. Problem: The council do not need to do more flood protection work.Explain: Over the last one hundred years river flow has never gone above 750 cumecs. So it would not make economic sense to carry out flood protection work to prevent a 750 cumec flood.Over the one hundred year period 3 months of the year have had flows of around 650 or above and months 1 and 12 have had flows of 700 cumecs at least once..Improvements: From box and whisker plots it is not possible to tell how frequently events (such as river flow reaching 700 cumecs) occur. The council needs to look at the data presented in another form that will give a more accurate indication of the frequency of the river flows reaching 650 cumecs to see if flood protection work needs to be done to prevent the more minor flooding that occurs at 650 cumecs.The council may also want to see the data presented in a form which gives river flow trends as factors like removal of bush cover, farmers using water for irrigation upstream and climate change may be causing changes in river flow over time. They also need to check out the cost of flood protection work in relation to the cost of a flood. (Excellence)

c. Problem: The council does not need to look at alternative water supplies.Explain: Over the last one hundred years, in only one month (Month 3) has the river flow gone below 20 cumecs but like (b) above, it is not possible to tell the frequency of this event.Months 2, 4 and 9’s lowest flows are getting close to 20 cumecs.Improvements: To get a better idea of how to progress the council needs to get the data presented in a different form. Then they can get a more accurate indication of the frequency of the river flow dropping below 20 cumecs. The data also needs to be presented in a form that enables the trend in river flow to be seen and trends in population of the area and water use per person need to be investigated. If the population or water use per person has an upward trend or the river flow has a downward trend, this would increase the overall water use and thus lower the river making it necessary to look at other water sources. (Excellence)

CHANCE & DATA

Study Tip:

Understand NCEAYou are sitting NCEA so:UNDERSTAND the marking system• Each question is marked out of 8 so attempt all three ques-

tions in each topic.• You can get up to 4 marks with Achieved parts. Merit parts

give 5 - 6 marks and Excellence gives 7 - 8.• If you are aiming for Excellence, do the Excellence parts of the

question first. This gives you the full 7 - 8 marks straight away for each question. Easier parts of the question do not give you any extra marks (but they are a safety net if you’ve made a mistake)

PROBABILITY

OLD NCEA QUESTIONS1. Alice’s local supermarket is running a competition.

On the back of each docket is printed one of the letters of the word ANKARA, a city in Turkey.If Alice can collect the six letters needed to spell Ankara, she will go in the draw for a holiday to Turkey. a. On each of the 5 weekdays for 5 weeks Alice

finds a discarded docket as she passes the supermarket. In the order that she collects them, the letters collected are:

N K K R R N A K A R N N A K K N R N A A A K K R R

i. Complete the table to summarise her data:

Letter FrequencyAKN 6R

ii. Using her data, what is the probability of Alice getting a K on the next docket?

iii. How valid is this probability? Give at least 2 statistical reasons for your answer.

Alice wonders how many dockets she would have to collect, on average, to be able to spell the word ANKARA.b. Using her collection of dockets in (a), how

many dockets did Alice collect before she had the whole word of ANKARA?

c. Alice realises that it will take too long to find an answer by collecting actual dockets. Instead, she takes a dice and puts the six letters of A, N, K, A, R, A on it. Alice wants to find out, on average, how many times she must roll the dice to spell the word ANKARA. She rolls the dice and whatever letter is on top, she imagines is the letter she has found on the back of a docket. Once she has all the letters she needs to spell Ankara, she begins again. She stops her experiment when she has spelt the word Ankara 10 times.R, K, N, A, A, K, K, K, AA, A, N, A, R, A, KA, N, N, K, A, N, R, AK, A, K, K, R, A, A, A, NR, A, N, A, A, R, A, KN, A, R, R, A, A, A, KA, A, A, A, R, A, R, A, N, R, A, A, A, A, R, A, A, N, R, N, A, KN, K, A, A, N, A, A, N, A, K, N, A, RN, K, A, N, N, A, N, A, A, RA, A, K, A, K, A, N, N, A, RAlice then uses her results to find out how many dockets she needed to spell the whole word of ANKARA. Her results are: 9, 7, 8, 9, 8, 8, 22, 13, 10, 10i. Using Alice’s data, give Alice an answer to her

question: “How many dockets would she have to collect, on average, to be able to spell the word ANKARA?” Give at least TWO averages.

ii. Explain which average you would suggest Alice uses and why.

iii. The supermarket says that each letter, A, N, K and R, is equally likely to be found. Explain why Alice’s experiment is not valid.

