Answer:
[tex] E(X) = \frac{625+640}{2}=632.5[/tex]
[tex]Var(X)= \frac{(b-a)^2}{12}=\frac{(640-625)^2}{12}=18.75 \approx 18.8[/tex]
[tex] \mu= 632.5, \sigma^2 = 18.8[/tex]
Step-by-step explanation:
Let X the random variable that represent "The wavelengths of photosynthetically active radiations PAR ". And we know that the distribution of X is given by:
[tex]X\sim Uniform(a=625 ,b=640)[/tex]
The density function is given by:
[tex] f(x) = \frac{1}{b-a} , a \leq x \leq b[/tex]
And [tex] f(x)=0[/tex] for other case.
The mean for this case is given by:
[tex] E(X)=\frac{a+b}{2}[/tex]
And if we replace we got:
[tex] E(X) = \frac{625+640}{2}=632.5[/tex]
For the variance of the random variable X we can use the following formula:
[tex]Var(X)= \frac{(b-a)^2}{12}=\frac{(640-625)^2}{12}=18.75 \approx 18.8[/tex]
So then the answer would be:
[tex] \mu= 632.5, \sigma^2 = 18.8[/tex]
