The variance of the data \( 2, 4, 6, 8, 10 \) is:
Step 1: To calculate the variance, we use the formula: \[ {Variance} = \frac{1}{n} \sum_{i=1}^{n} (x_i - \mu)^2, \] where \( \mu \) is the mean of the data and \( n \) is the number of data points.
Step 2: Calculate the mean: \[ \mu = \frac{2 + 4 + 6 + 8 + 10}{5} = \frac{30}{5} = 6. \]
Step 3: Compute the squared differences from the mean: \[ (2 - 6)^2 = 16, \quad (4 - 6)^2 = 4, \quad (6 - 6)^2 = 0, \quad (8 - 6)^2 = 4, \quad (10 - 6)^2 = 16. \]
Step 4: Compute the variance: \[ {Variance} = \frac{16 + 4 + 0 + 4 + 16}{5} = \frac{40}{5} = 8. \]
Based upon the results of regular medical check-ups in a hospital, it was found that out of 1000 people, 700 were very healthy, 200 maintained average health and 100 had a poor health record.
Let \( A_1 \): People with good health,
\( A_2 \): People with average health,
and \( A_3 \): People with poor health.
During a pandemic, the data expressed that the chances of people contracting the disease from category \( A_1, A_2 \) and \( A_3 \) are 25%, 35% and 50%, respectively.
Based upon the above information, answer the following questions:
(i) A person was tested randomly. What is the probability that he/she has contracted the disease?}
(ii) Given that the person has not contracted the disease, what is the probability that the person is from category \( A_2 \)?