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Simple questions part 1 4. (1) Assume that the readings on the thermometers are normally distributed with a mean of 0o and standard deviation of 1.00o C. A thermometer is randomly selected and tested. Draw a diagram and find the probability in degrees. Between -1.18 and 2.15 Answer: Mean = 0oCnd SD = 1oC Probability of temperature between -1.18o and 2.15o P( -1.18 < X < 2.15) = P(z1 < Z < z2) 1.18 0 1.18 1 2.15 0 z2 2.15 1 P(1.18 Z 2.15) 0.3810 0.4842 0.8652 z1 2) The lengths of pregnancies are normally distributed with a mean of 268 days and a standard deviation of 15 days. A woman claims to give birth 308 after conception. a. Find probability of a pregnancy lasting 308 days or longer. What does the result suggest? b. If we stipulate that a premature birth is in the lowest 4%, find the length that separates premature from full term births. Answer: a) Mean duration of pregnancy:= 268 days, with SD = 15days P(X 308) = P(Zz) 308 268 2.6667 15 P( Z 2.667) 0.5 0.4962 0.0038 z P( X 308days ) 0.0038 The result suggests that there is very little possibility that the pregnancy duration will be more than or equal to 308days. b) P(Z = z) = 4% = 0.04 Where z = 0.10 X 268 0.10 1.5 X 268 15 X 268 0.10 268.10 Hence the actual number of days = 268.10 and the expected number of days = 268. Their difference gives the desired value. (3) Ages 62, 46, 68, 64, 57. Assume the samples of size 2 are randomly selected with replacement from the population of five ages. a. after identifying the 25 different possible samples, find the mean of each of them. b. Describe sampling distribution of means. c. Find the mean of the sampling distribution d. Is the mean of the sampling distribution (from c.) equal to the mean of the five ages above? Are those always equal? Answer: a) probability of selecting 2 samples = 2/5 (with replacement) Hence probability of selecting a sample = 1/5 b) Since replacement is allowed, the repetition of choosing a sample is possible with samples n = 25, and p = 1/5 Such that mean = np = 25/5 = 5 c) Mean for the given samples = 59.4 d) No the means are not equal. However, if the samples are repeated and replaced every-time, then there is chance for the two samples to be equal. (4) Use the Central Limit Theorem - Assume normal distribution with a mean of 172 lb and standard deviation of 29 lb a. 4 men are randomly selected, what is the probability that they have a mean weight between 160 and 180 lb. b. Why can the central limit theorem be used in (a.), even though the sample size does not exceed 30? Answer: Mean = 172lb, SD = 29lb P( 160lbs X 180lbs) P( z1 Z z 2) 160 172 0.8275 29 / 4 180 172 0.5517 a) z 2 29 / 4 P(0.8275 Z 0.5517) z1 0.295 0.2088 0.5038 b) Since the distribution of data is assumed to be normally distributed, we can use the central limit theorem, with the fact that for any X ~ N(mean, variance), the corresponding Z~N(0,1) with z = (X-mean)/SD. (5) Building a bench to seat 18 men, normal distribution with a mean of 14.4 in and std deviation of 1.0 in hip breadth a. What is the minimum length of the bench if you want a 0.975 probability that it will fit the combined hip breadths of 18 randomly selected men? b. What would be wrong with actually using the result form part (a.) as the bench length? Answer: n = 18men, mean = 14.4inches, with SD = 1.0inch a) P(X = x) = P(Z =z) = 0.975 xx P(2.24 ) 0.975 s/ n x 14.4 P(2.24 ) 0.975 1 / 18 Thus we have: x 14.4 2.24 = = 2.24 *1.451 = x – 14.4 1 / 18 3.25 = x – 14.4 => x = 3.25 +14.4 = 17.65 Hence the minimum length of the bench should be 17.65inches. b) If we wrongly use the value 17.65, the deviation in length will be more and hence the required probability will be small. (6) Estimate the probability of getting at least 65 girls in 100 births. Assume boys and girls are equally likely. Is it unusual to get at least 65 girls