Question

(Uniform) Suppose X follows a continuous uniform distribution from 4 to 11. (a) Write down the PDF of X. (b) Find P(X ≤ 7). Round your answer to 4 decimals. (c) Find P(5 < X ≤ 7). Round your answer to 4 decimals. (d) Determine the conditional probability P(X > 5 | X ≤ 7). Round your answer to 4 decimals.

Answer 1

Answer:

a) $f(x)= \frac{1}{11-4}= \frac{1}{7}, 4 \leq x \leq 11$

b) $P(X\leq 7) = F(7) = \frac{7-4}{11-4}= 0.4286$

c) $P(5 < X \leq 7)= F(7) -F(5) = \frac{7-4}{7} -\frac{5-4}{7}= 0.2857$

d) $P(X >5 | X \leq 7)$

And we can find this probability with this formula from the Bayes theorem:

$P(X >5 | X \leq 7)= \frac{P(X>5 \cap X \leq 7)}{P(X \leq 7)}= \frac{P(5$

Step-by-step explanation:

For this case we assume that the random variable X follows this distribution:

$X \sim Unif (a=4, b =11)$

Part a

The probability density function is given by the following expression:

$f(x) = \frac{1}{b-a} , a \leq x \leq b$

$f(x)= \frac{1}{11-4}= \frac{1}{7}, 4 \leq x \leq 11$

Part b

We want this probability:

$P(X \leq 7)$

And we can use the cumulative distribution function given by:

$F(x) = \frac{x-a}{b-a}= \frac{x-4}{11-4}$

And replacing we got:

$P(X\leq 7) = F(7) = \frac{7-4}{11-4}= 0.4286$

Part c

We want this probability:

$P(5 < X \leq 7)$

And we can use the CDF again and we have:

$P(5 < X \leq 7)= F(7) -F(5) = \frac{7-4}{7} -\frac{5-4}{7}= 0.2857$

Part d

We want this conditional probabilty:

$P(X >5 | X \leq 7)$

And we can find this probability with this formula from the Bayes theorem:

$P(X >5 | X \leq 7)= \frac{P(X>5 \cap X \leq 7)}{P(X \leq 7)}= \frac{P(5$