Question

A random variable X is uniformly distributed over the interval (48, 96).
Find the following probabilities.
a) P(50 < X < 70)
b) P(X < 75)
c) P(X > 90)
d) P(X < 60 or X > 80)

Answer 1

Answer:

a) $P(50 < X<70)= P(X<70) -P(X<50) = \frac{70-48}{48} -\frac{50-48}{48} = 0.417$

b) $P(X<75)= \frac{75-48}{48} = 0.5625$

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

d) $P(X<60)$

And using the CDF we got:

$P(X<60)= \frac{60-48}{48} = 0.25$

$P(X>80)$

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:

$P(50 < X<70)$

And we can find this probability with this difference:

$P(50 < X<70)= P(X<70) -P(X<50)$

And replacing we got:

$P(50 < X<70)= P(X<70) -P(X<50) = \frac{70-48}{48} -\frac{50-48}{48} = 0.417$

Part b

We want this probability:

$P(X<75)$

And using the CDF we got:

$P(X<75)= \frac{75-48}{48} = 0.5625$

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

$P(X<60)$

And using the CDF we got:

$P(X<60)= \frac{60-48}{48} = 0.25$

$P(X>80)$

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