Question

a sample of 100 clients of an exercise facility was selected. let x = the number of days per week that a randomly selected client uses the exercise facility.

x	frequency
1	13
2	30
3	29
4	10
5	7
6	8

find the 80th percentile.

Explanation:

Step1: Calculate cumulative frequency

First, we calculate the cumulative frequency. The cumulative frequency for each \( X \) is the sum of frequencies up to that \( X \).
- For \( X = 0 \): Cumulative Frequency (\( CF \)) = \( 3 \)
- For \( X = 1 \): \( CF = 3 + 13 = 16 \)
- For \( X = 2 \): \( CF = 16 + 30 = 46 \)
- For \( X = 3 \): \( CF = 46 + 29 = 75 \)
- For \( X = 4 \): \( CF = 75 + 10 = 85 \)
- For \( X = 5 \): \( CF = 85 + 7 = 92 \)
- For \( X = 6 \): \( CF = 92 + 8 = 100 \)

Step2: Find the position of the 80th percentile

The formula for the position of the \( p \)-th percentile is \( i=\frac{p}{100}\times n \), where \( n \) is the total number of observations. Here, \( p = 80 \) and \( n = 100 \). So, \( i=\frac{80}{100}\times100 = 80 \).

Step3: Identify the class with cumulative frequency ≥ 80

We look for the smallest \( X \) such that the cumulative frequency is at least 80.
- For \( X = 3 \), \( CF = 75 < 80 \)
- For \( X = 4 \), \( CF = 85 \geq 80 \)

So the 80th percentile is the smallest \( X \) where the cumulative frequency is at least 80, which is \( X = 4 \). Wait, but let's check again. Wait, the cumulative frequency at \( X = 3 \) is 75, at \( X = 4 \) is 85. The position \( i = 80 \) falls within the class where \( X = 4 \) (since 75 < 80 ≤ 85). But wait, actually, in percentile calculation for discrete data, sometimes we use the formula: the \( p \)-th percentile is the smallest value \( x \) such that \( \text{Cumulative Frequency}(x)\geq\frac{p}{100}\times n \). Here, \( \frac{80}{100}\times100 = 80 \). The cumulative frequency at \( X = 3 \) is 75, at \( X = 4 \) is 85. So the smallest \( x \) with \( CF(x)\geq80 \) is \( x = 4 \)? Wait, no, wait. Wait, let's recalculate the cumulative frequencies again:

Wait, the frequencies are:

\( X = 0 \): 3, \( CF = 3 \)

\( X = 1 \): 13, \( CF = 3 + 13 = 16 \)

\( X = 2 \): 30, \( CF = 16 + 30 = 46 \)

\( X = 3 \): 29, \( CF = 46 + 29 = 75 \)

\( X = 4 \): 10, \( CF = 75 + 10 = 85 \)

\( X = 5 \): 7, \( CF = 85 + 7 = 92 \)

\( X = 6 \): 8, \( CF = 92 + 8 = 100 \)

