Question

A random variable X is uniformly distributed over the interval (48, 96).

Answer 1

Answer:

a) $P(50 < X$

b) $P(X$

c) $P(X>90)=1- \frac{90-48}{48} = 1-0.875 = 0.125$

d) $P(X$

And using the CDF we got:

$P(X$

$P(X>80)$

And using the CDF and the complement rule we got:

$P(X>80)=1- \frac{80-48}{48} = 1-0.667 = 0.333$

So then the probability required would be:

P(X < 60 or X > 80) = 0.25+0.333= 0.583

Step-by-step explanation:

For this case we define the random variable of interest X, and we know the distribution given by:

$X \sim Unif (a= 48, b=96)$

The density function is given by:

$f(x) = \frac{1}{96-48} = \frac{1}{48} , 48 \leq X \leq 96$

And the cumulative distribution function is given by:

$F(x) = \frac{x-48}{96-48} , 48 \leq X \leq 96$

Part a

We want this probability:

$P(50 < X$

And we can find this probability with this difference:

$P(50 < X$

And replacing we got:

$P(50 < X$

Part b

We want this probability:

$P(X$

And using the CDF we got:

$P(X$

Part c

We want this probability:

$P(X>90)$

And using the CDF and the complement rule we got:

$P(X>90)=1- \frac{90-48}{48} = 1-0.875 = 0.125$

Part d

$P(X$

And using the CDF we got:

$P(X$

$P(X>80)$

And using the CDF and the complement rule we got:

$P(X>80)=1- \frac{80-48}{48} = 1-0.667 = 0.333$

So then the probability required would be:

P(X < 60 or X > 80) = 0.25+0.333= 0.583