SUMMARY • Probability is a measure of the likelihood or chance of an event occurring• Events have a probability between 0 and 1 (0 is impossible, 1 is certain)• Probability of an event E is written P(E)• There are two ways of measuring probability:

▶ Theoretical Calculations ▶ Experimentally, using the proportion of times the event occurs

The experimental probability of an event E is For a complete tutorial on this topic visit www.learncoach.co.nz

P(E) Number of times E occursTotal number of events

=

CHANCE & DATA

2. There are 45 plants in flower in Anne’s garden.10 of the plants have blue flowers, and 11 of the plants have white flowers.A plant is chosen at random from Anne’s garden.a. What is the probability that it has neither blue

flowers nor white flowers?b. The heights of the 45 plants in Anne’s garden

were measured 3 weeks after planting.The heights were plotted on the histogram.

A plant is chosen at random from Anne’s garden 3 weeks after planting.What is the probability that the plant chosen is less than 12 cm high?

3. A health survey asked a group of 220 adults about their age and levels of exercise.The table shows some of the data from the survey.

Age Under 50 Over 50

Regular Exercise 98 38No Regular Exercise 62 22

a. Find the probability that an adult picked at random from the whole group is 50 years or older, and does no regular exercise.

b. From the group of adults 50 years or older, a person was picked at random.What is the probability that this person exercised regularly?

4. Three different airlines fly from Christchurch to different destinations in New Zealand each week, as shown in the table below.Flights are chosen at random from the table below and awarded to winners of a competition.

AirlineNumber of flights to each destination

Auckland Queenstown Mt Cook Rotorua TOTALFly NZ 34 21 16 4Mountain Air 12 17 6 3Blue Skys 15 10 1 1

Total 140

a. Anne wins a flight. What is the probability that her flight is to Queenstown?

b. Jan wins a flight to Auckland. What is the probability that her flight is with FlyNZ?

PRACTICE QUESTIONS5. A new anti acne cream was undergoing trials. It was

found that 3750 of the 5000 trialists had significant improvement in their acne. What is the probability the cream will work.

6. Eastern Rugby Club annually play United for the Magpie Trophy. Over the last 60 years Eastern have recorded the wind strength if they won or drew the game

Wind Win Draw GamesGale or above 14 2 31Moderate 13 1 14Calm 2 4Total 60

a. What is the probability that it was calm during the game?

b. What is the probability that the result was a loss for Eastern Rugby Club?

7. The table shows results for Southern High from the regional interschool sports meeting

1st 2nd 3rd TotalTrack 13 8 17 38Field 3 12 3 25Swimming 10 8 2 13Total 26 28 22 76

a. What is the probability that a 1st place is in the field?

b. What is the probability that a place (1st, 2nd or 3rd) is in swimming?

8. a. Using the data below, what is the probability that someone from Waiputa passed their restricted license at age 18?

b. What is the probability that a girl from Waiputa passed her Restricted license at age 17?

Age at which Restricted Driving License is Passed in Waiputa:

Age 17 18 19 20 Over 20 TotalBoys 109 89 22 49 9 278Girls 41 86 42 32 25 226Total 150 175 64 81 34 504

9. During youth week at Waipua College every time a student was observed doing something good the student was given one of the letters from the word Waipua. They received a prize once they spelt Waipua. Each letter was equally likely to be received. a. What is the probability of a student receiving a

card with a vowel (A, E, I, O, U) on itb. Callum has already received a P, U,U, P and A.

What is the probability that the next letter he receives is one he needs?

