Bradley's Maths

Solution

a) The mean is the sum of each set of scores divided by the number of scores.

Annabelle:

$$\bar{x}_A = \dfrac{150 + 155 + 148 + 152 + 158 + 160 + 151 + 143 + 158 + 153}{10} = \dfrac{1528}{10} = 152.8$$

Ben:

$$\bar{x}_B = \dfrac{120 + 180 + 141 + 192 + 119 + 163 + 127 + 165 + 191 + 130}{10} = \dfrac{1528}{10} = 152.8$$

b) To find the median, you should always put the data in numerical order. With an even number of data points (10), the median is the mean of the middle two (the 5th and 6th terms).

Annabelle: $143, 148, 150, 151, \mathbf{152, 153}, 155, 158, 158, 160$. $$M_A = \dfrac{152+153}{2} = 152.5$$

Ben: $119, 120, 127, 130, \mathbf{141, 163}, 165, 180, 191, 192$. $$M_B = \dfrac{141+163}{2} = 152$$

c) The range is the difference between the largest and smallest data items.

Annabelle: $R_A = 160 - 143 = 17$

Ben: $R_B = 192 - 119 = 73$

d) The Interquartile Range (IQR) is $Q_3 - Q_1$. For 10 data points, $Q_1$ is the median of the lower half and $Q_3$ is the median of the upper half.

Annabelle: $Q_3 = 158, Q_1 = 150$. $$IQR_A = 158 - 150 = 8$$

Ben: $Q_3 = 180, Q_1 = 127$. $$IQR_B = 180 - 127 = 53$$

e) Comparison: While both bowlers have the same mean, their consistency is vastly different. Annabelle’s low IQR of 8 shows her scores are tightly clustered around the middle. Ben’s high IQR of 53 shows he is much more erratic; he is capable of very high scores but also very low ones.