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You can't average means with standard deviations. What are you trying to do with the two sets of data?
The Bank, itself does not have a standard deviation. The number of branches, the number of customers, lending, profits, CEO's pay are all variables which will have standard deviations but none of them are mentioned. It is not possible to guess which one you are interested in!
Mean is the average, sum total divided by total number of data entries. Standard deviation is the square root of the sum total of the data values divided by the total number of data values. The standard normal distribution is a distribution that closely resembles a bell curve.
A z-score gives the distance (specifically number of standard deviations) from the mean so when you compare z-scores, it gives a direct comparison of how far from the mean the values are.
No. It cannot be. Remember that when you square a negative number it becomes a positive number. Thus all squared deviations are positive and their sum must be positive.
You cannot have a standard deviation for 1 number.
You can't average means with standard deviations. What are you trying to do with the two sets of data?
z-score or standard score... tells you how many standard deviations away from the mean a particular number is in relations to all numbers in a population (or sample)
Standard deviations are measures of data distributions. Therefore, a single number cannot have meaningful standard deviation.
z
The absolute value of the z-score.
The Bank, itself does not have a standard deviation. The number of branches, the number of customers, lending, profits, CEO's pay are all variables which will have standard deviations but none of them are mentioned. It is not possible to guess which one you are interested in!
16.5 is 1 standard deviation from the mean. If you add the mean of 14 to the 1 standard deviation of 2.5, the result is 16.5.
Mean is the average, sum total divided by total number of data entries. Standard deviation is the square root of the sum total of the data values divided by the total number of data values. The standard normal distribution is a distribution that closely resembles a bell curve.
A z-score gives the distance (specifically number of standard deviations) from the mean so when you compare z-scores, it gives a direct comparison of how far from the mean the values are.
z = (x - u)/(standard dev)The z score expresses the difference of the experimental result x from the most probable result u as a number of standard deviations. The probability can then be calculated from the cumulative standard normal distribution. ie sigma(z)
Suspect you've made a mistake in your calculations.Looking at the Normal curve, the area under it between the mean and 3.09 standard deviations is [approx] 0.4990, ie the probability that the data could exceed 3.09 standard deviations from the mean is 2 x (0.5-0.4990) = 0.002 = 0.2% [using a half-tail table], ie it is quite unlikely that a data point is much further away from the mean than the tables' limit of 3.09.Beyond 3[.09] standard deviations away from the mean, the area under the curve changes very little in the first 4 dp, so [most] tables are going to not be of much help anyway - when 4 standard deviations away are reached, it is almost all the distribution and rounds to 1.So if you are looking at a point greater than 3 standard deviations away from the mean it is either a very unusual event that has caused it, or (more likely) you've made a mistake in your calculations.