ANSWERSNCEA1. a.

i. Letter Frequency

A 6K 7N 6R 6

(Achieved)

ii. P(K on next docket) = =725

0 28. (Achieved)

iii. Assuming that each letter is equally likely to be found, the theoretical probability of getting a K on a randomly selected docket is 1/4 = 0.25

The frequency table in (i) suggests there is a fairly even distribution of the four letters despite the small sample, and the letters printed on previous dockets have no influence on the letter printed on the next docket. Therefore, the experimental probability calculated in (ii) is close to the theoretical probability and seems to be valid. (Excellence)

b. Alice had to collect 13 dockets before she could spell ANKARA. (Merit)

c. i. Mean = + + + + + + + + +

= =

9 7 8 9 8 8 22 13 10 1010

10410

10 4 .

Median of {7,8,8,8,9,9,10,10,13,22} = 9 (Merit)

ii. Alice should use the median because the mean might have been influenced by the data entry 22, which seems significantly higher than all the other data entries and should be treated as an outlier. The median ignores this outlier and will provide a more reliable average. (Merit)

iii. Alice’s experiment is not valid because her dice has A on three of its sides. This gives a 50% chance of rolling an A and does not reflect the equal likelihood (25%) of finding any of the four letters on the dockets.

(Excellence)

2. a. P = = = =2445

815

0 533 53. % (Achieved)

b. P = = = =3645

45

0 8 80. % (Achieved)

3. a. P = = =22220

110

10%

(Achieved)b. P = = = =

3860

1930

0 6333 63. %

4. a. 48140

0 34= . (2dp) (Achieved)

b. 3461

0 56= . (2dp) (Achieved)

PRACTICE5. P(CreamWorking) .= =

37505000

0 75 (Achieved)

6. a. Number of games at where it was calm:= − + =60 14 31 15( )Probability of it being calm at a game:

P(calm) = = =1560

14

0 25. (Merit)

b. Number of losses:= − + + + + + =60 14 13 2 2 1 4 24( )Probability of losing:

P(loose) = = =2460

25

0 4. (Merit)

7. a. P(1 in Field)st = =326

0 12. (Achieved)

b. P(Swimming) = =1376

0 17. (Achieved)

8. a. P(Passed at 18) = = =175504

2572

0 35. (Achieved)

b. P(Girl Passed at 17) = =41

2260 18.

(Achieved)

9. a. P(Vowel) P(A) P(I) P(U)= + +

= + + = = =15

15

15

35

0 6 60. % (Merit)b. P(W or I or A) P(W) P(I) + P(A)= +

= + + = = =15

15

15

35

0 6 60. % (Merit)

CHANCE & DATA

MULTI LEVEL PROBABILITYSUMMARY

This is when more than one event is occurring• Addition Law

Probability of E OR F = +P(E) P(F)• Multiplication Law

Probability of E AND F = ×P(E) P(F)• Probability Trees

▶ A tool for working out probabilities ▶ Multiply along branches ▶ Probabilities of branches from the same point

add to 1e.g. Probability of tails then heads:

For a complete tutorial on this topic visit www.learncoach.co.nz

OLD NCEA QUESTIONS1. Harley has a collection of unusual dice, which have

shapes on each side.

Each dice has 6 faces, and all of the dice are fair. Harley rolls a blue dice and a green dice 100 times and records the results. His results are summarised in the table below.

a. Estimate the probability that on the next roll he will get:

i. a and a .ii. ONE and ONE .

b. Estimate the probability that Harley will not get a on the next roll.

c. Estimate the probability that the blue dice shows a if the green dice shows a .

d. Estimate the probability that if there are no s showing, there is a showing.

e. Deduce how many s there are on the green dice.Use probability methods to find your answer and explain your reasoning clearly.

Harley has an orange dice which has only s and some s on its faces.He rolls this orange dice with the green dice 120 times and records the combinations that show up. His results are summarised in the table below. (The order of the shapes does not matter.)

f. Using a ratio in its simplest form, express the probability that he gets .

g. On the orange dice, what is the most likely number of faces with a marked on them?Use probability methods to find your answer and explain your reasoning clearly.

2. Nice Flight airline gives away flights from Wellington as prizes.Three different destinations are available.The probability of flights going to each of the destinations is shown in the table.

Destination ProbabilityGreymouth 0.1Kerikeri 0.4Queenstown 0.5

John wins two separate prizes with NiceFlight.All flights are chosen at random according to the probabilities in the table.Calculate the probability that John wins flights to two different destinations

P(TH) 0.5= ×=

0 50 25

..

