Question

A vaccine, when it is administered to an individual, produces no side effects with probability \(\frac{4}{5}\), mild side effects with probability \(\frac{2}{15}\) and severe side effects with probability \(\frac{1}{15}\). Assume that the development of side effects is independent across individuals. The vaccine was administered to 1000 randomly selected individuals. If X₁ denotes the number of individuals who developed mild side effects and X₂ denotes the number of individuals who developed severe side effects, then the coefficient of variation of X₁ + X₂ equals __________ (round off to 2 decimal places)

Answer 1

Correct Answer: 0.05

1. Identify the Given Probabilities

Let the possible outcomes for an individual be:

No side effects ($E_0$): $P(\text{No side effects}) = p_0 = \frac{4}{5}$

Mild side effects ($E_1$): $P(\text{Mild side effects}) = p_1 = \frac{2}{15}$

Severe side effects ($E_2$): $P(\text{Severe side effects}) = p_2 = \frac{1}{15}$

Verify the sum of probabilities:

$$\frac{4}{5} + \frac{2}{15} + \frac{1}{15} = \frac{12}{15} + \frac{2}{15} + \frac{1}{15} = \frac{15}{15} = 1$$

(The probabilities are valid).

2. Define the Random Variable of Interest

We are interested in the random variable $Y = X_1 + X_2$.

$X_1$: Number of individuals with mild side effects.

$X_2$: Number of individuals with severe side effects.

Therefore, $Y$ represents the number of individuals who developed any side effects (mild OR severe).

The probability of an individual developing any side effect ($p$) is the sum of the probabilities of mild and severe effects:

$$p = p_1 + p_2 = \frac{2}{15} + \frac{1}{15} = \frac{3}{15} = \frac{1}{5}$$

Alternatively, this is $1 - P(\text{No side effects}) = 1 - \frac{4}{5} = \frac{1}{5}$.

3. Identify the Distribution

Since the vaccine is administered to $n = 1000$ independent individuals, and we are counting the number of "successes" (where success = developing a side effect), the random variable $Y = X_1 + X_2$ follows a Binomial Distribution:

$$Y \sim B(n, p)$$

where $n = 1000$ and $p = 0.2$ ($\frac{1}{5}$).

4. Calculate Mean and Variance

For a Binomial distribution $B(n, p)$:

Mean ($E[Y]$):

$$\mu = n \cdot p = 1000 \cdot \frac{1}{5} = 200$$

Variance ($Var(Y)$):

$$\sigma^2 = n \cdot p \cdot (1 - p)$$

$$\sigma^2 = 1000 \cdot \frac{1}{5} \cdot \left(1 - \frac{1}{5}\right)$$

$$\sigma^2 = 1000 \cdot \frac{1}{5} \cdot \frac{4}{5}$$

$$\sigma^2 = 1000 \cdot \frac{4}{25} = 40 \cdot 4 = 160$$

Standard Deviation ($\sigma$):

$$\sigma = \sqrt{160} \approx 12.649$$

5. Calculate the Coefficient of Variation (CV)

The Coefficient of Variation is defined as the ratio of the standard deviation to the mean:

$$CV = \frac{\sigma}{\mu}$$

Substitute the values:

$$CV = \frac{\sqrt{160}}{200}$$

$$CV = \frac{4\sqrt{10}}{200}$$

$$CV = \frac{\sqrt{10}}{50}$$

Using $\sqrt{10} \approx 3.1622$:

$$CV \approx \frac{3.1622}{50} \approx 0.06324$$

6. Final Answer

Rounding off to two decimal places:

$$CV \approx 0.06$$

Answer: 0.06