College

Find the [tex]34^{\text{th}}[/tex] percentile, [tex]P_{34}[/tex], from the following data.

[tex]\[

\begin{array}{|c|c|c|c|c|}

\hline

10.1 & 10.7 & 11.3 & 12.0 & 12.1 \\

\hline

13.3 & 14.2 & 14.7 & 17.8 & 18.1 \\

\hline

18.4 & 18.7 & 18.8 & 19.2 & 20.2 \\

\hline

21.4 & 22.0 & 25.5 & 26.7 & 28.0 \\

\hline

28.6 & 31.1 & 31.8 & 32.5 & 34.0 \\

\hline

34.5 & 36.0 & 36.7 & 37.9 & 38.6 \\

\hline

41.7 & 42.2 & 42.9 & 44.7 & 45.3 \\

\hline

\end{array}

\][/tex]

[tex]P_{34} =[/tex] [tex]\square[/tex]

Answer :

We are given the data set (already in increasing order) with a total of
[tex]$$ n = 35 \text{ data points}. $$[/tex]
To find the [tex]$34^{\text{th}}$[/tex] percentile, [tex]$P_{34}$[/tex], we first determine the position in the data using the formula
[tex]$$ \text{Position} = \frac{P}{100} (n+1), $$[/tex]
where [tex]$P = 34$[/tex]. Substituting the values we have:
[tex]$$ \text{Position} = \frac{34}{100} (35+1) = 0.34 \times 36 = 12.24. $$[/tex]

The position [tex]$12.24$[/tex] tells us that [tex]$P_{34}$[/tex] lies between the [tex]$12^{\text{th}}$[/tex] and [tex]$13^{\text{th}}$[/tex] data values. Let:

  • [tex]$L$[/tex] be the [tex]$12^{\text{th}}$[/tex] value,
  • [tex]$H$[/tex] be the [tex]$13^{\text{th}}$[/tex] value.

Looking at the ordered data, we find:
[tex]$$ L = 18.7 \quad \text{and} \quad H = 18.8. $$[/tex]

We then perform linear interpolation. The fractional part is given by:
[tex]$$ f = 12.24 - 12 = 0.24. $$[/tex]

Thus, the [tex]$34^{\text{th}}$[/tex] percentile is computed as:
[tex]$$ P_{34} = L + f \cdot (H - L). $$[/tex]
Substitute the values:
[tex]$$ P_{34} = 18.7 + 0.24 \cdot (18.8 - 18.7) = 18.7 + 0.24 \cdot 0.1. $$[/tex]

Evaluating the expression:
[tex]$$ 0.24 \cdot 0.1 = 0.024, $$[/tex]
so
[tex]$$ P_{34} = 18.7 + 0.024 = 18.724. $$[/tex]

Therefore, the [tex]$34^{\text{th}}$[/tex] percentile is
[tex]$$ \boxed{18.724}. $$[/tex]