Question

A uniform probability density function of a random variable is defined in the interval [a.
9) and the value for the density function in this interval is 0.5. Find the value of a
​

Answer 1

Given:

The value for the density function in the interval is 0.5

To Find: The value of a

Solution:

The uniform distribution function is defined by

$f(s) = \begin{cases}\dfrac{1}{b-a} \ &;\ a \leq x \leq b\\\ 0\ \ & ; \ otherwise \end{cases}$

The given interval is $[a,b]=[a,9]$

The Value of the density function is $f(x)=0.5$

$Then\ f(x)=\dfrac{1}{b-a}\\0.5=\dfrac{1}{9-a}\\\\0.5(9-a)=1\\4.5-0.5a=1\\a=\dfrac{3.5}{0.5}\\=7.$

The value of a is $7.$

Answer 2

Answer:

The value for the density function in the interval is 0.5

To Find: The value of a

Solution:

The uniform distribution function is defined by

\begin{gathered}f(s) = \begin{cases}\dfrac{1}{b-a} \ &;\ a \leq x \leq b\\\ 0\ \ & ; \ otherwise \end{cases}\end{gathered}

f(s)=

⎩

⎨

⎧

b−a

1

0

; a≤x≤b

; otherwise

The given interval is [a,b]=[a,9][a,b]=[a,9]

The Value of the density function is f(x)=0.5f(x)=0.5

\begin{gathered}Then\ f(x)=\dfrac{1}{b-a}\\0.5=\dfrac{1}{9-a}\\\\0.5(9-a)=1\\4.5-0.5a=1\\a=\dfrac{3.5}{0.5}\\=7.\end{gathered}

Then f(x)=

b−a

1

0.5=

9−a

1

0.5(9−a)=1

4.5−0.5a=1

a=

0.5

3.5

=7.

The value of a is 7.7.