Statistical and Probability Reasoning and their Application in Real Life

The ability to accurately predict the winner of an election in a democracy is big business. Obviously, the ability to correctly predict the winner of an election can save political parties a lot of time and resources. Manufacturers would also like to know how ready a market is for a new product. In such a case, market research can be conducted by the manufacturers to determine how best to advertise and sell the product.

In this section, you will apply your knowledge of statistics to solve problems in your community. By designing tools to collect data and carefully analysing and interpreting the data collected, stakeholders will receive reliable statistics to make informed decisions. In this way, statistics can help you to study local and national issues. With the data collected, analysed and interpreted, you can predict and prepare for future events.

KEY IDEAS

• Event is any subset of the sample space.

• Experiment is performing an action and studying what happens.

• Outcome is the result of an experiment. Each experiment can have multiple possible outcomes.

• Probability is the possibility or likelihood that a given event will occur.

• Random experiment is an experiment whose outcome cannot be predicted.

• Sample space is the set of all the possible outcomes.

A close relative of statistics is probability. Probability is an important tool in scientific prediction. With probability you can make predictions about what is most likely to happen. Manufacturers use probability to determine the lifespan or quality of their products. Probability plays an important role in approving drugs for public consumption. In probability, we will study events which are equally likely, mutually exclusive, independent or dependent.

Imagine you want to study the height of learners in your school, you will need a tape measure or a metre rule. You will also need a piece of chalk to mark the height of students on the wall before measuring the height. In such an experiment or research, the measuring instruments, or tools, are readily available on campus.

In the Home Economics department, you can find a tape measure. From the Physics laboratory you can find a metre rule.

However, suppose you want to conduct an investigation into the attitude of students towards food served in the dining hall. Where can you find an instrument or tool to measure the attitude of students? In such a case, the best tool or measuring instrument is a questionnaire or interview.

As much as gathering information is important, until the data is processed very little sense can be made out of the raw data. This is where the analysis and interpretation of data comes in. Thus, tallying the data, drawing a frequency distribution table, finding the measures of central tendency or dispersion, as well as drawing charts and graphs to illustrate the data will assist stakeholders to understand the data at a glance and make informed decisions. This is the essence of statistics: gathering data, analysing the collected data and interpreting the data to make well-informed decisions.

Activity 8.1: Investigating the height of 50 students in your school Working in small groups carry out the following activity:

Step 1: Seek the permission of a teacher, guardian and the participants whose heights you are going to study in your research or experiment. Choose the participants at random, for example, the first 50 students going to the dormitory after classes.

Step 2: Mark the height of each participant on a wall, using a piece of chalk and measure it in centimetres or metres and record it.

Step 3: Choose appropriate class intervals for your data. Tally the heights of the participants and construct a frequency distribution table for the data.

Decide if you should split the data between males and females and defend your decision.

Step 4: From the frequency distribution table, find the measures of central tendency and dispersion and draw charts and graphs to illustrate the data.

Step 5: Based on the interpretation of data, summarise your observations and draw conclusions. Consider how you could improve the investigation if you were to carry it out again.

Activity 8.2: Investigating the percentage of heads or tails when a fair coin is tossed 50 times Working in small groups carry out the following activity:

Step 1: Toss a fair coin and record the outcome, using H for head and T for tail. Do this 50 times.

Step 2: Draw a table to show how many heads and tails were obtained as well as the percentage of each outcome.

Step 3: Draw a conclusion from the results.

Activity 8.3: Measuring students’ attitude towards food served in the dining hall Working in small groups carry out the following activity:

In this activity, we cannot simply measure and record the height of something or carry out a simple experiment. As you learnt in year one, to collect qualitative data requires the use of, for example, observations, interviews or questionnaires. You can choose which method to use. Below is an example of a questionnaire.

Step 1: Decide which method you will use to gather the data. In this example, we have used a questionnaire.

Step 2: Design the tool, remembering the rules guiding the construction of questionnaires. The most important question to ask is which dining hall food students like best. However, that information alone will not be sufficient for a thorough analysis. For this reason, details like the class, program, age, gender and the house of the participants will help to make your analysis more meaningful. Also, the reason behind the choice of food might be very helpful.

Step 3:Decide how many students from each form group and gender you will ask. Distribute the questionnaires through your chosen means, be it, for

example, handing out hard copies or using social media. Remember to give a date for the return deadline.

