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Example A random sample of 100 students is taken from the population of all part-time students in the U.S., for which the overall population of females is 0.6. Start by determining what is given in the original statement. 100 0.6 = Test to see if we can use the normal distribution. np = and nq = Can we use the normal approximation? 0.6, = Since the population proportion is p we know the mean of the sampling proportions is up (Hint: μ = p.) What is the standard deviation? o, op = (Round to the nearest thousandth.)

Example A random sample of 100 students is taken from the population of all part-time students in the U.S., for which the overall population of females is 0.6. Start by determining what is given in the original statement. 100 0.6 = Test to see if we can use the normal distribution. np = and nq = Can we use the normal approximation? 0.6, = Since the population proportion is p we know the mean of the sampling proportions is up (Hint: μ = p.) What is the standard deviation? o, op = (Round to the nearest thousandth.)

A First Course in Probability (10th Edition)
A First Course in Probability (10th Edition)
10th Edition
ISBN:9780134753119
Author:Sheldon Ross
Publisher:Sheldon Ross
Chapter1: Combinatorial Analysis
Section: Chapter Questions
Problem 1.1P: a. How many different 7-place license plates are possible if the first 2 places are for letters and...
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Example A random sample of 100 students is taken from the population of all part-time students in the U.S., for which the overall population of females is 0.6. Start by determining what is given in the original statement. 100 0.6 = Test to see if we can use the normal distribution. np = and nq = Can we use the normal approximation? 0.6, = Since the population proportion is p we know the mean of the sampling proportions is up (Hint: μp = p.) What is the standard deviation? σ, op . 0.4 Hint: Op Probability 1 There is a 95% chance that the sample proportion, p, falls between what two values? We are looking for two x-values between which 95% of the data lies. By the Empirical Rule, we know that 95% of the data in a normal distribution falls between 2 standard deviations. Therefore we need to find the x-values (the raw data scores) for x₁ = μ₂ 20 and ₂ = Mp + 20p x₁ = 0.6 — 2(0.049) = 0.502 X2 0.6 +2(0.049) = 0.698 95% of the data is between 0.502 and 0.698 We can write this as P(0.502 < p < 0.698) 0.95. The image below shows what this would look like on the normal curve. = = (Round to the nearest thousandth.) - pq n 0.5 0.6 = = 0.7 0.8
Transcribed Image Text:Example A random sample of 100 students is taken from the population of all part-time students in the U.S., for which the overall population of females is 0.6. Start by determining what is given in the original statement. 100 0.6 = Test to see if we can use the normal distribution. np = and nq = Can we use the normal approximation? 0.6, = Since the population proportion is p we know the mean of the sampling proportions is up (Hint: μp = p.) What is the standard deviation? σ, op . 0.4 Hint: Op Probability 1 There is a 95% chance that the sample proportion, p, falls between what two values? We are looking for two x-values between which 95% of the data lies. By the Empirical Rule, we know that 95% of the data in a normal distribution falls between 2 standard deviations. Therefore we need to find the x-values (the raw data scores) for x₁ = μ₂ 20 and ₂ = Mp + 20p x₁ = 0.6 — 2(0.049) = 0.502 X2 0.6 +2(0.049) = 0.698 95% of the data is between 0.502 and 0.698 We can write this as P(0.502 < p < 0.698) 0.95. The image below shows what this would look like on the normal curve. = = (Round to the nearest thousandth.) - pq n 0.5 0.6 = = 0.7 0.8
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