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Here is a sample data set.270.4 276.1 279.2 305.1 307.8 332.3338.6 346.9 381.5 403.6410.5412.4452.7 452.7 452.7 454.2454.5456.9458.4 459.3 459.8460.8464.1466.3466.4 466.9 468.3468.3 472.3479.5507.8511.4 527.3490.9 495.7 500.7529.2 534.3 535.9539.5 539.5539.5540.6 541.8 541.9 546.8551552.4554567.7 573.1 575.9 596.1617.6I need the first quartile and the 3rd

Here Is A Sample Data Set2704 2761 2792 3051 3078 33233386 3469 3815 4036410541244527 4527 4527 4542454545694584 4593 45984608464146634664 4669 46834683 4723479 class=

Sagot :

Explanation

To find the 1st and 3rd quartile, we must follow the following steps:

1) We order the data from lowest to highest values ✓

2) We look for the median, which gives us the 2nd quartile:

[tex]Q_2=\frac{468.3+468.3}{2}=468.3.[/tex]

To compute the median, we use the following rule:

• If the size of the data set is ,odd,, do not include the median when finding the first and third quartiles.

,

• If the size of the data set is ,even,, the median is the average of the middle 2 values in the data set. Add those 2 values, and then divide by 2. The median splits the data set into lower and upper halves and is the value of the second quartile Q2.

In this case, the number of elements is n = 54, so it is even.

The median splits the data into two halves.

3) The lower quartile Q1 is the median of the lower half of the data:

[tex]Q_1=452.7.[/tex]

4) The upper quartile Q3 is the median of the upper half of the data.

[tex]Q_3=539.5.[/tex]Answer

First quartile: Q_1 = 452.7

Third quartile: Q_3 = 539.5

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