Step 4: Go through the completed questionnaires to tally the results. A spreadsheet works well for this. After sorting out, you can discover how many first-year learners like breakfast or how many girls in form 3 like jollof rice etc.

Step 5: Use tables, appropriate charts and graphs to interpret and illustrate the data collected.

Thank you for taking the time to complete this questionnaire. The information will be kept confidential.

Date: ……………………………………………………………… Tick in the appropriate bracket.

1. Gender: Male ( ) Female ( )

2. Level: Form 1 ( ) Form 2 ( ) Form 3 ( )

3. House: House 1 ( ) House 2 ( ) House 3 ( ) House 4 ( )

4. How many times do you go to the dining hall every day?

None ( ) once ( ) twice ( ) thrice ( )

5. What is your favourite meal in the dining hall?

Breakfast ( ) Lunch ( ) Supper ( )

6. What is your favourite food in the dining hall?……………………....

7. Give one reason for your choice. …………………………………

Figure 8.1: A sample questionnaire to measure the attitude of students towards food served in the dining hall See the Review Questions where you will carry out your own mini-project. It is good if you choose topics to investigate which are important to you.

Probability is a tool for making scientific guesses. It gives us methods with which we can make predictions or estimates as accurate as possible. This means that probability has wide applications in many fields.

Application of probability in pharmacy Before a new drug is approved for human use, the drug goes through rigorous tests. Usually, the new drug is administered to a sample of willing participants.

The positive and negative effects of the drug are noted, as well as the percentage of participants experiencing positive or negative side effects. It is for this reason that the side effects of drugs can be seen inside or outside the packaged drug, along with the instructions and dosage. For example, this could be why consumers of certain drugs are advised not to drive after taking the drug.

If the research proves the efficacy and safety of the drug beyond reasonable doubt, it is approved for public use. That is, if the positive effects outweigh the negative effects, the drug is approved. For example, a drug like chloroquine can cause severe itching in certain users while others did not experience the negative effects, so it was deemed that the positives outweigh the negatives and it is approved for human use.

Application of probability in computer science When typing on your phone, the keyboard can offer suggestions of what you are going to type, to save your typing it. This is known as predictive text. For

example, immediately you type t, the keyboard may suggest the or thanks. The computer will have learnt your usual style of typing and use of language and, using probability, decide what you are likely to be wanting to type and give you help in this. In this way, probability helps speed up your communications.

Application of probability in weather forecast Have you ever seen the meteorologists giving the weather forecast on the television? Probability is an important tool in these predictions. For example, if in the last 10 years, some conditions in the clouds produced rain in the Axim area in June, then the probability that the same conditions will produce rain in the Axim area in June this year is very high.

Probability always lies between zero and one We are going to learn how to measure the likelihood that a given event will occur.

To do this we will assign it a numerical value. For example, what percentage will you assign to the probability that a particular classmate will come to school today?

For those who come to school regularly, the probability will approach 100%. For those who are less reliable, the probability that they will come to school today will be less. Thus, probability is always a rational number lying between zero and one. The lowest probability is zero. This means that the event will never happen.

The highest probability is one. This means that the event will definitely happen.

If you calculate a probability to be below 0, or above 1 means that you have gone wrong somewhere!

Random experiments Experiments are conducted to study the effect of a particular action. If the results of the experiment depend on chance or luck, you have a random experiment.

Tossing a fair coin or die is a classic example of a random experiment. Drawing identical balls from a bag or box at random is another example of a random experiment. Predicting who will win the football match or horse race is also a random experiment. One performance of a random experiment is a trial. So, tossing a fair coin once is one trial of the random experiment.

Sample space When a fair coin is tossed once, the result is either a head (H) or a tail (T). Thus, H and T are the only outcomes of this random experiment. The set of all possible outcomes is known as the sample space, S. Accordingly, when a fair coin is tossed once, the sample space or universal set is S = {H, T}. When a fair die is tossed once, the sample space, S = {1, 2, 3, 4, 5, 6}.

Event Any subset of the sample space is called an event. Thus, rolling a 4 is an event when a fair die is rolled once. In that case, the event E = {4} is the subset of the sample space. In the same way, getting a head is an event when a fair coin is tossed once. Getting a tail is another event.

