5 Week Modular Course In Statistics & Probability Strand 1
5 Week Modular Course in Statistics & ProbabilityStrand 1Module 5
Bernoulli TrialsBernoulli Trials show up in lots of places. There are 4 essential features:1. There must be a fixed number of trials, n2. The trials must be independent of each other3. Each trial has exactly 2 outcomes called success or failure4. The probability of success, p, is constant in each trialWhere do we see this occurring? tossing a coin looking for defective products rolling off an assembly line shooting free throws in a basketball gameWhenever we are dealing with a Bernoulli trial there is a discrete random variable X. Thisrandom variable needs to be identified because all probability questions will involvefinding the probability of different values of this variable. For example if you toss a coin ntimes, the random variable X could be the number of heads occurring in 3 tosses e.g. Xcan take on the values 0, 1, 2, 3.We will look at three different types of Problems:1. calculating the probability of first success after n repeated Bernoulli trials2. calculating the probability of k successes in n repeated Bernoulli trials3. calculating the probability until the kth success in n trials. Project Maths Development Team – Draft (Version 2)Module 5.1First Success After n Repeated Bernoulli TrialsA basketball player has made 80%, of his foul shots during the season. Assuming the shots areindependent, find the probability that in tonight's game he:(a)(b)misses for the first time on his fifth foul shotmakes his first basket on his fourth foul shot(c)makes his first basket on one of his first 3 foul shotsSolutionLet X the number of shots until the first missed shotLet Y the number of shots until the first made shot(a)Four shots made followed by a miss:[p 0.8, q 0.2][p 0.2, q 0.8]P(X 4) (0.8)4 (0.2) 0.08192(b)Three misses, then a shot made:P(Y 3) (0.2)3 (0.8) 0.0064first basket(c)1 miss, first basket2 misses, first basketP(Y 0) P(Y 1) P(Y 2) (0.8) (0.2)(0.8)1 (0.2)2 (0.8) 0.992 Project Maths Development Team – Draft (Version 2)Module 5.2
k Successes in n Repeated Bernoulli Trials: Binomial DistributionProblemProblemA die is tossed 10 times. What is the probabilityof getting four sixes ?A die is tossed n times. What is the probabilityof getting r sixes ?Solution1P(6) 6SolutionS,S,S,S P(not 6) Sr sucesses56F,F,F,F,F,F n r failuresThis is only one arrangement.F n But there are ways success can occur. r n r n r P(r successes) ( S ) (F ) r S S S S F F F F F F1 1 1 1 5 5 5 5 5 56 6 6 6 6 6 6 6 6 6This is only one arragement.How many arrangements overall?Now replace S with p and F with q.10! 10 10 C4 ways4! 6! 4 4 10 1 5 P(4 sixes in 10 goes) 4 6 6 6 n P(r successes) (p)r (q)n r r Project Maths Development Team – Draft (Version 2)Module 5.3Example 1A coin is tossed six times, what is the probability of getting four heads.?We can apply the Binomial Distribution to this question because:1.There must be a fixed number of trials, n2.3.The trials must be independent of each otherEach trial has exactly 2 outcomes called success or failure4.The probability of success, p, is constant in each trialThe Binomial Distrution n P(X r) (p)r (q)n r r p probability of successq 1 p probability of a failuren total no. of trialsr number of successes in n trialsSolutionLet X number of heads41 1 p 2 , q 2 2 6 1 1 15P(X 4) 0 2344 4 2 2 64 Project Maths Development Team – Draft (Version 2)Module 5.4
Graph of the Distribution0.35 6 33 (0.5) (0.5)3 6 51 (0.5) (0.5) 5 6 15 (0.5) (0.5) 1 6 60 (0.5) (0.5) 6 6 06 (0.5) (0.5) 0 0.25Probability 6 42 (0.5) (0.5) 4 6 24 (0.5) (0.5) 2 0.300.200.150.100.050123456No. of heads Project Maths Development Team – Draft (Version 2)Module 5.5Example 23In a game of chess against a particular opponent, the probability that Sean wins is .5He plays 6 games against his opponent. What is the probability that Sean will:(i)lose the second game and the 4th game and win the others?(ii)win exactly four games ?(iii)lose at least four games?Solution(i)The formula does not apply here it is P(w, l, w, l, w, w)3243 2 3 2 3 3P(w, l, w, l, w, w) 5 5 5 5 5 5 1562532In the next two parts a Binomial model is appropriate, where p , q and n 6.55Let X the number of games won4(ii)(iii)2 6 3 2 972P(X 4) 4 5 5 3125P(at least 4 losses) P(no more than 2 wins)P(X 2) P(X 0) P(X 1) P(X 2)061524 6 3 2 6 3 2 6 3 2 P(X 2) 0.1792 0 5 5 1 5 5 2 5 5 Project Maths Development Team – Draft (Version 2)Module 5.6