3. A cat has four kittens. Assume that each kitten is equally likely to be a female or a male.Explain fully and carefully, by finding and comparing probabilities, which of the following is most likely for the four kittens:

• There are 4 female kittens.• There are 3 female kittens and 1 male kitten.• There are equal numbers of female and male

kittens.

4. At a certain garage the probability of a customer buying diesel is 1

9.

The other customers all buy petrol.15

of both diesel and petrol customers pay cash for their fuel.All others pay by credit card.Some of the information is shown on the diagram.

a. What is the probability that a customer chosen at random bought petrol and paid by credit card?

b. What is the probability that the next two customers both bought petrol?

5. The oil pressure in a vintage car is noted by a warning light.We know that the oil pressure is too low 1% of the time.If the pressure becomes too low, the light should come on.However, the light is not always reliable.When the oil pressure is too low, the light is on 98% of the time.When the oil pressure is okay, the light is on 0.1% of the time.Find the probability that the light is giving an incorrect reading.

6. a. A research scientist plants eleven seeds in Plot A and nine seeds in Plot B.The probability that a seed germinates in Plot A is 0.7, and in Plot B it is 0.8

Find the probability that a seed chosen at random germinates.

b. The scientist then plants seeds in Plots D, E and F in the ratio 8 : 10 : 7The probability that a seed germinates in Plot D is 0.7 and in Plot E is 0.8.This scientist expects 390 seeds to germinate from the 600 he planted.What is the probability that a seed in Plot F germinates?

7. A circular spinner is divided into two quarters and one half, with the three sectors labelled with scores of “10”, “20” and “30”, as shown in the diagram.

a. Ngaire spins the arrow twice.What is the probability that the total of the two scores is 30?

b. Sue is also playing with the spinner.She spins the arrow twice and adds the two scores.She does this 80 times.How many times would she expect the total to be at least 40?

CHANCE & DATA

PRACTICE QUESTIONS8. Children are running round a playing field.

When the whistle blows, the children have to freeze.

A B

C

a. What is the chance Mia and Hone are both in area C as shown in the diagram?

b. What is the probability that both Mia and Hone are in the same area?

9. Joseph, Hayley and James have two coins and toss these to see who will cook dinner. Heads = H, Tails = T.a. If the coins land with 2 heads

up James will cook.What is the probability of this?

b. If the coins are: TT - Hayley will cookHH - Joseph will cookH and a T - James will cookWhat is the probability one of the boys will cook?

10. At a fundraising sausage sizzle you were offered onions and tomato sauce as extrasIt was found 70% had onions and of those 80% has tomato sauce as well.Of those who didn’t have onions, 40% didn’t have tomato sauce either. A probability tree may help.a. What is the probability of a randomly chosen

person who bought a sausage having tomato sauce with it?

b. What is the probability of a randomly chosen person having a sausage with just one extra (either sauce or onions)?

11. Natalie has made 10 batches of cupcakes for her little brother’s birthday party and decorated them. There are:

• 2 choices of icing – yellow or green

• 2 choices of edible animal decorations – spider or shark

• 2 choices of sprinkles – gold or silver. A probability tree has been started.

a. What is the probability that any random cupcake chosen

i. Will have yellow icing and a shark on the top?

ii. Will have a shark on the top?b. Natalie runs out of silver sprinkles quite quickly

so the probability of silver sprinkles is only 0.2i. Troy only eats cupcakes with yellow icing.

What is the probability he gets a cake with a shark and gold sprinkles?

ii. Mei is scared of spiders and will only eat cupcakes with sharks. What is the probability her cupcake is green with gold sprinkles?

NCEA1. a. i. 7

1000 07 7= =. % (Achieved)

ii. 14 9100

23100

0 23 23+= = =. %

(Achieved)

b. 19 16 7

10042

1000 42 42+ +

= = =. % (Achieved)

c. 1749

0 35= . (2 dp) (Merit)

d. 1735

0 49 49= =. % (Merit)

e. The results for the and on the green dice are almost the same for any symbol on the blue dice, so it can be said that there are equal numbers of each of and on the green dice – i.e. 3 of each. (Excellence)

f. 20:120 = 1:6 (Merit)g. If there were 3 each of and we would

expect the frequencies to be approx 1:2:1. They are not, and the lowest result for suggests that is only on 1 or 2 sides.The shaded row in the table most closely matches the experimental results.

