Let’s go through each part of this problem one by one.

Given Data

The random variable $X$ follows an exponential distribution with $\beta = 3$.

For an exponential distribution:

• The mean $E(X) = \beta$
• The standard deviation $SD(X) = \beta$
• The probability density function is $f(x) = \frac{1}{\beta} e^{-x / \beta}$ for $x \geq 0$.

Solution

(a) Complete the sentence:

The distribution of $X$ is skewed to the right (this is a property of the exponential distribution, which is right-skewed).

For the exponential distribution with $\beta = 3$:

• $E(X) = \beta = 3$
• $SD(X) = \beta = 3$

So, we have:

• Answer: The distribution of $X$ is skewed to the right, with $E(X) = 3$ and $SD(X) = 3$.

(b) Find $P(1 \leq X \leq 4)$

For an exponential distribution, the cumulative distribution function (CDF) is given by: $P(X \leq x) = 1 - e^{-x / \beta}$

To find $P(1 \leq X \leq 4)$, we calculate: $P(1 \leq X \leq 4) = P(X \leq 4) - P(X \leq 1)$

Using $\beta = 3$:

1. $P(X \leq 4) = 1 - e^{-4 / 3}$
2. $P(X \leq 1) = 1 - e^{-1 / 3}$

Calculating these values:

• $P(X \leq 4) \approx 1 - e^{-1.3333} \approx 1 - 0.2636 = 0.7364$
• $P(X \leq 1) \approx 1 - e^{-0.3333} \approx 1 - 0.7173 = 0.2827$

Thus, $P(1 \leq X \leq 4) \approx 0.7364 - 0.2827 = 0.4537$

Answer: $P(1 \leq X \leq 4) \approx 0.4537$

(c) Find the 25th percentile of $X$

The $p$-th percentile for an exponential distribution is given by: $X_p = -\beta \ln(1 - p)$

For the 25th percentile, $p = 0.25$ and $\beta = 3$: $X_{25} = -3 \ln(1 - 0.25) = -3 \ln(0.75)$

Calculating this: $X_{25} \approx -3 \times (-0.2877) = 0.8631$

Answer: $X_{25} \approx 0.86$

(d) Find $P(X > 9.5)$

The probability $P(X > x)$ for an exponential distribution is: $P(X > x) = e^{-x / \beta}$

Using $x = 9.5$ and $\beta = 3$: $P(X > 9.5) = e^{-9.5 / 3} = e^{-3.1667}$

Calculating this: $P(X > 9.5) \approx 0.0424$

Answer: $P(X > 9.5) \approx 0.0424$

Summary of Answers

1. (a) The distribution of $X$ is skewed to the right, with $E(X) = 3$ and $SD(X) = 3$.
2. (b) $P(1 \leq X \leq 4) \approx 0.4537$
3. (c) $X_{25} \approx 0.86$
4. (d) $P(X > 9.5) \approx 0.0424$

Would you like further details on any of these parts, or have any questions?

Additional Questions

1. How would the results change if $\beta$ was different?
2. Can the exponential distribution ever be symmetric?
3. What is the median of an exponential distribution with $\beta = 3$?
4. How do we interpret the 25th percentile in terms of the distribution?
5. Why is the exponential distribution often used to model waiting times?

Tip

For exponential distributions, the mean and standard deviation are always equal to $\beta$, which simplifies calculations when describing the spread of data.