Example 320% of the items produced by a machine are defective. Four items are chosen at random.Find the probability that none of the chosen items are defective.SolutionLet X number of items that are not defective[p 0.8 (not defective), q 0.2 (defective)] 4 256 0.4096P(X 4) (0.8)4 (0.2)0 625 4 Sample Space Project Maths Development Team – Draft (Version 2)Module 5.7Example 4Five unbiased coins are tossed.(i)Find the probability of getting three heads and two tails.(ii)The five coins are tossed eight times. Find the probability of getting three heads andtwo tails exactly four times.Solution(i)1 1 p 2 , q 2 3 heads (and 2 tails) from 5 coinsLet X number of heads32 5 1 1 5P(X 3) 3 2 2 16(ii)The probabilities for this part of the question are got from part (i)Let X number of times, 3 heads (and 2 tails) occur511 p ,q 1616 4 times out of 8 tries44 8 5 11 P(X 4) 0 149 4 16 16 Project Maths Development Team – Draft (Version 2)Module 5.8
Example 5During a match Owen take a number of penalty shots. The shots are independent of each other and his(i)4.5Find the probability that Owen misses each of his four penalty shots(ii)Find the probability that Owen scores exactly three of his first four penalty shots(iii)If Owen takes ten penalty shots during the match, find the probability that he scoresat least eight of themprobability of scoring with each shot isSolution4 1 p 5 , q 5 41 p 5 , q 5 Let X number of missesLet Y number of scores(i)(ii)Misses 4 out of 4 shots40 4 1 4 1P(X 4) 4 5 5 625(iii)Scores 3 out of 4 shots31 4 4 1 256P(Y 3) 625 3 5 5 Scores at least 8 out of 10 shotsP(Y 8) P(Y 8) P(Y 9) P(Y 10)8291100 10 4 1 10 4 1 10 4 1 P(Y 8) 0 678 8 5 5 9 5 5 10 5 5 Project Maths Development Team – Draft (Version 2)Module 5.9Example 6 (HL)Ronald is St. Patrick's College best basketball shooter. He is a 70% free throw shooter.Therefore the probability of him scoring on a free throw is 0.7.What is the probability that Ronald scores his third free throw on his fifth shot?SolutionHis last throw has to be success as we stop when he has 3 free throws after 5 shots.Let X number of baskets scored2 baskets out of first 4 5th is a basket 4 22P(X 3) (0.7) (0.3) (0.7) 2 4 P(X 3) (0.7)3 (0.3)2 0.18522 2 Project Maths Development Team – Draft (Version 2)Module 5.10
Example 7What is the probability that Ronald above from St. Patrick's College scores his first free throw on his fifthshot? (This has now become an OL question)SolutionThe only possibility is F F F F SLet X number of shots until a score[p 0.3, q 0.7]P(X 4) (0.3)4 (0.7) 0.00567This is quite low, lower than the last answer because Ronald is quite a sharp shooter and you expect himto have his first score before the 5th shot.If the probability of him scoring was 20% what would you expect the probability to be that his first free scoreis on the fifth shot?Let X number of shots until a score[p 0.8, q 0.2]P(X 4) (0.8)4 (0.2) (0.8)4 (0.2) 0.08192 Project Maths Development Team – Draft (Version 2)Module 5.11Normal DistributionDiscrete data (golf scores, dice scores) are generally represented by bar charts.In a bar chart we compare the heights of the bars.Continuous data (height, weight, physical characteristics) are represented by histograms.In a histogram we compare the areas of the columns.FrequencyHistogram Showing the Height of SeedlingsHeight (mm)The histogram shows that a large quantity of the data is clustered at the centre. Project Maths Development Team – Draft (Version 2)Module 5.12
