use the accompanying radiation levels (in $\frac{w}{kg}$) for 50 different cell phones. find the quartile…

use the accompanying radiation levels (in $\frac{w}{kg}$) for 50 different cell phones. find the quartile $q_{3}$\n$q_{3}=square\frac{w}{kg}$ (type an integer or decimal rounded to two decimal places as needed.)

use the accompanying radiation levels (in $\frac{w}{kg}$) for 50 different cell phones. find the quartile $q_{3}$\n$q_{3}=square\frac{w}{kg}$ (type an integer or decimal rounded to two decimal places as needed.)

Answer

Explanation:

Step1: Sort the data

The data is already sorted in ascending - order as given in the problem.

Step2: Calculate the position of Q3

The formula to find the position of the third - quartile $Q_3$ for a data set of size $n$ is $i=\frac{3(n + 1)}{4}$. Here, $n = 50$. So, $i=\frac{3\times(50+1)}{4}=\frac{3\times51}{4}=38.25$.

Step3: Interpolate to find Q3

Since the position $i = 38.25$, the third - quartile is located between the 38th and 39th ordered data values. Let $x_{38}$ and $x_{39}$ be the 38th and 39th ordered data values. The 38th value is $1.31$ and the 39th value is $1.32$. $Q_3=x_{38}+(i - \lfloor i\rfloor)(x_{39}-x_{38})$, where $\lfloor i\rfloor$ is the greatest integer less than or equal to $i$. $\lfloor38.25\rfloor = 38$. $Q_3=1.31+(38.25 - 38)(1.32 - 1.31)=1.31 + 0.25\times0.01=1.31+0.0025 = 1.3125\approx1.31$.

Answer:

$1.31$