z = (80 - 100)/20 = -20/20 = -1
Look up a table of z-scores: Probability = 15.87%
With a mean score of 110 and a standard deviation of 18, what percentage of scores would be less than 155?
In general, a mean can be greater or less than the standard deviation.
99.6% for
Yes. The standard deviation and mean would be less. How much less would depend on the sample size, the distribution that the sample was taken from (parent distribution) and the parameters of the parent distribution. The affect on the sampling distribution of the mean and standard deviation could easily be identified by Monte Carlo simulation.
standard deviation only measures the average deviation of the given variable from the mean whereas the coefficient of variation is = sd\mean Written as "cv" If cv>1 More variation If cv<1 and closer to 0 Less variation
With a mean score of 110 and a standard deviation of 18, what percentage of scores would be less than 155?
In general, a mean can be greater or less than the standard deviation.
Yes, a standard deviation can be less than one.
No, if the standard deviation is small the data is less dispersed.
99.6% for
About 98% of the population.
An acceptable standard deviation depends entirely on the study and person asking for the study. The smaller the standard deviation, the more acceptable it will be because the less likely there are to be errors.
Assuming a normal distribution, the proportion falling between the mean (of 8) and 7 with standard deviation 2 is: z = (7 - 8) / 2 = -0.5 → 0.1915 (from normal distribution tables) → less than 7 is 0.5 - 0.1915 = 0.3085 = 0.3085 x 100 % = 30.85 % (Note: the 0.5 in the second sum is because half (0.5) of a normal distribution is less than the mean, not because 7 is half a standard deviation away from the mean, and the tables give the proportion of the normal distribution between the mean and the number of standard deviations from the mean.)
0.9699
The larger the value of the standard deviation, the more the data values are scattered and the less accurate any results are likely to be.
z=-20/12 = -1.667 Assuming normal distribution, P(Z < -1.667) = 0.04779 or 4.8% of the scores should be less than 50. You can get the probabilities by looking them up on a table or use Excel, where +Normdist(50,70,12,true). My normal table has only 2 digit accuracy so for -1.67 = 0.0475.
The standard deviation (SD) is a measure of spread so small sd = small spread. So the above is true for any distribution, not just the Normal.