Probability AreaFrequencyHistogram Showing the Height of SeedlingsHeight (mm)If a seedling is chosen at random it has approximately a 77.88% chance of havinga height within the yellow area shown on the histogram i.e. between 12 mm and 26 mm Project Maths Development Team – Draft (Version 2)Module 5.13Sampling DistributionIf another batch of seedlings were taken the picture might look slightly different.It is likely that all batches will follow a common pattern with most of the dataclustered around the centre of the histogram. This pattern is common to mostmeasurements in nature. It peaks in the middle and tails at the beginning and end.To get a perfect model we would need to1. Increase the sample size to infinity2. Take measurements to an infinite number of decimal places3. Have the widths of the columns approach zeroThis is impossible to achieve. We can only create a mathematical model of it.This model is called the NORMAL DISTRIBUTION. Project Maths Development Team – Draft (Version 2)Module 5.14
What does a Normal Distribution curve look like?Mean, µ 19.62 Standard Deviation, σ 5.46 of seedlings2 (x µ )1If the graph of y e 2 σ2 is plotted for the seedling we would get the graph below.σ 2πµ 3σµ 2σµ σµµ σµ 2σµ 3σ Project Maths Development Team – Draft (Version 2)Module 5.15Normal Distribution to Standard Normal DistributionDifferent sets of data have different means and standard deviations but any that are normallydistributed have the same bell-shaped normal distribution type of curves.Normal Distribution CurveStandard Normal CurveIn order to avoid unnecessary calculations and graphing the scale a Normal Distribution curve isconverted to a standard scale called the z score or standard unit scale.Normal DistributionsStandard Normal Distributionµ 13µ 0σ 34σ 171013161922µ 278σ 12242–3254266278290302 Project Maths Development Team – Draft (Version 2)–2–10123314Module 5.16
Standard Normal Distribution1 12 z2If µ 0 and σ 1 we would plote2πThis graph gives the Standard Normal Graph with a standardised scale.Total area under the curveP( z ) µ 3σ 312π e1 z22 dz 1µ 2σ 2µ σ 1µ0µ σ1µ 2σ2µ 3σ3z scoresThe area between the Standard Normal Curveand the z axis between and is 1. Project Maths Development Team – Draft (Version 2)Module 5.17Empirical RuleEmpirical Rule Part 1About 68% of the area is within 1standard deviation of the mean.Empirical Rule Part 2About 95% of the area is within 2standard deviations of the mean.Empirical Rule Part 3About 99.7% of the area is within 3standard deviations of the mean.µ 3σ 3µ 2σ 268%µ σ 1µ0µ σ1µ 2σ2µ 3σ3z scores95%99.7% Project Maths Development Team – Draft (Version 2)Module 5.18
Standard Units (z – scores)x µσx is a data pointµ is the population meanz σ is the standard deviation of the populationz – scores define the position of a score in relation to the mean using the standard deviation as aunit of measurement.z – scores are very useful for comparing data points in different distributions.The z – score is the number of standard deviations by which the score departs from the mean.This standardises the distribution. Project Maths Development Team – Draft (Version 2)Module 5.19Why do we standardise?In the 2004 Olympics, Austra Skujte of Lithuania put the shot 16.4 meters, about 3 meters farther thanthe average of all contestants. Carolina Kluft won the long jump with a 6.78 m jump, about a metrebetter than the average. Which performance deserves more points for a heptathlon event?Long Jump Shot PutMean6.16 m13.29 mSD0.23 m1.24 mn2628Kluft6.78 m14.77 mSkjyte6.30 m16.40 mBoth won one event, but Kluft's shot put was secondbest, while Skujyte's long jump was seventh.(all contestants)SolutionStandardise the scores, the z scores can then be added together.KluftLong JumpShot PutTotal z scores for 2 events:6.78 m14.77 mKluft : 2.70 1.19 3.89Skjyte : 0.61 2.51 3.1214.77 13.29z score 6.78 6.16 2.70 1.190.231.24Skjyte6.30 m16.40 mz score 6.30 6.1616.40 13.29 0.61 2.510.231.24 Project Maths Development Team – Draft (Version 2)The z scores measure how far each result isfrom the event mean in standard deviation unitsModule 5.20
Reading z – values From TablesExample 1Using the tables find P(Z 1 31).12π z1 te 2 dt Pg. 37P(Z z) Pg. 36For a given z, the table gives–3–2–101231.31P(Z 1 31) can be read from the tables directlyP(Z 1 31) 0 9049 90.49% Project Maths Development Team – Draft (Version 2)Module 5.21Pg. 37Pg. 36Example 2Using the tables find P(Z 1 32)–3–2–1011.3223The table only gives value to the left of z, butP(Z z) is equal to 1 P(Z z)P(ZP(Z 1 32) 1 P(Z 1 32)the fact that the total area under the curveequals 1, allows us to use, P(Z z) 1 P(Z z)P(Z 1 32) 1 0 9066 0 0934 9.34%P(Z z)P(Z z)1 P(Z z)0 Project Maths Development Team – Draft (Version 2)zModule 5.22
