Question

use the round-trip airfare table below to determine the percentile rank of $1,133.
round-trip airfare
chicago to orlando
2 adults, 2 children | frequency | relative frequency | cumulative frequency | relative cumulative frequency
$919 | 4 | 0.050 | 4 | 0.050
$951 | 2 | 0.025 | 6 | 0.075
$999 | 3 | b. | 9 | 0.113
$1,005 | 12 | 0.150 | 21 | 0.263
$1,053 | 7 | 0.088 | d. | 0.350
$1,079 | 2 | 0.025 | 30 | 0.375
$1,133 | 4 | 0.050 | 34 | f.
$1,157 | 20 | c. | 54 | 0.675
$1,205 | 5 | 0.063 | 59 | 0.738
$1,209 | 7 | 0.088 | e. | 0.825
$1,213 | 2 | 0.025 | 68 | 0.850
$1,265 | 12 | 0.150 | 80 | g.
total | a. | ... | ... | ...

Explanation:

Step1: Recall Percentile Rank Formula

The formula for percentile rank of a value \( x \) is:
\( \text{Percentile Rank} = \frac{\text{Number of values less than } x + 0.5 \times \text{Number of values equal to } x}{\text{Total number of values}} \times 100 \)

Step2: Find Total Number of Values

Sum all frequencies. From the table, total frequency \( a = 4 + 2 + 3 + 12 + 7 + 2 + 4 + 20 + 5 + 7 + 2 + 12 \).
Calculating: \( 4+2=6 \); \( 6+3=9 \); \( 9+12=21 \); \( 21+7=28 \); \( 28+2=30 \); \( 30+4=34 \); \( 34+20=54 \); \( 54+5=59 \); \( 59+7=66 \); \( 66+2=68 \); \( 68+12=80 \). So total \( n = 80 \).

Step3: Count Values Less Than $1,133$

Check cumulative frequency before $1,133$. The cumulative frequency for $1,079$ is $30$. So values less than $1,133$: \( 30 \).

Step4: Count Values Equal to $1,133$

Frequency for $1,133$ is $4$.

Step5: Apply Percentile Rank Formula

Substitute into the formula:
\( \text{Percentile Rank} = \frac{30 + 0.5 \times 4}{80} \times 100 \)
Simplify numerator: \( 30 + 2 = 32 \)
Then: \( \frac{32}{80} \times 100 = 0.4 \times 100 = 40 \)

Answer:

The percentile rank of $1,133$ is \( 40 \).