Options (No. of each

shape)Expected Numbers

1 5 10 60 502 4 20 60 403 3 30 60 304 2 40 60 205 1 50 60 10

Therefore the orange die is most likely to have four sides with a marked on them. (Excellence)

2. 2 0 1 0 4 0 1 0 5 0 4 0 5 0 58. . . . . . .× + × + ×[ ] = (Merit)

3. P F

P F

P F

( )

( )

( )

4 116

3 416

2 616

=

=

=

Equal male and female most likely. (Merit)

4. a. P(P) P(CC)× = × =89

45

3245

(Achieved)

b. P(P) P(P)× = × =89

89

6481 (Achieved)

5.

P = × + ×= +=

0 01 0 02 0 99 0 0010 0002 0 000990 00119

. . . .

. .

. (Excellence)

6. a.

P = × + ×

= + = =

1120

0 7 920

0 8

0 385 0 36 0 745 75

. .

. . . %ORP(Plot A) (8 plants)P(Plot B) (7 plan

= × == × =

11 0 7 7 79 0 8 7 2

. .. . tts)

P = = =1520

0 75 75. %

(Merit)

b. Let g be the germination of plot F.

P(Total) P(D) P(E) P(F)= + +

= × + × + ×

=

390600

825

0 7 1025

0 8 725

0 65 0

. .

.

g

.. . .. . .. ..

224 0 32 0 280 65 0 544 0 280 106 0 280 379

+ += +==

gg

gg

ORtotal seeds seeds germinated

Plot D: 192 0.7 134.4 (134)Plot

× =EE: 240 0.8 192

Plot F: 168 g 63.6 (64) × =× =

==

168 63 60 379g

g.

. (Excellence)

ANSWERS

CHANCE & DATA7. a. P(30) P(1 is 10, 2 is 20)

P(1 is 20, 2 is 10)

st nd

st nd=

+

= ×14

144

14

14

216

18

+ × = =

(Merit)b. P(T 40) P(total is 20 or 30)

P(total 20 P(total 30)≥ = −

= − −

=

11 )

11 14

14

216

1 116

216

1316

1316

80

6

− × −

= − −

=

= ×

=

(from a.)

Expected No

55 (Excellence)

PRACTICE8. a. P(Hone in C) = 1

2

P(Mia in C) = 12

P(Both in C) P(Hone in C) P(Mia in C)

=

= ×

× = =12

12

14

0 25. (Merit)

b. P(Both in same area)P(Both in A) P(Both in B) P(Both in C

=+ + ))

= ×

+ ×

+ ×

= + + = = =

14

14

14

14

12

12

116

116

14

616

38

0..375 (Merit)

9. a. Equal possible outcomes are: HH, TT, TH, HT.

P(HH) = =14

0 25. (Achieved)

b. P(TT)

P(Boys Cooking)

= =

= − =

14

0 25

1 0 25 0 75

.

. .

Alternative MethodBoys will cook if result is HH or HT or TH.

P(HH or HT or TH) = + + = =14

14

14

34

0 75.

(Merit)

10.

a. P(Sauce) P(O S) P(No O S)= + + += × + ×= + =

0 7 0 8 0 3 0 60 56 0 18 0 74. . . .. . . (Merit)

b. P(1 Extra) P(O No S) P(No O S)= + + += × + ×= + =

0 7 0 2 0 3 0 60 14 0 18 0. . . .. . ..32 (Excellence)

11. a. i. P(Yellow & Shark) P(Yellow P(Shark)= ×

= × =)

. . .0 6 0 3 0 18 (Merit)

ii. P(Shark) P(Y & S P(G & S)= += + × =

). . . .0 18 0 4 0 6 0 42

(Merit)b.

i. P(Shark&Gold) P(Shark P(Gold)= ×= × =

). . .0 3 0 8 0 24

(Excellence)ii. P(Green&Gold) P(Green P(Gold)= ×

= × =)

. . .0 4 0 8 0 32 (Excellence)

PROBABILITY ANALYSIS

OLD NCEA QUESTIONS1. Evaluate the claims in the advertisement below. In your answer, you may find it helpful to comment

on:• What the variables of interest are• The purpose of the advertisement• The question the advertisement is posing• If the claims in the advertisement are valid,

correct and sensible• Any other information that would be helpful.