Example 3–3–20–112Pg. 37Pg. 36Using the tables find P(Z 0 74).3–0.74The tables only work for positive values but asthe curve is symmetrical about z 0P(Z 0 74) P(Z 0 74)Both areas are the same and henceboth probabilities are equal as the curveis symmetrical about the mean, 0.P(Z 0 74) 1 P(Z 0 74)P(Z 0 74) 1 0 7704 0 2296 22.96%P(Z z)P(Z z)0–z Project Maths Development Team – Draft (Version 2)zModule 5.23–3–20–11–1.322Pg. 37Pg. 36Example 4Using the tables find P( 1 32 z 1 29)31.29–3–2–10121.293–3–2 –1–1.320123P( 1 32 z 1 29) Area to the Left of 1 29 Area to the left of 1.32 P(z 1 29) [1 P(z 1 32)] 0 9015 [1 0 9066] 0 8081 80.81% Project Maths Development Team – Draft (Version 2)Module 5.24
Example 5The amounts due on a mobile phone bill in Ireland are normally distributed with a mean of 53 and astandard deviation of 15. If a monthly phone bill is chosen at random, find the probability that theamount due is between 47 and 74.Solutionx µz1 σ47 53z1 15z 1 0 4x µσ74 53z2 15z2 1 4z2 v8–323–238 47 53–1 –0.4 068 741 1.4832983P( 0 4 Z 1 4)P( 0 4 Z 1 4) P(Z 1 4) [1 P(Z 0 4)]P( 0 4 Z 1 4) 0 9192 [1 0 6554]P( 0 4 Z 1 4) 0 5746 Project Maths Development Team – Draft (Version 2)Module 5.25Example 6The mean percentage achieved by a student in a statistic exam is 60%. The standard deviation of theexam marks is 10%.(i)What is the probability that a randomly selected student scores above 80%?(ii)(iii)What is the probability that a randomly selected student scores below 45%?What is the probability that a randomly selected student scores between 50% and 75%?(iv)Suppose you were sitting this exam and you are offered a prize for getting a mark which isgreater than 90% of all the other students sitting the exam?What percentage would you need to get in the exam to win the prize?Solution(i)x µ 80 60 2σ10P(Z 2) 1 P(Z 2)z P(Z 2) 1 0.9772 0.0228 2.28%(ii)30–340–250–160070180290330–340 45 50–2 –1.5 –1600701802903x µ 45 60 1.5σ10P(Z 1.5) P(Z 1.5) 1 P(Z 1.5)z P(Z 1.5) 1 0.9332 0.0668 6.68% Project Maths Development Team – Draft (Version 2)Module 5.26
(iii)x µσ50 60z1 10z1 1z1 x µσ75 60z2 10z2 1.5z2 P( 1 Z 1 5) P(Z 1 5) [1 P(Z 1)]30–3P( 1 Z 1 5) 0.9332 [1 0.8413]P( 1 Z 1 5) 0.7745(iv)40–250–160070 75 801 1.5 2903From the tables an answer for an area of 90% (0.9) 1.28 Z 1.28x µσx 601.28 x 72.8 marks10z 30–340–250–160070 72.8 801 1.28 2903 Project Maths Development Team – Draft (Version 2)Module 5.27Hypothesis TestingOften we need to make a decision about a population based on a sample.1. Is a coin which is tossed biased if we get a run of 8 heads in 10 tosses?Assuming that the coin is not biased is called a NULL HYPOTHESIS (H0)Assuming that the coin is biased is called an ALTERNATIVE HYPOTHESIS (H1)2. During a 5 minute period a new machine produces fewer faulty parts than an old machine.Assuming that the new machine is no better than the old one is called a NULL HYPOTHESIS (H0)Assuming that the new machine is better than the old one is called an ALTERNATIVE HYPOTHESIS (H1)3. Does a new drug for Hay-Fever work effectively?Assuming that the new drug does not work effectively called a NULL HYPOTHESIS (H0)Assuming that the new drug does work effectively called an ALTERNATIVE HYPOTHESIS (H1) Project Maths Development Team – Draft (Version 2)Module 5.28
