Alright, let's solve the problem step-by-step:
First, we need to find the total number of people with iron deficiency.
From the table:
- Ears
- 20-30 years: [tex]\( 37 \)[/tex]
- Above 30 years: [tex]\( 24 \)[/tex]
- Total: [tex]\( 102 \)[/tex]
- 09
- 20-30 years: [tex]\( 43 \)[/tex]
- Above 30 years: [tex]\( 46 \)[/tex]
- Total: [tex]\( 198 \)[/tex]
- 50
- 20-30 years: [tex]\( 80 \)[/tex]
- Above 30 years: [tex]\( 70 \)[/tex]
- Total: [tex]\( 300 \)[/tex]
First, let's calculate the overall total of people with iron deficiency:
[tex]\[ 102 + 198 + 300 = 600 \][/tex]
Next, we need to determine the number of people with iron deficiency who are 20 years or older. This includes both those in the 20-30 years category as well as those above 30 years.
We sum the values for ages 20-30 and above 30 across all categories:
[tex]\[ (37 \, \text{[20-30]} + 24 \, \text{[Above 30]}) + (43 \, \text{[20-30]} + 46 \, \text{[Above 30]}) + (80 \, \text{[20-30]} + 70 \, \text{[Above 30]}) \][/tex]
[tex]\[ = (37 + 24) + (43 + 46) + (80 + 70) \][/tex]
[tex]\[ = 61 + 89 + 150 \][/tex]
[tex]\[ = 300 \][/tex]
Now, we need to calculate the probability that a person with an iron deficiency is 20 years or older. This probability is given by the ratio of the number of people 20 years or older with an iron deficiency to the total number of people with an iron deficiency.
[tex]\[ \text{Probability} = \frac{\text{Number of persons 20 years or older}}{\text{Total number of persons with iron deficiency}} \][/tex]
[tex]\[ = \frac{300}{600} \][/tex]
[tex]\[ = 1.0 \][/tex]
Therefore, the probability that a person with an iron deficiency is 20 years or older is:
[tex]\[ \boxed{1.0} \][/tex]
