Find the median the first and third quartiles the interquartile range and my outliers for set of data 81,69,90,79,59,125,95,116,183,97,92 ,50,122,55,92

Step-by-step explanation:

Find the median of the first and third quartiles

81, 69 ,90, 79, 59, 125, 95, 116, 183, 97, 92, 50, 122, 55, 92

50, 55, 59, 69, 79, 81, 90,   92,     92, 95, 97, 116, 122, 125, 183.

The first quartile: 50, 55, 59, 69, 79, 81, 90: median of the first quartile is 69

The third quartile: 92, 95, 97, 116, 122, 125, 183: median of the third quartile is 116.

The interquartile range is the difference between the median of both the first and third quartile:  which is 116 - 69 = 47.

There are no outliers in this set of data