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Socy 601 Homework # Responses 1 [15 points] a.) Probability of Knowing Someone Who Committed Suicide in Last 12 Months Num ber of people know w ho com m itted suicide w ithin last 12 m onths Probability P(Y) 0 0. 1 0. 2 0. Total 1.
Probability Distribution for the Number of People
One has Known that have Committed Suicide in the
Previous 12 Months
0
1 0 1 2 Number of People that have Committed Suicide Probability 1b.) Yi p(Yi) = 0(.90) + 1(.08) + 2(.02) =. The mean of the distribution of Y is .12, meaning that the average person knows. people who committed suicide in the past 12 months.
3 [10 points] a) Probability Distribution of Number of Female Students Admitted Number of Female Students Admitted Probability 0 0. 1 0. 2 0. Total 1. 3b) E(Y) = p(Yi) = 0(.3) + 1(.6) + 2(.1) = .8 The expected value of Y is .8, i.e. the expected number of female students admitted is .8, or ~ 1. 7 [16] a) For a normal distribution, the probability that an observation is at least one standard deviation above the mean is .1587. 1.00 standard deviation above the mean = +1.00 z-score =. 7b) For a normal distribution, the probability that an observation is least one standard deviation below the mean is .1587. 1.00 standard deviation below the mean = -1.00 z-score =. 7c) For a normal distribution, the probability that an observation is least .67 standard deviations above the mean is .2514. .67 standard deviation above the mean = +.67 z-score =. 7d) For a normal distribution, the probability that an observation is least 2.33 standard deviations below the mean is .0099. 2.33 standard deviation below the mean = -2.33 z-score =. 9 [24] (For 9a-f, refer to the table at the back of Agresti and Finlay and convert p-values to z scores.)
a) The z-value for which the probability that a normal variable exceeds + z()
equals .01 is 2.33.
9b) The z-value for which the probability that a normal variable exceeds + z() equals
.025 is 1.96.
A population parameter you are studying has a mean of μ = 6.0 and a standard deviation of σ = 2.0. A.) What is the z-score for person 16, with a score of Y 16 = 3? z = (Yi – μ) / σ = (3-6.0)/2.0 = -1. The z score of Y 16 is -1.5. B.) What percent of the population has a higher score than person 16? (Refer to page 668, and assume that the population parameter is normally distributed.) Checking the Table A on page 668, we get P value for Z= -1.5 is .0668, which means about 6.7% of the population has a lower score than person 16. Therefore, 1-. = .9332, or about 93.3 % of population has a higher score than person 16. C.) What is the expected mean, standard deviation, and standard error of a sample of size 100, drawn from the above population? For a random sample of size 100, the expected mean should be the population mean μ if it is a probability sample. In this case, it will be 6.0. The expected standard deviation of a sample will be the population standard deviation, which is σ = 2.0. The standard error of a sampling distribution of size 100 is equal to σ / √(n) =2.0/ √(100) = 0.2. Therefore, the expected standard error will be .2. D.) When you take one sample of size 100, the sample mean is Ybar = 5.7 and the sample standard deviation is s = 2.2. If you were working only from this sample, what would be your estimate of the standard error? Based on this sample, my best guess of the population standard deviation σ will be the sample standard deviation, in this case S=2.2. Therefore, my estimate of the standard error from this sample = σ / √(n) =2.2/ √(100) = 0.
