Question

x∽U(-3,3) , P(x > k)=1/3 ആണെങ്കിൽ k എത്ര ?

• A. 1
• B. -1
• C. 0
• D. 2

Answer 1

Answer:

A. 1

$x∽U(-3,3)$

$f(x)=\frac{1}{b-a}=\frac{1}{3+3}=\frac{1}{6}$

P(x > k)=1/3

$\int_k^3f(x)dx=\int_k^3\frac{1}{6}dx=\frac{1}{3}$

$=\frac{1}{6} [x]_k^3 = \frac{1}{6}(3-k)=\frac{1}{3}$

$3-k =2$

$k=3-2=1$