Question

The length of time X, needed by an examinee of competition to complete a 1-hour exam, is a random variable with
PDF \(f(x)=\dfrac{6}{5}(x^2+x);0 \le x \le 1.\), The value of F(0.5) is:
1. \(\frac{4}{5}\)
2. \(\frac{2}{5}\)
3. \(\frac{3}{5}\)
4. \(\frac{1}{5}\)

Answer 1

Correct Answer - Option 4 : \(\frac{1}{5}\)

Given

F(x) = 6/5(x² + x); 0 ≤ x ≤ 1

Calculation

F(x) = \(\mathop \smallint \nolimits_0^x f\left( x \right)dx\)

⇒ \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaaeaaaaaaaaa8 % qadaqfWaqabSWdaeaapeGaaGimaaWdaeaaieWapeGaa8hEaaqdpaqa % a8qacqGHRiI8aaaaaa!3A60! \mathop \smallint \nolimits_0^x \)6/5(x² + x)dx

After integration we get,

⇒ 6/5(x³/3 + x²/2)

⇒ f(0.5) = (6/5)((0.5)³/3 - (0.5)²/2)

⇒ (6/5)(0.25/6 + 0.75/6)

⇒ (6/5)(1/6)

⇒ 1/5

∴ The value of F(0. 5) is:1/5