S1181 Assignment 4 S24 (1)

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Apr 3, 2024

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STAT 1181 Assignment 4 (33 marks) Due March 21, 2024 1. A bag contains three red balls, four green balls, and five blue balls. To play the game, you pay $8 to randomly draw a ball from the bag. If a red ball is drawn, you win $12. If a green ball is drawn, you win $9. If a blue ball is drawn, you win nothing. Let X be the amount you win or lose per game on your $8 bet. (7 marks) a) What is the probability that you win on a single play of the game? (1 mark) Answer:- favorable outcome/total ball 7/12 is the probability of winning. b) Construct the probability distribution for the random variable X . (3 marks) Answer:- Probability of winning $4 (drawing a red ball): 3/12. Probability of winning $1 (drawing a green ball): 4/12. Probability of winning $8 (drawing a blue ball): 5/12 c) Find the expected value of the random variable X and interpret this value in the context of the question. (3 marks) Answer:- The expected value of X is -$2.00. 2. It is reported that 30% of adults are overweight. A random sample of eight adults was drawn across the country to investigate this. (9.5 marks) a) Does this process of selecting a random sample of eight adults satisfy the “independent trials” requirement to be a binomial experiment? Justify your answer in the context of the question. (2 marks) Answer:- Yes, selecting a random sample of eight adults generally satisfies the “independent trials” requirement , assuming the population is large enough that selecting one adult does not significantly affect the probability of being overweight for the others. b) Find the mean of the binomial random variable and interpret this value in the context of the question. (2 marks) Answer:- Mean of the Binomial random variable is 2.4(n * p = 8 * 0.3 = 2.4). We can expect 2.4 out of 8 adults in the sample to be overweight. c) What is the probability that at least one but less than four adults are overweight in the sample? Give your answer to four decimal places. (2 marks) Answer:- P(X = 1),P(X = 2),P(X = 3) using using binomial formula The probability that at least one but fewer than four adults are overweight in the sample is 0.7482. d) Suppose one hundred adults are randomly selected. What is the probability that more than 35% of them are overweight? (3.5 marks) Answer:- z-score = 35.5(0.5 adding for threshold for more accurate correction as n is large) 1– CDF(Z) = 0.1150. The probability 0.1150 ,that more than 35% of 100 randomly selected adults are overweight.

3. A recent study found that the average lifetime of smartphones is 27 months. Assume the population standard deviation is 5.8 months. In a random sample of 32 smartphones, what is the probability that the sample mean lifetime would be more than 24 months but less than 29 months? (4.5 marks) Answer:- Calculating standard error = 5.8/√32 Calculating Z 1 for 24 months = 24-27/ SE Calculating Z 2 for 29 months = 29-27/ SE P(24 < x < 29) = 0.9727 is probability that sample mean will fall within the specified range. 4. The average credit score of 150 randomly selected Canadians is 660. Assume the population standard deviation is 85. (6 marks) a) Define the parameter of interest. (1 mark) Answer:- average credit score of Canadians. b) Find the 88% confidence interval for the parameter. Check the required condition(s) to validate the interval. (4 marks) Answer:- mean = 660 Population standard deviation = 85 Sample size = 150 Confidence level = 88% So, z score for 88% confidence level = 1.5548. Standard Error = 85/ √ 150. Margin Error = ME = 1.5548 * SE Confidence interval: mean + ME or mean - ME =649.21,670.79. c) Interpret the interval in the context of the question. (1 mark) Answer :- I am 88% confidence that average credit score of Canadians falls between 649.21 and 670.79. 5. In a survey of 300 randomly selected smartphone users, 240 reported that they use Android operating system. (6 marks) a) Describe the parameter of interest. (1 mark) Answer:- The proportion of all smartphone users who use the Android operating system.

b) Obtain a 92% confidence interval for the parameter. Check the required condition(s) to validate the interval. (4 marks) Answer:- sample proportion = n = 240/300 = 0.8. z-score for 92% of confidence level is 1.75. Standard error = √(p(1-p)/n) = √(0.8(0.2)/300). Margin error = ME = 1.75 * SE CI = p + ME or p – ME = 0.76 to 0.84. c) Interpret the interval in the context of the question. (1 mark) Answer:- I am 92% of confident that probability of android smartphone users lies between 76% and 84%. 1

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