in 100 births? Answer: p = 65/100 = 0.65, q = 0.35, n = 100 Hence P = 0.50 = Q (Boys and girls are equally likely) P( X 65 girls ) P( Z z ) X np 0.65 0.5 where : z 3.144 npq 0.35 * 0.65 / 100 P( Z 3.144) 0.5 P(0 Z 3.144) 0.5 0.4992 0.0008 Hence the percentage of at-least 65% of female birth is very small, hence it is it unusual to get at least 65 girls in 100 births. (7) Construct Normal Quantile Plots, identify if the corresponding z scores that are used for a normal quantile plot the construct the normal Qplot and determine whether the data appear to be from a population with normal distribution. a. LA Lakers heights, use the sample in inches 85, 79, 82, 73, 78 Answer: fi = (i -3/8) / (n+1/4) i xi fi 1 2 3 4 5 73 78 79 82 85 0.5 0.722 0.8076 0.8529 0.8809 Quantile plot: Data values, x Quantile - Plot 86 84 82 80 Series1 78 76 74 72 0 0.2 0.4 0.6 0.8 1 fraction , f Normal quantile Plot: Normal Q-Q Plot Expected Normal Value 86 84 82 80 78 76 74 72.5 75.0 77.5 80.0 82.5 85.0 Observed Value 5. (1) Find critical value zo/2 that corresponds with the given confidence level of 92% Answer: For 92% confidence level, the critical z-value = 1.75 i.e. when =0.08level, z-value =1.75 (Look for the z-value from Normal tables). (2) Find the margin of error; assume the sample is used to estimate a population proportion. Find the margin for E. a. n=1200, x=400, 99% confidence Answer: Margin of error = Z-score* standard error For 99% confidence level, Z-score = 2.575 Proportion p = x/n = 400/1200 = 1/3 q = 1- p = 1-1/3 = 2/3 Hence standard error for p: = SE (p) = pq = n 1/ 3 * 2 / 3 0.0136 1200 So, the margin of error: 2.575 * 0.0136 = 0.03502 (3) Of 491 randomly selected adults 65% favor the death penalty a. find the point estimate of the % of adults who are in favor of the death penalty. Answer: N = 491, x = 65% of 491 => 319.15, I.e., p = x/N = 319.15/491 = 0.65, q = 1-p = 1-0.65 = 0.35 Then the mean => np = 491*0.65 = 319.15 Hence the point estimate is 319.65 for 65% of adults who are in favor of the death penality. b. find a 95% confidence interval estimate of the % of adults who are in favor. Answer: 95% confidence interval: pq 0.65 * 0.35 0.65 1.96 * n 491 0.65 1.96 * 0.0215 0.65 0.04214 (0.60786,69214) This is the required confidence interval estimate for 65% of adults who are in favor. p Z * c. can we safely conclude that the majority of adults are in favor? Explain why. Answer: Since the percentage of adults who are in favor of death penalty is 65% than that of young ones, we conclude that the majority of adults are in favor for the capital punishment.(death penalty) (4) Women heights are normally distributed with a mean of 63.6 in. and a std deviation of 2.5 in. a. how many women must be surveyed if we want to estimate the % who are taller than 5ft? assume that we want 98% confidence that the error is no more that 2.5 % points. (answer is substantially smaller than 2712) Answer: We have to compute the sample size ‘n’. To find this n, we have the formula given as: Z2 *S2 n 2 E Where Z 2.33 S 2.5 S 2 6.25 And E 2.5% Then, Z 2 * S 2 (2.33) 2 * (6.25) 5.4289 * 6.25 2712.45 ~ 2712 E2 (2.5%) 2 (0.025) 2 Hence the number of women to be surveyed should be not more than 2712. n (5) Margin of error and confidence Int., a. ages of drivers in the passing lane while driving 25 mi/h with left signal flashing: 99% confidence; n=50, mean=80.5 yrs and std is 4.6 yrs Answer: M arg in of error : s 4.6 Z * 2.575 * 1.675 n 50 Confidence int erval : mean M arg in of error 80.5 1.675 (78.825,82.175) This is the required 99% confidence interval for the true mean age for drivers. (6) Find confidence Int., a. 99% confidence; n=15, mean=496, s=0.07in Answer: M arg in of error : s 0.07 t ( n1, / 2) * 2.977 * 0.0538 n 15 Confidence int erval : mean M arg in of error 496 0.0538 (495.9462,496.0528) This is the required 99% confidence interval for the true mean.