-Chapter 6 Help The mean of a normal distribution is 530 kg. The standard deviation is...
16. The mean of a normal probability distribution is 400 pounds. The standard deviation is 10 pounds. a. What is the area between 415 pounds and the mean of 400 pounds? . a. .4332 b. .1915 c. .3085 b" What is the area between the mean and 395 pounds? .с. What is the probability of selecting a value at random and discovering that it has a value of less than 395 pounds?
help Assume a random variable Z has a standard normal distribution (mean 0 and standard deviation 1). Use all decimal places from the Normal Table. Your final answers to 4 decimal places. a) The probability that Z lies between 1.55 and 1.86 is Select b) What is the value of Z if only 1.5% of all possible Z values are larger? Select]
Consider a normal distribution with mean 25 and standard deviation 5. What is the probability a value selected at random from this distribution is greater than 25? (Round your answer to two decimal places.) Assume that x has a normal distribution with the specified mean and standard deviation. Find the indicated probability. (Round your answer to four decimal places.) μ = 14.9; σ = 3.5 P(10 ≤ x ≤ 26) = Need Help? Read It Assume that x has a...
A normal distribution has a mean of 60 and a standard deviation of 10. Refer to the table in Appendix B.1. Determine the value above which 80 percent of the values will occur. (Round z-score computation to 2 decimal places and the final answer to 2 decimal places.) X
Given a standardized normal distribution (with a mean of O and a standard deviation of 1), complete parts (a) through (d). 5 Click here to view page 1 of the cumulative standardized normal distribution table. E: Click here to view page 2 of the cumulative standardized normal distribution table. The probability that Z is less than 1.51 is 0.9344. (Round to four decimal places as needed.) b. What is the probability that Z is greater than 1.89? The probability that...
A distribution of values is normal with a mean of 238 and a standard deviation of 93.7. Find the probability that a randomly selected value is between 219.3 and 481.6. P(219.3<x< 481.6) = Enter your answer as a number accurate to 4 decimal places. Answers obtained using exact z-scores or z-scores rounded to 3 decimal places are accepted.
Given a standardized normal distribution (with a mean of O and a standard deviation of 1), complete parts (a) through (d) below. Click here to view page 1 of the cumulative standardized normal distribution table Click here to view page 2 of the cumulative standardized normal distribution table. a. What is the probability that Z is between - 1.54 and 1.88? The probability that Z is between - 1.54 and 1.88 is .9061. (Round to four decimal places as needed.)
a. Consider a normal distribution with mean 20 and standard deviation 4. What is the probability a value selected at random from this distribution is greater than 20? (Round your answer to two decimal places. b. Assume that x has a normal distribution with the specified mean and standard deviation. Find the indicated probability. (Round your answer to four decimal places.) μ = 14.3; σ = 3.5 P(10 ≤ x ≤ 26) = c. Assume that x has a normal...
Given a standardized normal distribution (with a mean of 0 and a standard deviation of 1) complete parts (a) through (d) below. Click here to view page 1 of the cumulative standardized normal distribution table, Click here to view page 2 of the cumulative standardized normal distribution table. a. What is the probability that Z is between 1.57 and 1.83? - The probability that Z is between 1.57 and 1.83 is (Round to four decimal places as needed.) particular train...
Question 22 A distribution of values is normal with a mean of 28.8 and a standard deviation of 20.8. Find the probability that a randomly selected value is greater than -31.5. P(X > -31.5) - Enter your answer as a number accurate to 4 decimal places. Submit Question Question 23 A distribution of values is normal with a mean of 71.9 and a standard deviation of 68. Find the probability that a randomly selected value is between - 138.9 and...