Abe plays for his school's cricket team. The number of runs scored by Abe in the eight games that he batted in, is shown below. (Abe was given out in all of the games.) 21 8 19 7 15 32 14 12 1.1 Determine the variance of the data. 1.2 Calculate the standard deviation 1.3 Interpret what the standard deviation tells us about the spread of the data.

Here are the solutions for the given data: $21, 8, 19, 7, 15, 32, 14, 12$. The number of data points is $n=8$.

1.1.1 Determine the variance of the data.

Step 1: Calculate the mean ($\bar{x}$) of the data. $\bar{x} = \frac{\sum x_i}{n}$ $\bar{x} = \frac{21 + 8 + 19 + 7 + 15 + 32 + 14 + 12}{8}$ $\bar{x} = \frac{128}{8}$ $\bar{x} = 16$

Step 2: Calculate the squared difference of each data point from the mean $(x_i - \bar{x})^2$. $(21 - 16)^2 = 5^2 = 25$ $(8 - 16)^2 = (-8)^2 = 64$ $(19 - 16)^2 = 3^2 = 9$ $(7 - 16)^2 = (-9)^2 = 81$ $(15 - 16)^2 = (-1)^2 = 1$ $(32 - 16)^2 = 16^2 = 256$ $(14 - 16)^2 = (-2)^2 = 4$ $(12 - 16)^2 = (-4)^2 = 16$

Step 3: Sum the squared differences. $\sum (x_i - \bar{x})^2 = 25 + 64 + 9 + 81 + 1 + 256 + 4 + 16 = 456$

Step 4: Calculate the sample variance ($s^2$). $s^2 = \frac{\sum (x_i - \bar{x})^2}{n-1}$ $s^2 = \frac{456}{8-1}$ $s^2 = \frac{456}{7}$ $s^2 \approx 65.142857$ Rounding to two decimal places: $s^2 \approx 65.14$ The variance of the data is $\boxed{\frac{456}{7} or 65.14}$.

1.1.2 Calculate the standard deviation.

Step 1: Calculate the standard deviation ($s$) by taking the square root of the variance. $s = \sqrt{s^2}$ $s = \sqrt{\frac{456}{7}}$ $s \approx \sqrt{65.142857}$ $s \approx 8.071112$ Rounding to two decimal places: $s \approx 8.07$ The standard deviation is $\boxed{8.07}$.

1.1.3 Interpret what the standard deviation tells us about the spread of the data.

The standard deviation of $8.07$ runs indicates the average amount of variation or spread of Abe's scores around his mean score of $16$ runs. A higher standard deviation would mean the scores are more spread out, while a lower standard deviation would mean the scores are clustered closer to the mean. In this case, a standard deviation of $8.07$ suggests a moderate spread in Abe's cricket scores.