Margin of Error for Population ProportionsA sample of 60 students in a school were asked to work out how much money they spent on mobilephone calls over the last week. If the mean of this sample was found to be 5 80.Can we say that the mean amount of money spent by the students in the school (population) was 5 80?The answer is no, (unless the sample size was the same as the population size), we can’t say forcertain.However we could say with a certain degree of confidence, if the sample was largeenough and representative then the mean of the sample was approximately equal tothe mean of the population.How confident we are is usually expressed as a percentage.We already saw (from the empirical rule) that approximately 95% of the area of a Normal Curve lieswithin 2 standard deviations of the mean.This means that we are 95% certain that the population mean is within 2 standard deviations of thesample mean. 2 standard deviations is our margin of error and the 2 standard deviations dependson the sample size.If n 1000 the percentage margin of error of 3%At 95% percent level of confidence1vMargin of Error nwhere n, is the sample size Project Maths Development Team – Draft (Version 2)Module 5.29Some Notes on Margin of Error As the sample size increases the margin of error decreases A sample of about 50 has a margin of error of about 14% at 95% level of confidence1 14.14%50A sample of about 1000 has a margin of error of about 3% at 95% level of confidence1 3.16%1000 The size of the population does not matter If we double the sample size (1000 to 2000) we do not get do not half the margin of error Margin of error estimates how accurately the results of a poll reflect the “true” feelingsof the population Project Maths Development Team – Draft (Version 2)Module 5.30
Example 1A survey is carried out on 900 randomly selected people and the result is that 40% are in favour of achange of government. The confidence level is cited as 95%.(i)Calculate the margin of error.(ii)The following month another survey was carried out on 900 randomly selected people to seeif there was a change in support for the government.The result is that 42% are now in favour of a change of government. State the null hypothesis.According to this new survey would you accept or reject the null hypothesis?Give a reason for your conclusion.Solution11 0.03 3%n900(i)Margin of Error (ii)Null hypothesisH0 : "There is no change in the support for the government"Accept H0 the null hypothesis.Reason : The result of the first survey was 40% with a margin of error of 3 or 3.The results of the second survey was 42% which is within or 3% of the first survey sothere is no need for the government to be concerned. Project Maths Development Team – Draft (Version 2)Module 5.31Example 2In a survey I want a margin of error of or 5% at 95% level of confidence.What sample size must I pick in order to achieve this?SolutionMargin of Error 0.051 0.05 n1( 0.05)2 n1n 0.0025n 400 Project Maths Development Team – Draft (Version 2)Module 5.32
Bernoulli Trials Bernoulli Trials show up in lots of places. There are 4 essential features: 1. There must be a fixed number of trials, n 2. The trials must be independent of each other
(prorated 13/week) week 1 & 2 156 week 3 130 week 4 117 week 5 104 week 6 91 week 7 78 week 8 65 week 9 52 week 10 39 week 11 26 week 12 13 17-WEEK SERIES* JOIN IN MEMBER PAYS (prorated 10.94/week) week 1 & 2 186.00 week 3 164.10 week 4 153.16 week 5 142.22 week 6 131.28 week 7 120.34
Week 3: Spotlight 21 Week 4 : Worksheet 22 Week 4: Spotlight 23 Week 5 : Worksheet 24 Week 5: Spotlight 25 Week 6 : Worksheet 26 Week 6: Spotlight 27 Week 7 : Worksheet 28 Week 7: Spotlight 29 Week 8 : Worksheet 30 Week 8: Spotlight 31 Week 9 : Worksheet 32 Week 9: Spotlight 33 Week 10 : Worksheet 34 Week 10: Spotlight 35 Week 11 : Worksheet 36 .
28 Solving and Graphing Inequalities 29 Function and Arrow Notation 8th Week 9th Week DECEMBER REVIEW TEST WEEK 7,8 and 9 10th Week OCTOBER 2nd Week 3rd Week REVIEW TEST WEEK 1,2 and 3 4th Week 5th Week NOVEMBER 6th Week REVIEW TEST WEEK 4,5 and 6 7th Week IMP 10TH GRADE MATH SCOPE AND SEQUENCE 1st Week
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Introduction to the SOC and the Tier 1 Analyst Role Operating System Security Analysing Advanced Threats Week 1 Week 2 Week 3 Week 4 Week 5 Incident Response . Certification Quiz Week 6 DUE 1st day of Week 8 Week 9 Week 12 Cyber Security Analyst Course Outline. Cyber Security Analyst Course Outline TECHNICAL REQUIREMENTS CAREERS Hardware and .
MMWR Week. Week 10: 3/29-4/4 3/1-3/7 Week 11: 3/8-3/14 Week 12: 3/15-3/21 Week 13: 3/22-3/28 Week 14: Week 15: 4/5-4/11 Week 16: 4/12-4/18 Week 17: 4/19-4/25 Week 18: 4/26-5/2 Week 19: 5/
