By empirical rule, about 68% of the values lie within 1 standard deviation of the mean. So 68% of the observations lie between 1500-150=1350 and 1500+150=1650 B About 95% of the observations lie between what two values?
Suppose a large sample is selected (instead of just 5). About 68% of the predictions would be between what two values? e. Determine the .95 confidence interval for the mean predicted when X = 7. f.
Calculate mean for the data. Mean = 27 Calculate standard deviation for the data. SD=61.9 Rule 68%. 68% data should lie within +/- 1SD from mean.
In contrast to the normal distribution, the binomial distribution describes the behavior of a count variable X if the following conditions apply: 1: The number of observations n is fixed. 2: Each observation is independent. 3: Each observation
188270 Mean, Standard Deviation Mean, Standard Deviation 1. What does it mean for a sample to have a standard deviation of zero? Describe the scores in such a sample.
(n+1)/2. d. none of the above. Answer: c. (n+1)/2. 14 According to Empirical Rule, what percent of the observations lie within plus and minus 1 standard deviations of the mean? a. About 99 percent b.
If k = 2, we have P(|X - μ| >= 2σ) <= 1/4. This means that less than 25% of the data points will lie beyond 2 std. dev. Then more than 75% of the data will be within 2 standard deviations. If k = 3, P(|X - μ| >= 3σ) <= 1/9.
If we assume a normal distribution of Systolic blood pressures, then between what two values can we be assured 99.7% of all Systolic blood pressures will lie? according to empirical rule, 99.7% corresponds to 3 standard deviation of the mean.
Empirical rule states that for a normal distribution, About 68% (precisely 68.27%) of the values lie within 1 standard deviation of the mean (or between the mean minus 1 times the standard deviation, and the mean plus 1 times the standard deviation).
If 0.05 of the values are less than X, then 0.45 lie between X and the mean (0.5 - 0.05), see the following graph. Refer to the standard normal distribution table and search the body of the table for 0.45.