IS Project -Suraj Mishra .docx

INFERENTIAL STATSTICS Project Report By – Suraj Mishra

Contents - Problem 1 - An independent research organization is trying to estimate the probability that an accident at a nuclear power plant will result in radiation leakage. The types of accidents possible at the plant are, fire hazards, mechanical failure, or human error. The research organization also knows that two or more types of accidents cannot occur simultaneously. Problem 2 - Grades of the final examination in a training course are found to be normally distributed, with a mean of 77 and a standard deviation of 8.5. Problem 3- E-news Express, an online news portal, aims to expand its business by acquiring new subscribers. With every visitor to the website taking certain actions based on their interest, the company plans to analyze these actions to understand user interests and determine how to drive better engagement. The executives at E-news Express are of the opinion that there has been a decline in new monthly subscribers compared to the past year because the current web page is not designed well enough in terms of the outline & recommended content to keep customers engaged long enough to make a decision to subscribe.

Problem 1 - An independent research organization is trying to estimate the probability that an accident at a nuclear power plant will result in radiation leakage. The types of accidents possible at the plant are, fire hazards, mechanical failure, or human error. The research organization also knows that two or more types of accidents cannot occur simultaneously. According to the studies carried out by the organization, the probability of a radiation leak in case of a fire is 20%, the probability of a radiation leak in case of a mechanical 50%, and the probability of a radiation leak in case of a human error is 10%. The studies also showed the following.  The probability of a radiation leak occurring simultaneously with fire is 0.1%.  The probability of a radiation leak occurring simultaneously with a mechanical failure is 0.15%.  The probability of a radiation leak occurring simultaneously with a human error is 0.12%. On the basis of the information available, answer the questions below: 1.1 What are the probabilities of a fire, a mechanical failure, and a human error respectively? 1.2 What is the probability of a radiation leak? 1.3 Suppose there has been a radiation leak in the reactor for which the definite cause is not known. What is the probability that it has been caused by: a) a fire? b) a mechanical failure? c) a human error? Solution – Defining the events F = Fire M = Mechanical Error H = Human Error R = Radiation leak

N = No accident Given Probabilities: P(R/F) = 0.2 P(R/M) = 0.5 P(R/H) = 0.1 P (R ∩ F) = 0.001 P (R ∩ M) = 0.0015 P (R ∩ H) = 0.0012 1.2 What are the probabilities of a fire, a mechanical failure, and a human error respectively? P(F) = P (P (R ∩ F)/ P(R/F) = 0.001/0.2 = 0.005 P (M) = P (P (R ∩ M) / P(R/M) = 0.0015/0.5 = 0.003 P(H) = P (P (R ∩ H) / P(R/H) = 0.0012/0.1 = 0.012 1.2 What is the probability of a radiation leak? Since the types of accidents possible at the plant are fire hazards, mechanical failure, or human error. = P(N) = 1 – (0.005 + 0.003 + 0.012) = 0.98 P(R/N) = 0 P (R ∩ N) = P(R/N) P(N) = 0 By Probability Theorem: P(R) = P (R ∩ F) + P(R ∩ M) + P (R ∩ H) + P (R ∩ N) P(R) = 0.001+0.0015+0.0012+0 P(R) = 0.0037 1.3 Suppose there has been a radiation leak in the reactor for which the definite cause is not known. What is the probability that it has been caused by: a) a fire? b) a mechanical failure?

c) a human error? = If there has been a radiation leak in the reactor for which the definite cause is not known. The probability that it has been caused by a) The probability of a fire radiation is: P(F/R) = P (P (R ∩ F) / P(R) = 0.001 / 0.0037 = 0.270 b) The probability of the mechanical failure radiation leak P(M/R) = P (P (R ∩ M) / P(R) = 0.0015 / 0.0037 = 0.405 c) The probability of the Human Error radiation leak P(H/R) = P (P (R ∩ H) / P(R) = 0.0012 / 0.0037 = 0.324 Problem 2 Grades of the final examination in a training course are found to be normally distributed, with a mean of 77 and a standard deviation of 8.5. Based on the given information answer the questions below. 2.1 What is the probability that a randomly chosen student gets a grade below 85 on this exam? 2.2 What is the probability that a randomly selected student score between 65 and 87? 2.3 What should be the passing cut-off so that 75% of the students clear the exam? 2.1 What is the probability that a randomly chosen student gets a grade below 85 on this exam? = To find the probability that a randomly chosen student gets a grade below 85, we need to calculate the cumulative probability up to the value of 85. Using the Z -score formula: Z = (X - μ) / σ Where: X = the value we want to find the probability for (85 in this case) μ = the mean (77) σ = the standard deviation (8.5)

Z = (85 -77) /8.5 Z = 0.941176 Now, we can use a standard normal distribution table or a calculator to find the cumulative probability corresponding to the Z-score of 0.941176. From the standard normal distribution table, the cumulative probability (area under the curve) for a Z-score of 0.941176 is appr 0.8264. Therefore, the probability that a randomly chosen student gets a grade below 85 is approximately 0.8264 or 82.64%. 2.2 What is the probability that a randomly selected student score between 65 and 87? To find the probability that a randomly selected student scores between 65 and 87, we need to calculate the cumulative probability for both values and subtract the smaller probability from the larger probability. Using the Z-score formula: For 65: Z1 = (65 – 77) / 8.5 Z1 = -1.411765 For 87: Z2 = (87 – 77) / 8.5 Z2 = 1.76471 Using the standard normal distribution table or a calculator, we find the cumulative probabilities corresponding to Z1 and Z2. For Z1 = -1.411765, the cumulative probability is approximately 0.0793. For Z2 = 1.176471, the cumulative probability is approximately 0.8790. The probability of scoring between 65 and 87 is the difference between the cumulative probability: 0.8790 – 0.0793 = 0.7997 Therefore, the probability that a randomly selected student score between 65 and 87 is approximately 0.7997 or 79.97 %.

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