So \( i = 80 \). We need to find the value of \( X \) such that the cumulative frequency is at least 80. The cumulative frequency at \( X = 3 \) is 75, which is less than 80. The cumulative frequency at \( X = 4 \) is 85, which is greater than 80. So the 80th percentile is the smallest \( X \) where \( CF(X)\geq80 \), which is \( X = 4 \)? Wait, but wait, maybe I made a mistake. Wait, another way: the 80th percentile is the value where 80% of the data is less than or equal to it. So we need to find the smallest \( x \) such that at least 80% of the data is ≤ \( x \). So 80% of 100 is 80. So we need the smallest \( x \) where the number of data points ≤ \( x \) is at least 80. The number of data points ≤ 3 is 75, ≤ 4 is 85. So 85 ≥ 80, so the smallest \( x \) with ≤ \( x \) having at least 80 data points is 4? Wait, but let's check the cumulative frequency again. Wait, maybe I messed up the cumulative frequency. Wait, \( X = 0 \): 3, \( X = 1 \): 13 (total 16), \( X = 2 \): 30 (total 46), \( X = 3 \): 29 (total 75), \( X = 4 \): 10 (total 85), \( X = 5 \): 7 (total 92), \( X = 6 \): 8 (total 100). So when \( X = 4 \), the cumulative frequency is 85, which means 85% of the data is ≤ 4. So the 80th percentile is the smallest \( x \) where the cumulative frequency is at least 80. Since 75 (at \( X = 3 \)) is less than 80, and 85 (at \( X = 4 \)) is at least 80, so the 80th percentile is 4? Wait, but wait, maybe the correct answer is 4? Wait, no, wait, let's check with another approach. The formula for percentile in discrete data: the \( p \)-th percentile is the value \( x \) such that \( \text{Frequency}(x) \) and \( \text{Cumulative Frequency}(x) \) satisfy \( \text{Cumulative Frequency}(x - 1) < \frac{p}{100}n \leq \text{Cumulative Frequency}(x) \). Here, \( \frac{p}{100}n = 80 \). \( \text{Cumulative Frequency}(x - 1) \) when \( x = 4 \) is \( \text{Cumulative Frequency}(3) = 75 \), and \( \text{Cumulative Frequency}(4) = 85 \). Since \( 75 < 80 \leq 85 \), so \( x = 4 \).

Wait, but let's check the cumulative frequency again. Wait, maybe I made a mistake in the cumulative frequency. Let's recalculate:

- \( X = 0 \): frequency 3, cumulative 3
- \( X = 1 \): 13, cumulative 3 + 13 = 16
- \( X = 2 \): 30, cumulative 16 + 30 = 46
- \( X = 3 \): 29, cumulative 46 + 29 = 75
- \( X = 4 \): 10, cumulative 75 + 10 = 85
- \( X = 5 \): 7, cumulative 85 + 7 = 92
- \( X = 6 \): 8, cumulative 92 + 8 = 100

Yes, that's correct. So the 80th percentile is the smallest \( x \) where cumulative frequency ≥ 80. So 75 (at x=3) < 80, 85 (at x=4) ≥ 80. So x=4. Wait, but wait, another thought: maybe the question is using the method where we interpolate? But for discrete data, interpolation is not usually done. So in discrete data, the percentile is the smallest value with cumulative frequency ≥ (p/100)*n. So here, (80/100)*100 = 80. So the smallest x with CF(x) ≥ 80 is x=4, since CF(3)=75 < 80 and CF(4)=85 ≥ 80.

Wait, but let's check with an example. Suppose we have data: 1,2,3,4,5 with frequencies 1 each. The 80th percentile: n=5, i=0.8*5=4. The cumulative frequencies: 1,2,3,4,5. So the 4th position (i=4) corresponds to x=4, since cumulative frequency at x=4 is 4 ≥ 4, and x=3 has cumulative frequency 3 < 4. So that's correct. So in our case, i=80, n=100, so i=80. The cumulative frequency at x=3 is 75 < 80, at x=4 is 85 ≥ 80. So the 80th percentile is 4.

Wait, but wait, the cumulative frequency at x=4 is 85, which is the number of data points ≤4. So 85% of the data is ≤4, so the 80th percentile is 4, because 80% of the data is ≤4 (since 85 ≥ 80). So yes, the 80th percentile is 4.

Wait, but let me check again. Wait, the frequency for x=4 is 10, so cumulative frequency at x=4 is 75+10=85. So 85 clients use the facility 4 or fewer days. So 85% of the clients use it 4 or fewer days. So the 80th percentile is the smallest x where at least 80% of the clients use it x or fewer days. Since 75% (at x=3) use it 3 or fewer days, and 85% (at x=4) use it 4 or fewer days, the 80th percentile is 4.

Answer:

\( \boxed{4} \)