Question

The time it takes to preform a task has a continuous uniform distribution between 43 min and 57 min. What is the the probability it takes between 52.3 and 54.7 min. Round to 4 decimal places.

Answer 1

Answer:

The values is  $P(52.3 < X < 54.7 ) = 0.17142$

Step-by-step explanation:

From the question we are told that

  The lower limit of the interval is  a = 43 minutes

   The upper limit of the interval is  b  =  57 minutes

Generally the probability distribution function of  uniform distribution is mathematically represented as

          $F(x) = \left \{ {{\frac{1}{(b - a)}\ \ \ for \ x \ \epsilon\ | a, b | } \atop {0 } \ \ \ \ \ \ otherwise} \right.$

Generally the probability it takes between 52.3 and 54.7 min is mathematically represented as

       $P(52.3 < X < 54.7 ) = \int\limits^{52.4}_{54.7} {F(x)} \, dx$

=>  $P(52.3 < X < 54.7 ) = \int\limits^{52.4}_{54.7} {\frac{1}{57- 43} } \, dx$

=>   $P(52.3 < X < 54.7 ) = \int\limits^{52.4}_{54.7} {\frac{1}{14} } \, dx$

=>   $P(52.3 < X < 54.7 ) = {\frac{1}{14} } \int\limits^{52.4}_{54.7} \, dx$

=>   $P(52.3 < X < 54.7 ) = {\frac{1}{14} } [x] \ | \left 54.7} \atop {52.3}} \right.$

=>   $P(52.3 < X < 54.7 ) = \frac{54.7 -52.3}{14}$

=>   $P(52.3 < X < 54.7 ) = 0.17142$