Compute the range, variance and standard deviation of the two sets of scores.
Jade 10, 11, 11, 13 , 15
Jane 9, 10, 12, 12, 17
To compute the range, variance, and standard deviation of the two sets of scores, you need to follow these steps:
1. Calculate the range:
- The range of a data set is the difference between the largest and smallest values.
- For Jade's scores, the largest value is 15, and the smallest value is 10.
- So, the range for Jade's scores is 15 - 10 = 5.
- For Jane's scores, the largest value is 17, and the smallest value is 9.
- So, the range for Jane's scores is 17 - 9 = 8.
2. Calculate the variance:
- Variance measures how spread out the data is compared to the mean.
- To calculate variance, you need to find the mean and then calculate the differences between each score and the mean.
- Let's start with Jade's scores:
- The mean for Jade's scores is (10 + 11 + 11 + 13 + 15) / 5 = 12.
- Subtract the mean from each score and square the result:
- (10 - 12)^2 = 4
- (11 - 12)^2 = 1
- (11 - 12)^2 = 1
- (13 - 12)^2 = 1
- (15 - 12)^2 = 9
- Add up these squared differences: 4 + 1 + 1 + 1 + 9 = 16.
- Divide the sum of squared differences by the number of scores (5):
- 16 / 5 = 3.2.
- So, the variance for Jade's scores is 3.2.
- Now, let's calculate the variance for Jane's scores:
- The mean for Jane's scores is (9 + 10 + 12 + 12 + 17) / 5 = 12.
- Subtract the mean from each score and square the result:
- (9 - 12)^2 = 9
- (10 - 12)^2 = 4
- (12 - 12)^2 = 0
- (12 - 12)^2 = 0
- (17 - 12)^2 = 25
- Add up these squared differences: 9 + 4 + 0 + 0 + 25 = 38.
- Divide the sum of squared differences by the number of scores (5):
- 38 / 5 = 7.6.
- So, the variance for Jane's scores is 7.6.
3. Calculate the standard deviation:
- The standard deviation is the square root of the variance.
- For Jade's scores:
- The standard deviation is the square root of 3.2, which is approximately 1.79 (rounded to two decimal places).
- For Jane's scores:
- The standard deviation is the square root of 7.6, which is approximately 2.76 (rounded to two decimal places).
Therefore, the range for Jade's scores is 5, the variance is 3.2, and the standard deviation is 1.79.
For Jane's scores, the range is 8, the variance is 7.6, and the standard deviation is 2.76.