Answer:
(a) 3.75
(b) 2.00083
(c) 0.4898
Step-by-step explanation:
It is provided that X has a continuous uniform distribution over the interval [1.3, 6.2].
(a)
Compute the mean of X as follows:
[tex]\mu_{X}=\frac{a+b}{2}=\frac{1.3+6.2}{2}=3.75[/tex]
(b)
Compute the variance of X as follows:
[tex]\sigm^{2}_{X}=\frac{(b-a)^{2}}{12}=\frac{(6.2-1.3)^{2}}{12}=2.00083[/tex]
(c)
Compute the value of P(X < 3.7) as follows:
[tex]P(X < 3.7)=\int\limits^{3.7}_{1.3}{\frac{1}{6.2-1.3}}\, dx\\\\=\frac{1}{4.9}\times [x]^{3.7}_{1.3}\\\\=\frac{3.7-1.3}{4.9}\\\\\approx 0.4898[/tex]
Thus, the value of P(X < 3.7) is 0.4898.
