If x is uniformly distributed over the interval [ a , b ] then,
[tex]\begin{gathered} f(x)\text{ = }\frac{1}{b-a}\text{ , a }\leq\text{ x }\leq\text{ b} \\ f(x)\text{ = 0 , otherwise} \end{gathered}[/tex]
Also ,
[tex]\begin{gathered} \text{Mean = }\frac{a\text{ + b}}{2} \\ \text{Std deviation = }\sqrt[]{\frac{(b-a)^2}{12}} \end{gathered}[/tex]
It is given that ,
[tex]\begin{gathered} a\text{ = 9.25} \\ b\text{ = 9.75} \\ b\text{ - a = 9.75 - 9.25 = 0.5} \end{gathered}[/tex]
Therefore,
[tex]\begin{gathered} f(x)\text{ = }\frac{1}{0.5}\text{ 9.25 }\leq\text{ x }\leq\text{ 9.75} \\ f(x)\text{ = 0 otherwise} \end{gathered}[/tex]
(a)The distribution is as follows :
(b)The mean is calculated as,
[tex]\begin{gathered} \text{Mean = }\frac{a\text{ + b}}{2} \\ \text{Mean = }\frac{9.25\text{ + 9.75}}{2} \\ \text{Mean = 9.5} \end{gathered}[/tex]
Standard deviation is calculated as,
[tex]\begin{gathered} \text{Standard deviation = }\sqrt[]{\frac{(b-a)^2}{12}} \\ \text{Standard deviation = }\sqrt[]{\frac{(0.5)^2}{12}} \\ \text{Standard deviation }\approx\text{ 0.1443} \end{gathered}[/tex]
(c) The probability is calculated as,
[tex]\begin{gathered} P(\text{ atleast 9.4 points ) = P( x }\ge\text{ 9.4)} \\ P(\text{ atleast 9.4 points ) = }\int ^{9.75}_{9.4}(\frac{1}{0.5})dx \\ P(\text{ atleast 9.4 points ) = }\frac{9.75\text{ - 9.4}}{0.5} \\ P(\text{ atleast 9.4 points ) = 0.7} \end{gathered}[/tex]
(d) The probability is calculated as,
[tex]\begin{gathered} P(\text{between 9.45 and }9.75\text{ ) = P( 9.45 }\leq\text{ x }\leq\text{ 9.75 )} \\ P(\text{between 9.45 and }9.75\text{ ) = }\int ^{9.75}_{9.45}(\frac{1}{0.5})dx \\ P(\text{between 9.45 and }9.75\text{ ) =}\frac{9.75\text{ - 9.45}}{0.5} \\ P(\text{between 9.45 and }9.75\text{ ) = 0.6} \end{gathered}[/tex]