Simple and compound events If the event has a single outcome, you have a simple event. For example, rolling a six when a fair die is tossed once is a simple event: E = {6}. Or if you toss a fair coin once, the outcome is a simple event.

If the event contains two or more elements, you have a compound event. For

example, when you toss a fair coin and a fair die together, the outcome is a compound event. In this case, the events include {H, 1}, {H, 2}, {T, 6} and so on.

Equally Likely Events

How will your class select a class prefect in such a way that every student in the class is given an equal chance of becoming the class prefect? Discuss this with your classmates.

For example, to give every student in the class an equal chance, each student could write his or her name on a piece of paper. These are collected in and put in an opaque container. The teacher could then choose a paper at random. In this way, everyone in the class will have the same chance of becoming the class prefect.

If a fair die is tossed once, getting 1, 2, 3, 4, 5 or 6 are equally likely events. This is because all of them have got an equal chance of showing up. If all the events in a random experiment have an equal chance of occurring, you have a set of equally likely events. In this case, the probability that the event E will occur, denoted by P(E), is given by P(E) = n(E)____ n (S) , where S is the sample space. Frequently terms like at random, fair, unbiased are hints that the events under discussion are equally likely.

Example 8.1

A bag contains 26 identical balls labelled A to Z. If a ball is chosen at random from the bag, calculate the probability that it is a vowel.

Solution

Since items are being chosen at random, you are being told that you have equally likely events. Thus, the formula to use is P(E) = n(E)____ n (S). The event under consideration is E = {vowels} and the sample space S = {letters in the English alphabet}.

Thus, n(E) = n{a, e, i, o, u} = 5 and n(S) = n{English alphabet letters} = 26 .

Thus, P(E) = n(E)____ n (S) = 5/26.

Example 8.2

Identical cards in a box are numbered from 1 to 40. A card is chosen at random.

Calculate the probability that a prime number is chosen.

Solution

The sample space, S = {1, 2, 3… 40}. Thus, n(S) = 40.

Event E = {prime numbers} = {2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37}.

Thus, P(E) = n(E)____ n (S) = 12/40 = 2/5 = 0.4 = 40 % .

Example 8.3

A fair die was rolled once. Find the probability that a multiple of 3 was rolled.

Solution

The sample space, S = {1, 2, 3, 4, 5, 6} and the event C = {multiple of 3} = {3, 6}.

Thus, P(C) = n(C)____ n (S) = 2/6 = 1/3.

Mutually Exclusive Events

Two events C and F are mutually exclusive if they cannot occur at the same time.

For example:

1. When a fair die is tossed once, rolling a 1 and a 2 simultaneously are mutually exclusive events. That is, you can roll a 1 or a 2 but not both at the same time.

2. If there are red and blue identical balls in a bag and a single ball is picked at random, picking a red ball and picking a blue ball are mutually exclusive events.

3. The event that your mathematics teacher will come to school today and the event that they will not come to school today are mutually exclusive events.

These can never happen simultaneously, ie, your teacher cannot be both in school and not in school.

You can remember that in sets, n(C ∪ F) = n(C) + n(F) − n(C ∩ F). However, if C and F are mutually exclusive events, then C and F cannot occur at the same time.

Hence, n(C ∩ F) = 0.

Therefore, n(C ∪ F) = n(C) + n(F).

⟹ n(C ∪ F)_ n(S) = n(C)_ n(S) + n(F)_ n(S) ∴ P(C ∪ F) = P(C) + P(F).

This is the addition law of probability (also called the AND rule).

Independent Events

Two events C and F are independent, if the probability that C will happen does not affect the probability that F will happen. In such a case, both C and F can happen at the same time and the probability of this is denoted by P(C ∩ F) = P (C) × P(F).

This is the multiplication law of probability (also called the OR rule).

When you toss a fair coin twice, the outcomes of the first and second tosses will be independent. This is because the outcome of the first toss has no influence on the second outcome. Making sure that events are independent is important in decision-making. It is for this reason that in examinations the sitting arrangements and invigilators are employed to make the performance of the candidates as independent as possible. It is for the same reason that during a 100-metre race, competitors are not allowed to steal start or tracks.

The probability of an event and its complement Another close relative of probability is sets. You will have realised that the sample space is a universal set and the events are subsets of this sample space. If U is the universal set, A is a set and A′ is the complement of A, then:

A ∪ A′ = U ⟹ n(A) + n(A′) = n(U) ⟹ n(A)____ n(U) + n(A′)____ n(U) = n(U)____ n(U) ⟹ P(A) + P(A′) = 1.

