Questions and Answers>Mathematics_Probability Mathematics Practice Problems.

Problem 1 PART 1.1 ) The probability that a randomly chosen player would suffer an injury = 0.6170 Number of player who would suffer an injury = 145 Total number of players = 235 The probability that a randomly chos ...

Problem 1 PART 1.1 ) The probability that a randomly chosen player would suffer an injury = 0.6170 Number of player who would suffer an injury = 145 Total number of players = 235 The probability that a randomly chosen player would suffer an injury = 145 / 235 = 0.6170 PART 1.2 ) The probability that a randomly chosen player is a forward or a winger = 0.4511 Number of player who are a forward or a winger = 77 + 29 = 106 Total number of players = 235 The probability that a randomly chosen player is a forward or a winger = 106 / 235 = 0.4511 PART 1.3 ) The probability that a randomly chosen player plays in a striker position and has a foot injury = 0.1915 Number of player who plays in a striker position and has a foot injury = 45 Total number of players = 235 The probability that a randomly chosen player plays in a striker position and has a foot injury = 45 / 235 = 0.1915 PART 1.4 ) The probability that a randomly chosen injured player is a striker = 0.3103 Number of injured player who are striker = 45 Number of injured players = 145 The probability that a randomly chosen injured player is a striker = 45 / 145 = 0.3103 PART 1.5 ) The probability that a randomly chosen injured player is either a forward or an attacking midfielder = 0.5517 Number of injured player who are either a forward or an attacking midfielder = 56 + 24 = 80 Number of injured players = 145 The probability that a randomly chosen injured player is either a forward or an attacking midfielder = 80 / 145 = 0.5517 Problem 2 Probability of radiation leak happening, given that fire is there is .20. this is a Baye's theorem situation. According to Baye's theorem probability of an event A occurring given that B has already occurred is given by P(A/B)=P(AnB)/P(B) Let F be the event denoting fire M - mechanical failure H- human failure R - radiation leak From the problem P(R/F) =0.2(20%) P(R/M)=0.5 P(R/H)=0.1 Now given that probability of a fire and radiation leak occurring together is .0010 This means the probability of intersection of these events is .0010 P(FnR)=.0010 Like this P(MnR)=.0015 P(HnR)=.0012 a) Now we have P(R/F)=0.2 P(RnF)/P(F)=0.2 But we already know that P(RnF)=.0010 S P(F)=P(RnF)/P(R/F) =.0010/0.2 =.005 Like this P(M)=P(MnR)/P(R/M) =.0015/.5 =.003 P(H)=P(HnR)/P(R/H) =.0012/.1 =.012 b) Now probability that radiation leak occurred due to fire. So a radiation leak has happened and what is the probability that it happened due to fire Which is same as probability of fire occuring given that radiation leak is there Which is P(F/R) Now according to Baye's theorem P(A/B)=P(B/A)*P(A)/P(B) So P(F/R)=P(R/F)*P(F)/P(R) We have to find P(R) We have P(A/B)*P(B)=P(AnB) Here events F,M and H are mutually exclusive events. Tgey never occur together and if a radiation occurs then there is 100% probability that one of these occurred. So P(F/R)+P(M/R)+P(H/R)=1 Now we have P(F/R)=P(FnR)/P(R) P(F/R)+P(M/R)+P(H/R)= P(MnR)/P(R)+P(HnR)/P(R)+P(FnR)/P(R)=( P(MnR)+P(HnR) +P(FnR) )/P(R)=1 (.0010+.0015+.0012)/P(R)=1 P(R)=.0038 Now P(F/R)=P(FnR)/P(R) =.0010/.0037=.27 P(M/R)=P(MnR)/P(R) =.0015/.0037=.4054 P(H/R)=P(HnR)/P(R) =.0012/.0037=.3243 Step-by-step explanation c) we have already found probability of Radiation leak in b section P(R)=.0037 Problem 4: Part 4.1 What is the probability that a randomly chosen student gets a grade below 85 on this exam? 0.8264 Part 4.2 What is the probability that a randomly selected student scores between 65 and 87? 0.80173 Part 4.3 What should be the passing cut-off so that 75% of the students clear the exam? 82.695 Problem 5: 5.1 Yes, Zingaro is justified in thinking so. The confidence interval for the population mean of the unpolished stones is 134.11 +- 1.96*(33.04/sqrt(75)). This confidence interval does not contain 150, which is the minimum value of Brinell's hardness index required for optimum printing. Therefore, we can conclude that the population mean of the unpolished stones is not adequate for optimum printing. Explanation - The Zingaro stone printing company specializes in printing images or patterns on polished or unpolished stones. In order to achieve the optimum level of printing quality, the stone surface must have a Brinell's hardness index of at least 150. Recently, Zingaro received a batch of polished and unpolished stones from a client. Based on earlier experiences with this particular client, Zingaro has reason to believe that the unpolished stones in this batch may not be suitable for printing. To test this hypothesis, we can use a confidence interval approach. The confidence interval for the population mean of the unpolished stones is 134.11 +- 1.96*(33.04/sqrt(75)). This confidence interval does not contain 150, which is the minimum value of Brinell's hardness index required for optimum printing. Therefore, we can conclude that the population mean of the unpolished stones is not adequate for optimum printing. Based on this analysis, Zingaro is justified in thinking that the unpolished stones in this batch are not suitable for printing. 5.2 The null hypothesis would be that there is no difference between the two means. To test this, we can use a two-sample t-test. The t-statistic is -5.5019 and the p-value is 2.2e-08. Since the p-value is less than 0.05, we can reject the null hypothesis and conclude that there is a difference between the two means. The mean hardness of the polished and unpolished stones is not the same. Explanation - A two-sample t-test is a statistical test used to compare the means of two groups. In this case, we are interested in whether the mean hardness of polished and unpolished stones is the same.