Answer:
0.5
Explanation:
Sort the given data in ascending order:
47, 48, 50, 52, 53, 53, 55, 68
Possible outlier is number 68. Check whether this number is an outlier:
![Q_1=49\\ \\Q_2=52.5 \\ \\Q_3=54](https://img.qammunity.org/2020/formulas/mathematics/middle-school/7if9as1m2ow41sl31b99qgec9azzzc1cb8.png)
The interquartile range is
![Q_3-Q_1=54-49=5](https://img.qammunity.org/2020/formulas/mathematics/middle-school/i4yev5whbfhbkhrcurhc5s2zfwkoqiqlti.png)
Multiply it by 1.5:
![1.5\cdot 5=7.5](https://img.qammunity.org/2020/formulas/mathematics/middle-school/3fn4sslaf2ss3rf2jjxi2qzu214wl01d9v.png)
and add to third quartile:
![7.5+54=61.5](https://img.qammunity.org/2020/formulas/mathematics/middle-school/h94ts5wm0mzbouw892fqgwmt0qfmaw0gka.png)
Since
number 68 is an outlier.
The median of the sample with outlier is
![Q_2=(52+53)/(2)=52.5](https://img.qammunity.org/2020/formulas/mathematics/middle-school/narkpzxbamvh1cc1pf6c6shuz42vpzdc5c.png)
The median of the sample without outlier is 52
The difference between the median of the data, including the possible outlier and excluding the possible outlier is 52.5-52=0.5