Problem

Jse the accompanying radiation levels $\left(\right.$ in $\left.\frac{\mathrm{W}}{\mathrm{kg}}\right)$ for 50 different cell ohones. Find the percentile $P_{50}$.
\[
\begin{array}{llllllllll}
0.25 & 0.29 & 0.33 & 0.53 & 0.58 & 0.60 & 0.62 & 0.65 & 0.80 & 0.85 \\
0.86 & 0.91 & 0.92 & 0.94 & 0.96 & 0.97 & 0.98 & 1.05 & 1.06 & 1.07 \\
1.09 & 1.11 & 1.12 & 1.13 & 1.14 & 1.15 & 1.16 & 1.18 & 1.19 & 1.21 \\
1.24 & 1.24 & 1.24 & 1.25 & 1.26 & 1.28 & 1.29 & 1.31 & 1.32 & 1.32 \\
1.33 & 1.37 & 1.40 & 1.40 & 1.45 & 1.46 & 1.50 & 1.51 & 1.52 & 1.57
\end{array}
\]
$P_{50}=\square \frac{W}{\mathrm{~kg}}$ (Type an integer or a decimal. Do not round.)

Answer

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Answer

Final Answer: \(P_{50}=\boxed{1.145} \frac{W}{\mathrm{~kg}}\)

Steps

Step 1 :The percentile $P_{50}$ is also known as the median. This is the value that separates the higher half from the lower half of a data sample. In other words, 50% of the observations are below this value and 50% are above.

Step 2 :To find the median, we need to sort the data in ascending order and then find the middle value. If the number of observations is even, the median is the average of the two middle numbers.

Step 3 :Given radiation levels are already sorted in ascending order.

Step 4 :Since the number of observations is 50, which is an even number, the median is the average of the 25th and 26th observations.

Step 5 :The 25th and 26th observations are 1.14 and 1.15 respectively.

Step 6 :So, the median is the average of 1.14 and 1.15, which is 1.145.

Step 7 :Final Answer: \(P_{50}=\boxed{1.145} \frac{W}{\mathrm{~kg}}\)

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