∴ P(A) = 1− P(A′).

What conclusion can you draw from this? It means that if the probability of an event is known, you can easily calculate the probability of its complement. For

example, the probability that it will rain today is 0.1. Therefore, the probability that it will not rain today is 1 – 0.1 = 0.9.

Example 8.4

Ato tossed a fair die once. Calculate the probability that he got either 3 or 6.

Solution

For a fair die, the sample space, S = {1, 2, 3, 4, 5, 6}. The event T = {3} and Q = {6}. Since events T and Q cannot happen at the same time, they are mutually exclusive events. Thus, you must apply the addition law of probability.

That is, P(Q or T) = P(Q ∪ T) = P(Q) + P(T) = n(Q)____ n(S) + n(T)____ n(S) = 1/6 + 1/6 = 2/6 = 1/3.

Thus, the probability that he got either 3 or 6 is 1/3.

Example 8.5

Show that the total probability of all the possible outcomes is 1, when a fair:

1. coin is tossed once.

2. die is tossed once.

Solution

1. When a fair coin is tossed once, the possible outcomes are a head (H) or a tail (T). The probability of getting a head is half and the probability of getting a tail is also half; and half plus half is one. Thus, the total probability of all the possible outcomes in this case is one. That is, P(H ∪ T) = P(H) + P(T) = n(H)____ n(S) + n(T)____ n(S) = 1/2 + 1/2 = 2/2 = 1

2. Similarly, if P(1), P(2), and P(3) represent the probability of getting 1, 2 and 3 respectively, then the sum of all the possible outcomes when a fair die is tossed once is given by:

P(1) + P(2) + P(3) + P(4) + P(5) + P(6) = 1_ 6 + 1_ 6 + 1_ 6 + 1_ 6 + 1_ 6 + 1_ 6 = 1.

Note that all these outcomes are mutually exclusive.

Example 8.6

The probability that it will rain today is 0.3 and the probability that it will rain tomorrow is 0.6. Assume that these events are independent. Find the probability that:

1. it will not rain today;

2. it will not rain tomorrow;

3. it will rain today and tomorrow.

Solution

1. If the event that it will rain today is denoted by R then the event that it will not rain today is denoted by R′. Hence, P(R′) = 1− P(R) = 1 − 0.3 = 0.7.

Thus, the probability that it will not rain today is 0.7.

2. The probability that it will not rain tomorrow is given by 1 − 0. 6 = 0.4.

3. Let P(R) denote the probability that it will rain today and P(T) denote the probability that it will rain tomorrow. Since R and T are independent events:

P(R ∩ T) = P(R) × P(T) = 0.3 × 0. 6 = 0.18. Therefore, the probability that it will today and tomorrow is 0.18.

Example 8.7

The probabilities that Arango and Kabenla will score a goal during a match are 0.18 and 0.72 respectively. Assume that these events are independent. If both of them play, what is the probability that:

1. both of them will score;

2. only one of them will score;

3. at least one of them will score.

Solution

Let P(A) = 0.18 be the probability that Arango will score a goal and P(K) = 0.72 be the probability that Kabenla will score a goal.

Then P(A′) = 1 − 0.18 = 0.82 is the probability that Arango will not score a goal and P(K′) = 1 − 0.72 = 0.28 is the probability that Kabenla will not score a goal.

A and K are independent events so we can apply the multiplication law of probability.

1. P (both of them will score) = P(A ∩ K) = P(A) × P(K) = 0.18 × 0.72 = 0.1296.

2. There are two possibilities or events. That is, the event that Arango will score but Kabenla will not score OR the event that Arango will not score but Kabenla will score. These are mutually exclusive.

Note that in probability OR goes with addition or union and AND goes with multiplication or intersection.

