There is no information on how much time can be spent on a call.
No. It's the number that will help you work out the percentage.
A standard deviation in statistics is the amount at which a large number of given values in a set might deviate from the average. A percentile deviation represents this deviation as a percentage of the range.
100 x (standard deviation/mean)
In a normal distribution, approximately 68% of the data falls within one standard deviation of the mean. This means that around 34% of the data lies between the mean and one standard deviation above it, while another 34% lies between the mean and one standard deviation below it.
The standard deviation of the population. the standard deviation of the population.
No. It's the number that will help you work out the percentage.
4.55% falls outside the mean at 2 standard deviation
A standard deviation in statistics is the amount at which a large number of given values in a set might deviate from the average. A percentile deviation represents this deviation as a percentage of the range.
100 x (standard deviation/mean)
In a normal distribution, approximately 68% of the data falls within one standard deviation of the mean. This means that around 34% of the data lies between the mean and one standard deviation above it, while another 34% lies between the mean and one standard deviation below it.
For normally distributed data. One standard deviation (1σ)Percentage within this confidence interval68.2689492% (68.3% )Percentage outside this confidence interval31.7310508% (31.7% )Ratio outside this confidence interval1 / 3.1514871 (1 / 3.15)
The standard deviation is the standard deviation! Its calculation requires no assumption.
no
The standard deviation of the population. the standard deviation of the population.
Mean = 100 Standard Deviation = 25 (120 -100) / 25 = 0.80 th z value of .80 from the table = .2881 P(100-120) = 28.81% What percentage of calls are completed in less than 60 minutes ? (60 - 100) / 25 = - 1.6 from the table the z value is .4452 (one side), then : .5000 - .4452 = .0548 P(60) = 5.48%
The standard deviation is 0.
Information is not sufficient to find mean deviation and standard deviation.