contestada

777,498,619,379,895,1,256,1052 a. What is the mediam of the first half of data?(first quatile. b. What's the mediam of second half of the data?(third quartile) c. What is the interquartile range ?

Respuesta :

MrRoyal

Answer:

[tex]Q_1 = 498[/tex]

[tex]Q_3 =1052[/tex]

Step-by-step explanation:

Given

[tex]Data: 777,\ 498,\ 619,\ 379,\ 895,\ 1,256,\ 1052[/tex]

[tex]n = 7[/tex]

First, the data has to be arranged:

[tex]Data: 379,\ 498,\ 619,\ 777,\ 895,\ 1052,\ 1256[/tex]

Solving (a): Lower Quartile

This is calculated as:

[tex]Q_1 =\frac{1}{4} * (N + 1)th[/tex]

[tex]Q_1 =\frac{1}{4} * (7 + 1)th[/tex]

[tex]Q_1 =\frac{1}{4} * 8th[/tex]

[tex]Q_1 = 2nd[/tex]

The 2nd data is 498

Hence:

[tex]Q_1 = 498[/tex]

Solving (b): Third Quartile

This is calculated as:

[tex]Q_3 =\frac{3}{4} * (N + 1)th[/tex]

[tex]Q_3 =\frac{3}{4} * (7 + 1)th[/tex]

[tex]Q_3 =\frac{3}{4} * (8)th[/tex]

[tex]Q_3 =6th[/tex]

The 6th data is 1052

Hence:

[tex]Q_3 =1052[/tex]

Solving (c): Interquartile Range

This is calculated as:

[tex]IR = Q_3 - Q_1[/tex]

[tex]IR = 1052- 498[/tex]

[tex]IR = 554[/tex]

Otras preguntas