SUMMARY1. Problem - Make a Statement about whether the PROBABILITY in the question is correct

(It usually isn’t).2. Explain how you worked out the correct answer:

• Show all working• Give reasons for you working if possible

3. Improve - Give any improvements that are needed. List any possible circumstances where the situation could have been correct. For a complete tutorial on this topic visit www.learncoach.co.nz

PRACTICE QUESTIONS2. Cody sees the advertisement below and concludes

he has 1 in a 1000 chance of winning $1000 worth of electronics.

Evaluate his conclusion by commenting on the following:

• What the relevant variables are• Assumptions Cody is making• If his conclusion is valid• Any other information that would be helpful.

3. Talia stated ‘Redheads are becoming more common’ .This is because Talia has 5 people in her class with naturally red hair whereas when her Mum was growing up there were only 2 in her class and her Nana never had any. Evaluate Talia’s statement by commenting on:

• The variables• The assumptions that Talia has made• The nature of the sample (size, randomness)• Any other information that would be helpful

CHANCE & DATA4. Zoe wanted to do a tandem skydive for her 16th

birthday. Her father said she couldn’t because it was too dangerous. Zoe found out that 480 people died on the roads last year in New Zealand and of those who skydived, 1 person out of 25000 died. So she told her father she should be allowed to sky dive because that she had a 1 in 480 chance of dying during a car ride tomorrow compared to only 1 in 25000 chance of dying in a skydive accident. Evaluate Zoe’s statement. Exact calculations are not required. You may like to comment on

• The purpose of her statement• Assumptions or errors Zoe has made about the

risk of her dying in a car accident• Subsets of the national statistics she may be in

and the affect that has on relative risks

5. Jeremy and Sally decided to create a chance game for a family reunion. They told all their relatives that if they can get a total of 12 after two throws of a dice then they win a lolly. They told everyone that there was a 1 in 11 chance of winning. Discuss the claim that Jeremy and Sally made. You may like to include:

• If the claims are valid, correct and sensible• How they may have decided upon the 1 in 11

chance.• Any other information that would be helpful

6. A supermarket promotion included a scratch card if you spent $50 or more. There were three panels on the card, each with a picture of one of four different fruits. If all three fruits on the scratch card were the same, a $10 voucher was won. The supermarket stated ‘Match 3 apples, bananas, oranges or pears and WIN! 25% chance of winning!’ Discuss this promotion.Include:

• The purpose of the advertisement• If the claims in the promotion are valid, correct

and sensible• Any other information that would be helpful.

7. There was a chance game at a school fair that you pay to enter. The player first has to spin a wheel which is split up into 4 quarters each with a different colour on it: red, blue, green, and yellow. The player then had to reach into a box with 40 balls (10 of each colour) and pull one out. The player wins a small prize if the colour spun matches the colour of the ball. The sign advertising the game states there is a 1 in 8 chance of winning. Evaluate the sign and any repercussions on fund-raising.

Exam Tip:

Proofreading• Only takes a few minutes• Finds the silly mistakes• Gives you extra marks (especially if your teacher is super-me-

ticulous)SO many people miss out on Extra Marks by not proof-reading

ANSWERSNCEA1. Problem: The chance of getting a yellow circle is

actually, 112

which is only about 8%, not 30% as the

ad claims. Explain: This is because there are 3 types of token and each comes in 4 colours, making 12 different tokens in total, only 1 of which is a yellow circle.Improvements: However, this calculation assumes that there is the same number of each token, and colour, being put into the lamburgers so that the distribution is uniform, which you would expect but which may not be the case. ORIt could be that there are actually more yellow circles in circulation than any other token, so that there is actually a 30% chance of winning, you would have to check with the companyORPerhaps the 30% applies only to the change of getting the circle, since this is 33% and may have been rounded. (Excellence)

PRACTICE2. Problem & Explain: Cody is correct that he has a 1

in a 1000 chance of winning (10/10000). However:• He assumes all 10 prizes are $1000 worth of

electronics. • The purpose of the poster is to sell tickets and

make money.• The value of each prize is not specified, except

to say it has a maximum value of $1000. The poster also does not say that all the prizes are electronic goods.

