sample size, n = 140
standard deviation, s = 11.45
standard error of the mean, SE = s / n^1/2 = 11.45 / 140^1.2 = 0.9677
95% confidence interval => mean +- 1.96SE
95% CI = 86.5 - 1.96*0.9677; 86.5 + 1.96*0.9677
= 84.6; 88.4
at first, it was held randomly; but since the world war II, the chess olympiad has been held at a time interval of every 2 years.
That is the correct spelling of the word "randomly" (by chance).
If dealt from a randomly shuffled pack it is 0.0399, approx.If dealt from a randomly shuffled pack it is 0.0399, approx.If dealt from a randomly shuffled pack it is 0.0399, approx.If dealt from a randomly shuffled pack it is 0.0399, approx.
That means that a number is selected randomly from a certain set. For example, if you throw a die, you can randomly get any of the integers from 1 to 6. It is random, because you don't know the result in advance.That means that a number is selected randomly from a certain set. For example, if you throw a die, you can randomly get any of the integers from 1 to 6. It is random, because you don't know the result in advance.That means that a number is selected randomly from a certain set. For example, if you throw a die, you can randomly get any of the integers from 1 to 6. It is random, because you don't know the result in advance.That means that a number is selected randomly from a certain set. For example, if you throw a die, you can randomly get any of the integers from 1 to 6. It is random, because you don't know the result in advance.
The reason the standard deviation of a distribution of means is smaller than the standard deviation of the population from which it was derived is actually quite logical. Keep in mind that standard deviation is the square root of variance. Variance is quite simply an expression of the variation among values in the population. Each of the means within the distribution of means is comprised of a sample of values taken randomly from the population. While it is possible for a random sample of multiple values to have come from one extreme or the other of the population distribution, it is unlikely. Generally, each sample will consist of some values on the lower end of the distribution, some from the higher end, and most from near the middle. In most cases, the values (both extremes and middle values) within each sample will balance out and average out to somewhere toward the middle of the population distribution. So the mean of each sample is likely to be close to the mean of the population and unlikely to be extreme in either direction. Because the majority of the means in a distribution of means will fall closer to the population mean than many of the individual values in the population, there is less variation among the distribution of means than among individual values in the population from which it was derived. Because there is less variation, the variance is lower, and thus, the square root of the variance - the standard deviation of the distribution of means - is less than the standard deviation of the population from which it was derived.
The confidence interval for this problem can be calculated using the following formula: Confidence Interval = p ± z*√(p*(1-p)/n) Where: p = observed proportion (54%) n = sample size (80) z = z-score (1.96) Confidence Interval = 0.54 ± 1.96*√(0.54*(1-0.54)/80) Confidence Interval = 0.54 ± 0.07 Therefore, the confidence interval is 0.47 - 0.61, meaning that we can be 95% confident that the percentage of voters who prefer the referred candidate is between 47% and 61%.
Both are measures of the spread of the data around a mean or average. In fact the standard deviation == sqrt(variance). The standard deviation (SD) is defined to be the square root of the sample variance (V). EX: Some value X = M +/- 1 SD indicates that there is roughly a 2/3 probability that the value will fall randomly within the interval from minus one SD up to plus one SD around the mean M.
Approx 0.0027
Either he likes you and just worked up the confidence to talk to you, in which case talk back, or his confidence will shatter, or he was dared by his friends and he doesn't really know you.
.820=82.0%
0.9699
To analyze the variation of vitamin supplement tablets, calculate the mean and standard deviation of the weights of the 14 tablets. With a 90% confidence level, determine the margin of error using the t-distribution. This margin of error will help define the range within which the true mean weight of the tablets is likely to lie.
at first, it was held randomly; but since the world war II, the chess olympiad has been held at a time interval of every 2 years.
84% To solve this problem, you must first realize that 66 inches is one standard deviation below the mean. The empirical rule states that 34% will be between the mean and 1 standard deviation below the mean. We are looking for the prob. of the height being greater than 66 inches, which is then 50% (for the entire right side of the distribution) + 34%
The probability can be calculated by finding the proportion of the interval where the diameter is greater than 7.4 mm. In this case, the proportion of the interval greater than 7.4 mm is (8.5 - 7.4) / (8.5 - 6.8) = 0.55, so the probability is 55%.
Randomly
If the mean is less than or equal to zero, it means there has been a serious calculation error. If the mean is greater than zero and the distribution is Gaussian (standard normal), it means that there is an 84.1% chance that the value of a randomly variable will be positive.