Question

In a study on survival time for ten patients following a new treatment for AIDS. The time in months were given thus: 24, 12, 8, 20, 3, 18, 24, 25,and 27. Determine the variance for the given data.

a. 17.9
b. 6.8
c. 24
d. 8

Answer 1

In a study on survival time for ten patients following a new treatment for AIDS. The time in months were given thus: 24, 12, 8, 20, 3, 18, 24, 25, and 27. Determine the variance for the given data.

Solution

The variance of any given set of data containing $n$ values can be calculated as:

$$V = \frac {\left[ \left(x _ {1} - x\right) ^ {2} + \left(x _ {2} - x\right) ^ {2} + \left(x _ {3} - x\right) ^ {2} + \ldots + \left(x _ {n} - x\right) ^ {2} \right]}{n}$$

Where:

$x, x_{2}, x_{3} \ldots x_{n}$ represent the $n$ values, and

$$x = \text{mean of } n \text{ values} = \frac {x _ {1} + x _ {2} + x _ {3} + \ldots + x _ {n}}{n}$$

From the given values we calculate the mean $x$ as:

$$x = \frac {24 + 12 + 8 + 20 + 3 + 18 + 24 + 25 + 27}{9} = 17.9$$

And

$$(x _ {1} - x) ^ {2} = (24 - 17.9) ^ {2} = 37.21; (x _ {2} - x) ^ {2} = (12 - 17.9) ^ {2} = 34.81$$ $$(x _ {3} - x) ^ {2} = (8 - 17.9) ^ {2} = 98.01; (x _ {4} - x) ^ {2} = (20 - 17.9) ^ {2} = 4.41$$ $$(x _ {5} - x) ^ {2} = (3 - 17.9) ^ {2} = 222.01; (x _ {6} - x) ^ {2} = (18 - 17.9) ^ {2} = 0.01$$ $$(x _ {7} - x) ^ {2} = (24 - 17.9) ^ {2} = 37.21; (x _ {8} - x) ^ {2} = (25 - 17.9) ^ {2} = 50.41$$ $$(x _ {9} - x) ^ {2} = (27 - 17.9) ^ {2} = 82.81$$

And

$$V = \frac {\left[ \left(x _ {1} - x\right) ^ {2} + \left(x _ {2} - x\right) ^ {2} + \left(x _ {3} - x\right) ^ {2} + \ldots + \left(x _ {9} - x\right) ^ {2} \right]}{9}$$ $$V = \frac {37.21 + 34.81 + 98.01 + 4.41 + 222.01 + 0.01 + 37.21 + 50.41 + 82.81}{9} = 63$$

Answer: 63.