Continuity Correction Factor: You add or subtract .5, depending on the condition. 3) P(4.5 < x < 5.5) for x = 5 4) P(x > 5.5) for x > 5 5) P(x < 5.5) for x ≤ 5
Use the normal approximation to the binomial distribution. mean = np = 80 * .6 = ? standard deviation = √npq = √(80 * .6 * .4) = ? Calculate. (Note: q = 1 - p) Next use z-scores: z = (x - mean)/sd x = 80 * .5 = ? Once you have the z-score, find the probability usin...
Answered in a previous post.
7 STATS MATHGURU PLEASE HELP
2.58 = 99% confidence interval 1.645 = 90% confidence interval Use a formula to find sample size. Here is one: n = [(z-value * sd)/E]^2 ...where n = sample size, z-value = values above for the intervals needed, sd = 17.8, E = .02, ^2 means squared, and * means to multiply. Plu...
See previous post.
95% confidence = 1.96 You can determine these values by using a z-table. Make a note of the more commonly used confidence intervals so you don't have to constantly refer to the table. It will help you remember them too.
This is the same type of problem as your previous post on confidence intervals. Use the same formula with these values: n = 1003 mean = 18.9 sd = 19.9 I'll let you take it from here.
Formula: CI95 = mean ± 1.96 (sd/√n) ...where ± 1.96 represents the 95% interval using a z-table, sd = standard deviation, and n = sample size. n = 1015 mean = 12.3 sd = 16.6 Plug values into the formula and determine the upper bound (+) and lower bound (-)....
The mean is the middle, so take 15 + 27, then divide by 2. The margin of error will be the difference between the mean and the upper bound or the mean and the lower bound.
Formula to find sample size: n = [(z-value)^2 * p * q]/E^2 ... where n = sample size, z-value is found using a z-table for 90% confidence (which is 1.645), p = .48 (when no value is stated in the problem p = .50), q = 1 - p, ^2 means squared, * means to multiply, and E = .02. ...
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