We are given a frequency distribution table for the ages of a sample of 25 students. We need to complete the table by calculating the cumulative frequencies and relative frequencies for each age group. We are given the frequencies for the age groups 10-15, 16-21, 22-32, 33-43, and 44-60. The relative frequency for the age group 22-32 is also given as 0.32.

Probability and StatisticsFrequency DistributionCumulative FrequencyRelative FrequencyData Analysis
2025/5/3

1. Problem Description

We are given a frequency distribution table for the ages of a sample of 25 students. We need to complete the table by calculating the cumulative frequencies and relative frequencies for each age group. We are given the frequencies for the age groups 10-15, 16-21, 22-32, 33-43, and 44-
6

0. The relative frequency for the age group 22-32 is also given as 0.

3
2.

2. Solution Steps

First, let's define the variables:
fif_i = frequency of the ii-th age group
cficf_i = cumulative frequency of the ii-th age group
rfirf_i = relative frequency of the ii-th age group
nn = total number of students = 25
The formula for cumulative frequency is:
cfi=k=1ifkcf_i = \sum_{k=1}^{i} f_k
The formula for relative frequency is:
rfi=finrf_i = \frac{f_i}{n}
We are given n=25n = 25.
Age 10-15: f1=2f_1 = 2.
cf1=f1=2cf_1 = f_1 = 2.
rf1=f1n=225=0.08rf_1 = \frac{f_1}{n} = \frac{2}{25} = 0.08
Age 16-21: f2=4f_2 = 4.
cf2=cf1+f2=2+4=6cf_2 = cf_1 + f_2 = 2 + 4 = 6.
rf2=f2n=425=0.16rf_2 = \frac{f_2}{n} = \frac{4}{25} = 0.16
Age 22-32: rf3=0.32rf_3 = 0.32.
rf3=f3nrf_3 = \frac{f_3}{n}, so f3=rf3n=0.3225=8f_3 = rf_3 * n = 0.32 * 25 = 8.
cf3=cf2+f3=6+8=14cf_3 = cf_2 + f_3 = 6 + 8 = 14.
Age 33-43: Let the frequency be f4f_4.
We are given the frequency for Age 44-60 is f5=5f_5=5.
We know that the sum of frequencies is
2

5. So, $f_1 + f_2 + f_3 + f_4 + f_5 = 25$.

2+4+8+f4+5=252 + 4 + 8 + f_4 + 5 = 25.
19+f4=2519 + f_4 = 25.
f4=2519=6f_4 = 25 - 19 = 6.
cf4=cf3+f4=14+6=20cf_4 = cf_3 + f_4 = 14 + 6 = 20.
rf4=f4n=625=0.24rf_4 = \frac{f_4}{n} = \frac{6}{25} = 0.24.
Age 44-60: f5=5f_5 = 5.
cf5=cf4+f5=20+5=25cf_5 = cf_4 + f_5 = 20 + 5 = 25.
rf5=f5n=525=0.20rf_5 = \frac{f_5}{n} = \frac{5}{25} = 0.20.
Completed table:
Age(years) | Frequency | Cumulative Frequency | Relative Frequency
---|---|---|---
10-15 | 2 | 2 | 0.08
16-21 | 4 | 6 | 0.16
22-32 | 8 | 14 | 0.32
33-43 | 6 | 20 | 0.24
44-60 | 5 | 25 | 0.20

3. Final Answer

Here is the completed table:
Age(years) | Frequency | Cumulative Frequency | Relative Frequency
---|---|---|---
10-15 | 2 | 2 | 0.08
16-21 | 4 | 6 | 0.16
22-32 | 8 | 14 | 0.32
33-43 | 6 | 20 | 0.24
44-60 | 5 | 25 | 0.20

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