Convert to Standard Normal Distribution the values of 1.22 & 8.78. Z = (8.78-5)/1.41 = 2.68; and Z = (1.22-5)/1.41 = -2.68. Find the area between 2.68 & -2.68 from Table. Area @ 2.68 = .9963; Area @ -2.68 = .0037. Take the difference in the areas and that is the solution. 0.9963 - 0.0037 = .9926 or 99.26% of the scores fall between 1.22 & 8.78.
68 % is about one standard deviation - so there score would be between 64 and 80 (72 +/- 8)
If the standard deviation of 10 scores is zero, then all scores are the same.
It is 68.3%
67% as it's +/- one standard deviation from the mean
mean
68 % is about one standard deviation - so there score would be between 64 and 80 (72 +/- 8)
If the standard deviation of 10 scores is zero, then all scores are the same.
It is 68.3%
mrs.sung gave a test in her trigonometry class. the scores were normally distributed with a mean of 85 and a standard deviation of 3. what percent would you expect to score between 88 and 91?
67% as it's +/- one standard deviation from the mean
The variance and the standard deviation will decrease.
Within 1 stdev of the mean - between 40 and 60.
All the scores are equal
mean
The standard deviation is defined as the square root of the variance, so the variance is the same as the squared standard deviation.
Standard Deviation tells you how spread out the set of scores are with respects to the mean. It measures the variability of the data. A small standard deviation implies that the data is close to the mean/average (+ or - a small range); the larger the standard deviation the more dispersed the data is from the mean.
If a random variable X has a normal distribution with mean m and standard error s, then the z-score corresponding to the value X = x is (x - m)/s.