Find p from z: Find p > z = 1.28

U1.5 DA

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Find p from z:
1. Find p > z = 1.28
2. Find p < z = –.37 3. Find p between z = ±.93 Find X from p: For questions 4 and 5, use the following population parameters to find p from X: μ = 28.29, σ = 3.87. 4. What is the raw score (X) that corresponds to p = .309 in the lower tail of the distribution? 5. What are the raw scores (X’s) that corresponds to p = .92 in the two tails of the distribution combined? Find z from p: For questions 6 and 7, use the mean and standard deviation of a normal z-distribution to find p from z. 6. What is the z-score that corresponds to p = .115 in the upper tail of the distribution? 7. What are the z-scores that corresponds to p = .050 in the two tails of the distribution combined? Find p > z = 1.28
Based on the standard normal distribution table, when z = 1.28, the corresponding p-value is 0.1. Therefore, p > 1.28 is 0.1.
Find p < z = –.37 When z = -0.37, the corresponding p-value is 0.355. Therefore, p < -0.37 is 0.355. Find p between z = ±.93 When z = 0.93, p = 0.175 in the upper tail. When z = -0.93, p = 0.175 in the lower tail. Therefore, the total p between ±0.93 is 0.175 + 0.175 = 0.35. What is the raw score (X) that corresponds to p = .309 in the lower tail of the distribution? Given: μ = 28.29, σ = 3.87 To find the raw score X corresponding to a p-value in the lower tail, we use the inverse normal cumulative distribution function norminv(p, mean, standard deviation) norminv(0.309, 28.29, 3.87) = 25.5 Therefore, the raw score X corresponding to p = 0.309 in the lower tail is 25.5. What are the raw scores (X’s) that corresponds to p = .92 in the two tails of the distribution combined? To find the scores in the two tails corresponding to a combined p-value of 0.92, we calculate 0.92/2 = 0.46 for the p-value of each tail norminv(0.46, 28.29, 3.87) = 31.5 for the upper tail norminv(1-0.46, 28.29, 3.87) = 24.9 for the lower tail Therefore, the raw scores corresponding to a combined p=0.92 are 31.5 and 24.9. What is the z-score that corresponds to p = .115 in the upper tail of the distribution? When p = 0.115, the z-score in the upper tail from the standard normal distribution table is 1.28. What are the z-scores that corresponds to p = .050 in the two tails of the distribution combined? To find the z-scores corresponding to a combined p-value of 0.05 in the two tails, we calculate: 0.05/2 = 0.025 for each tail From the standard normal distribution table, z = 1.96 for the upper tail z = -1.96 for the lower tail Therefore, the z-scores corresponding to a combined p=0.05 are 1.96 and -1.96.