P (only one of them will score) = P(A ∩ K′)+ P(A′ ∩ K) = P(A) × P(K′) + P(A′) × P(K) = 0.18 × 0.28 + 0.82 × 0.72 = 0.0504 + 0 . 5904 = 0.6408

3. If at least one of them will score, then there are three events which are mutually exclusive: either Arango scores and Kabenla does not score or Arango does not score but Kabenla scores or both of them score. Consequently, the probability that at least one of them will score is given by;

P (at least one of them will score) = P(A ∩ K′)+ P(A′ ∩ K)+ P(A ∩ K) = P(A) × P(K′) + P(A′) × P(K) + P(A) × P(K) = 0.18 × 0.28 + 0.82 × 0.72 + 0.18 × 0.72 = 0.0504 + 0 . 5904 + 0.1296 = 0 . 7704 Alternatively, P (none will score) + P (at least one will score) = 1.

Thus, P (at least one will score) = 1 – P (none will score) = 1 − P(A′ ∩ K′) = 1 − [P(A′) × P(K′)] = 1 − (0.82 × 0.28) = 1 − 0.2296 = 0.7704.

When there are more than two events, this route is the best, the shortest.

Dependent Events

If two events cannot occur at the same time, they are mutually exclusive.

Consequently, if two events are not mutually exclusive, they can happen at the same time. In that case, they could be either dependent or independent. If they are independent, the probability that one will occur will not affect the probability that the other will occur. However, if they are dependent, then the probability that the first event will occur will determine the probability that the second event will occur.

Consider this scenario. There are 39 cards in a box. Only three of them are black.

In a class of nine boys and six girls, anyone who picks a black card at random is given Gh¢50.00. What conclusions can you draw if all the boys are allowed to choose a card before the girls and replacement is not allowed?

If you thought that the competition was unfair to the girls in the class, you are right. This is a classic example of a series of events which are dependent on each other.

If two events C and F are dependent on each other, then you can find the probability that C will occur given that F has already taken place. That is, if two events C and F are dependent, a condition can be placed on F while finding the probability that C will occur. This is called conditional probability. The probability that C will occur given that F has already taken place is denoted by P(C|F).

Now, P(C|^(F)) = P(C ∩ F)_______ P(F) . This is the formula for conditional probability.

We can rearrange this to give us another formula. This formula will help you to calculate the probability of events you will meet when drawing identical items at random without replacement. That is, P(F) × P(C|F) = P(C ∩ F).

In other words, the probability that events C and F will occur is product of the probability that F will occur and the probability that C will occur given that F has occurred.

Example 8.8

There are 6 red and 3 blue identical balls in a box. If two balls are drawn at random, one after the other without replacement, calculate the probability that:

1. both of them are red;

2. both balls have the same colour;

3. both balls have different colours.

Solution

n(red) = 6 and n(blue) = 3 and n(balls) = 9.

If P( R₁) is the probability that the first ball drawn is red and P( R₂) is the probability that the second ball drawn is red, then:

1. P(both are red) = P( R₁ ∩ R₂) = P( R₁) × P(R₂| ᴿ1) = 6/9 × 5/8 = 5/12.

Since the balls are drawn at random without replacement, P( R₁) = n(red)______ n(balls) = 6/9.

After the first red ball has been taken out, there are only five red balls out of the eight balls left in the box. Thus, the probability that the second ball is red is 5/8.

P(R₂| ᴿ1) represents the probability that the second ball is red, given that the first ball is red.

2. (ii) P(both balls have the same colour) = P(both are red) + P(both are blue) = P(R₁ ∩ R₂) + P(B₁ ∩ B₂) = P( R₁) × P(R₂) + P( B₁) × P(B₂) = 6_ 9 × 5_ 8 +3_ 9 × 2_ 8 = 30_ 72 + 6_ 72 = 36_ 72 = 1_ 2

3. (iii) P(both balls have different colours) = P(R₁ ∩ B₂) + P(B₁ ∩ R₂) = P( R₁) × P(B₂) + P( B₁) × P(R₂) = 6_ 9 × 3_ 8 + 3_ 9 × 6_ 8 = 18_ 72 + 18_ 72 = 36_ 72 = 1_ 2 Tree diagrams can be used to solve this problem (see below). Visit You Tube to find a video that will help you solve this problem using tree diagrams. Here is a link you could use, but there are plenty more available, https://www.youtube.com/watch?v=PYEvSuz1Dxo 1ˢᵗpick 2ⁿᵈpick Outcome Probability 5/8 R₂ R₁∩ R₂ 6_ 9 × 5_ 8 = 30_ 72 R₁ 6_ 9 3_ 8 B₂ R₁∩B₂ 6_ 9 × 3_ 8 = 18_ 72 3_ 9 6/8 R₂ B₁∩R₂ 3_ 9 × 6_ 8 = 18_ 72 B₁ 2_ 8 B₂ B₁∩ B₂ 3_ 9 × 2_ 8 = 6_ 72

Example 8.9

Kwam tosses a fair die once. Find the probability that the outcome:

1. is an odd number, given that it is a prime number;

2. is a prime number, given that it is odd.

