assignment#3

# assignment#3 - Elisa Chan Assignment#3 5.16 a X will be the...

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Unformatted text preview: Elisa Chan Assignment #3 5.16 a) X will be the length of time the bus is expected to be delayed and will be the values between 0 ( c) and 20 (d ). µ = (c+d)/2 µ = (0+20)/2 µ = 10 The expected length of time that a bus will be late is 10 minutes. b) The probability that the last bus on a given day will be more than 19 minutes late is 1 minus the probability that X is smaller than 19 minutes. So P(X>19) = 1‐ P(X<19) P(X<19) = (19‐0)/ (20‐0) = .95 1‐0.95 = 0.05 P(X>19) = 0.05 c) We can conclude that the director’s claim is wrong. This is because the probability of having to wait 19 minutes is 0.05, the probability of a rare/uncertain event. 5.42 a) X = maximum oxygen uptake and has normal distribution with a mean of 24.1 and standard deviation of 6.3. z = (x‐µ)/ σ We want to find P(x ≥ to 20). € b) Probability of a patient who exercises regularly with a maximum oxygen uptake of less than or equal to 10.5 = P (x ≤ 10.5) € c) The probability of a patient who exercises and has a maximum oxygen uptake of 10.5 is 0.0154 (part b). Because this number is less than 0.05, that indicates that this is a rare event meaning that it is not likely that the patient participates in sports or exercise programs. 5.44 a) µ = 37.9; σ = 12.4 Probability that x is between 40 and 50 milliseconds P( 40<x< 50) 40 − 37.9 x −υ )< < 50 − 37.9) 12.4 σ P (0.1694 < z < 0.9758) P (0 ≤ z ≤ 0.97) − P (0 ≤ z ≤ 0.17) 0.3340 − 0.0675 P( 0.2665 b) € Probability that x is < 30 milliseconds P(x<30) c) The probability that x is in the interval (µ ‐ (196)σ, µ _ (1.96)σ = 0.95 Therefore the required interval would be d) Assume 10% of the experimental subjects have startle responses above x0. This means that 10% of the experimental subjects have startle responses over 53.7968 ms. 5.50 The probability that the load is between 10 and 30 thousand pounds is 0.95. Therefore the standard deviation is 5.102 thousand pounds for the load distribution. 5.66 Driving distance Points are well within lines and no outliers. Driving distance is approximately normally distributed. Accuracy We can’t conclude that it’s approximately normally distributed because of the outliers and how the points are not well within the lines. Performance index We can’t conclude that it’s approximately normally distributed once again because of the outliers and the definitely out of the line border points. 5.80 a) The mean of x is 290. b) The standard deviation of x is 14.35. c) At x = 200.5, z score is ‐6.24. d) P( z ≤ ‐ 6.24) = 0.00003 (value is taken from the Normal table.) € 5.84 p= 1/3 µ = np = 150(1/3) = 50 Intervals lie within range so we use normal approximation. a) P(x > 75) = 1 – P(x ≤ 74) 74.5 − 50 1 – P(x ≤ 74) = 1 ‐ P ( z ≤ ) 5.7735 € 1‐P(z ≤ 4.24) € € 1‐.9997 = .0003 € Approximately 0.0003 probablity that more than half are victims of domestic abuse. b) P(x<50) approximate probability that less than 50 are victim of domestic abuse P(x<50) 49.5 − 50 )= 5.7735 P ( z < −0.866) P(z < = 0.3023 € c) No I would not expect to see fewer than 30 because of the normal distribution giving it a very low probability. 5.88 Intervals for µ +/‐ 3σ lie within range of 0 to 800. So normal probability distribution will be used. a) Probability that more than half will wait more than 20 minutes b) Probability that more than 85 will wait more than 20 minutes c) Probability that between 60 and 90 will wait more than 20 minutes 6,36 n = 45 µ = 18 σ = 5 Standard deviation of the sampling distribution : a) Probability that sample mean FNE score is more than 17.5 : Find standard normal z value for mean of 17.5 b) Probability for FNE score between 18 and 18.5 Z value for sample mean 18 Z value for sample mean 18.5 P ( 0 < z < 0.67) = 0.2486 c) Probability that FNE score is less than 18.5 P(z < 0.67) = 0.7486 6,40 n = 100; µ = \$ 2.835; σ = \$ 0.15 Central limit theorem tells us that the sample mean cost for regular unleaded fuel is approximately normally distributed; also the means are the same for the sampling distribution and the sampled population. a) Mean of the sampling distribution = µ = 2.835 Standard deviation of the sampling distribution b) Z value for sample mean 2.84 and 2.86 P ( 0.33 < z < 1.67) = 0.5+0.4524 – 0.5+0.1293 = 0.9525 – 0.6293 = 0.3232 Probability of 0.3232 that the sample has a mean fuel cost between 2.84 and 2.86 dollars. c) Z value for sample mean 2.855 P ( z > 1.33) = 0.5 – 0.4082 = 0.0918 The probability that the sample has a mean fuel cost over 2.855 is 0.0918. d) If we doubled n to 200, we would see that the sampling distribution of them mean will be more peaked and around the true population mean of 2.835. Also the spread of the distribution will lessen. If we increased n to 200, our standard deviation would decrease because n is the denominator for its equation. Then will a smaller standard deviation, we would get different z values thus resulting in different probabilities. If we calculated the values, we would find that the probabilities decreased after the sample size was doubled. 6,64 a) Use the central limit theorem to deduce that sampling distribution for the number of years a student is expected to work for an employer for a sample of 344 students is approximately normally distributed. Also, the means are the same for both sampling distribution and the sample population. b) The mean for both sampling distribution and the sample population are the same so the mean for the sampling distribution is 18.5. So: Z = P(z > 1.85) = 0.5 – 0.4678 = 0.0322 c) Z = P (z < 1.24) = 0.5 + 0.3925 = 0.8925 d) e) Therefore µ is less than 19.1 years. 6.68 a) Using central limit theorem, we deduce that the sampling distribution will be approximately normally distributed with the same mean. We do not need the shape of the distribution because we can use this central limit theorem where the only requisite is that the sample size is reasonably large. b) Z value P(z > 1.83) = 0.5 – 0.4664 = 0.0336. The approximate probability that the sample mean will be 110 or more is 0.0336. c) If we found that the juveniles with no record of delinquency had the same mean IQ or 110, we can assume that the mean and the standard deviation are both not the same as the ones of the juveniles with record of delinquency. This is because looking at part be, we see that the probability that the sample mean being 110 or more is 0.0336 which is less than 0.05 meaning it is a rare event. Therefore the even that we find the same mean is very unlikely if standard deviation and population mean remained the same. ...
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