Improvement: Cody should assume that he has a 1 in a 1000 chance of winning a prize. While it could be that all prizes are $1000 worth of electronics, for the reasons above, this is unlikely. (Excellence)

3. Problem: Talia’s statement is unlikely to be true. Explain: For Talia’s statement to be true the following needs to be considered:

• Were the samples the same size – Talia’s class size may be bigger than her mother’s or grandmother’s.

• Were all 3 at school in areas with typical mixes of nationality – for example her grandmother could have been at school in an area with a high proportion of Maori pupils.

• Was the proportion of redheads similar over different classes in the school and in different schools around the country?

Improvements: Talia’s statement may be true but she needs to do further investigations of different classes and schools now and in the time of her mother and grandmother. (Excellence)

4. Problem: Zoe’s statement is not true. Zoe wants to convince her father to allow her to skydive so the aim of her statements is to persuade her father.Explain: There are two errors in Zoe’s reasoning:• The value of 480 deaths is an annual figure,

not a daily figure. The 480 would need to be divided by 365 if she was to make it daily.

• It is incorrect to use 1 in 480 as the probability of dying. It is actually 480 deaths per 4 million people (population of New Zealand).

Improvement: A more realistic probability would be calculated by multiplying 1 in 365 by 480 in 4,000,000. This ends up giving an probability of around 1 in 3,000,000.Zoe could also investigate the sky diving statistics more thoroughly. It could be that tandem skydivers or people under 20 or females are more or less likely than average to have a fatal accident.Who is driving the car and at what time could also need to be taken into account. For example probably getting a ride with her father is lower risk than if one of her fellow students gives her a ride. Late at night may be more risky than during the day.Even without further investigation of skydiving statistics and driving habits there is still a much lower chance of her dying in a car crash when compared to skydiving. (Excellence)

5. Problem: Jeremy and Sally’s claim of a 1 in 11 chance of winning is incorrect. Explain: To get a 12 the player needs to throw two sixes each of which has a 1 in 6 chance. So to get a total of 12 there is actually a 1 in 36 chance.Improvements: They may have reached the 1 in 11 chance by just counting the possible totals (2-12). Jeremy and Sally should have accounted for the fact that some totals come up more often than others. For example, 7 comes up the most often, as there are 6 different combinations that can make 7. (Excellence)

6. Problem: The supermarket’s statement is not true.Explain: The aim of the promotion is to try and encourage people to shop at that supermarket and to spend more when they are there. By stating that there is a 25% chance of winning it encourages people to shop there and to spend at least $50 to try and win a $10 voucher.If the fruits all occur with equal frequency, stating that there is a 25% chance of winning is incorrect. There is a 25% chance of getting a particular fruit on one of the panels but the chance of having the same fruit on all 3 panels is 0.25 x 0.25 x 0.25 = 6.25%.Improvements: Their claim of 25% may be correct if they have made 25% of the scratch cards specifically to have three of the same fruits. (Excellence)

CHANCE & DATA7. Problem: The sign saying a 1 in 8 chance of winning

is incorrect. Explain: There is a 1 in 4 chance of getting a particular colour and its corresponding ball. It does not matter what colour is spun on the wheel, there is always a 1 in 4 chance of pulling out a matching ball. If they have worked out pricing based on winning 1 in 8 times instead of 1 in 4, this means their profits wild be lower and they may even loose money because of the cost of prizes. Improvements: In addition, the promoters need to change the advertisement to accurately reflect the higher chance of winning, meaning potentially increased numbers of paying participants leading to increased profits. (Excellence)

Study Tip:

RewardsReward yourself for passing a tough year!Small achievements (e.g. topping a quiz) also deserve small re-wards Maybe you could remind your parents about this one!

study tip: music - amazon s3 · graphing analysis p 129 chance & data 1. problem - make a...

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