Solution

Let E be the event that a prime number is chosen and F be the event that an odd number is chosen. The sample space S = {1, 2, 3, 4, 5, 6}, E = {2, 3, 5} and F = {1, 3, 5}.

Thus, n(F ∩ E) = n{3, 5} = 2.

1. P(it is odd|it is a prime number) = P(F|ᴱ) = n(F ∩ E)_______ n(E) = 2/3 ⋅

2. P(it is a prime number|it is odd) = P(E|^(F)) = n(F ∩ E)_______ n(F) = 2/3 ⋅

Example 8.10

The results of an election are shown in the table below. Use it to answer the following questions.

NCP ACP

Female votes 81 72 Male votes 108 99

1. If a voter is female, find the probability that she voted for ACP.

2. Find the probability that a voter voted for NCP, given that the voter is a male.

Solution

Let F be the event that a voter is a female, M be the event that the voter is a male, N be the event that the voter voted for NCP and A be the event that the voter voted for ACP. Now let us find the totals for the number of males, females and NCP and ACP voters.

NCP ACP Total

Females 81 72 1 5 3 Males 108 99 207 Total 189 171 3 6 0

1. P (voted for ACP|voter is female) = P(A|^(F)) = n(A ∩ F)_______ n(F) = 72/153 = 8/17 ⋅

2. P (voted for NCP|voter is male) = P(N|ᴹ) = n(N ∩ M)_______ n(M) = 108/207 = 12/23 ⋅

1. Given that P(C) = 0.27 and P(C ∪ F) = 0.81, find P (F), if the events C and F are:

a. mutually exclusive;

b. independent events.

2. A lottery machine has two compartments. The first compartment contains 3 red and 4 violet identical balls. The second compartment has 3 violet and 4 yellow balls. When a button is pressed, one ball falls from the first compartment and another one from the second compartment at random.

Calculate the probability that:

a. one is red and one is violet;

b. both of them are violet;

c. one is yellow and one is violet.

3. The probabilities that three strikers, A, B and C, will score a goal during any match are 1/2, 1/3 and 1/4 respectively. We can assume that these are independent events. If they all play a match, calculate the probability that:

a. only one of them scores a goal;

b. only two of them score a goal;

c. at least one of them scores a goal.

4. Two fair dice are tossed once. Find the probability that:

a. both numbers are prime;

b. sum of the numbers is 9;

c. product of the numbers is 6.

5. a. Find the probability that when a fair die is tossed once a number less than five is obtained, given that the toss resulted in an odd number.

b. If the toss of a fair die resulted in a number less than five, find the probability that it was an odd number.

6. Of the 6 0 students in 1 Language, 18 boys read Twi and 9 of the boys do not read Twi. In addition, 20 girls read Twi and 13 girls do not read Twi.

a. If a girl is chosen at random from the class, find the probability that she does not read Twi;

b. If a student reads Twi, find the probability that the student is a boy.

7. There are 5 blue and 4 white identical balls in a box. If two balls are drawn from the box at random, one after the other, calculate the probability that both balls have different colours, if replacement:

a. is allowed;

b. is not allowed.

Choose one or more of the topics below to carry out your own mini- project or choose a mini-project of your own which is important to you, your school or your local region.

8. You have been tasked by a micro finance institution to investigate the number of people in your community who save with microfinance institutions on a daily basis.

Conduct research and present your work to the micro finance institution.

9. A marketing agent has tasked your club to produce a report on the different languages spoken in your area. Conduct a small survey and use it to write your report.

10. According to health experts, fizzy drinks cause health problems.

Conduct a mini-project to determine whether students in your school are at risk.

11. Measure the height and mass of 40 people in your school.

Calculate the average body mass index.

Based on the average body mass index, what conclusions would you draw?

What recommendations would you give?

12. Drug abuse includes taking medication without a doctor’s prescription.

Provide the school authorities with reliable data that can help them to make well-informed decisions concerning this type of drug abuse.