1317 to 1705
1511+194= 1705
1511-194=1317
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?
The answer depends on the degrees of freedom (df). If the df > 1 then the mean is 0, and the standard deviation, for df > 2, is sqrt[df/(df - 2)].
68% of the scores are within 1 standard deviation of the mean -80, 120 95% of the scores are within 2 standard deviations of the mean -60, 140 99.7% of the scores are within 3 standard deviations of the mean -40, 180
78
About 98% of the population.
If the standard deviation of 10 scores is zero, then all scores are the same.
It is 68.3%
All the scores are equal
mean
68 % is about one standard deviation - so there score would be between 64 and 80 (72 +/- 8)
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.
67% as it's +/- one standard deviation from the mean
Since the standard deviation is zero, the scores are all the same. And, since their mean is 10, they must all be 10.
Because the standard deviation is a measure of the spread in scores. As individuals score more similarly, the spread gets smaller. Because the standard deviation is a measure of the spread in scores. As individuals score more similarly, the spread gets smaller. Because the standard deviation is a measure of the spread in scores. As individuals score more similarly, the spread gets smaller. Because the standard deviation is a measure of the spread in scores. As individuals score more similarly, the spread gets smaller.
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?
The answer depends on the degrees of freedom (df). If the df > 1 then the mean is 0, and the standard deviation, for df > 2, is sqrt[df/(df - 2)].