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 ?

Answers

Answer 1

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]


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