Question

The time (in minutes) for a lab assistant to prepare the equipment for a certain experiment is a random variable taking values between 25 and 35 minutes with p.d.f. $f(x) = \left\{ \begin{array}{ll} \frac{1}{10} & \quad ,\ \ \ 25\leq x \leq 35; \\ 0 & \quad ,\ \ \ otherwise. \end{array} \right$ What is the probability that preparation time exceeds 33 minutes ? Also find the c.d.f. of X.

Answer 1

Answer:

Probability that preparation time exceeds 33 minutes = $\mathbf{\frac{1}{5}}$

=> F(x) = $\mathbf{\frac{(x - 25)}{10}}$

Step-by-step explanation:

In this question,

We have been asked what is the probability that the preparation time exceeds 33 minutes

Therefore, According to the question,

$\int\limits^{35}_{33} {\frac{1}{10}} \, dx$

= $^{35}_{33}|\frac{x}{10}|$

= $\frac{35}{10}\ -\ \frac{33}{10}$

= $\frac{2}{10}$

= $\frac{1}{5}$

Probability that preparation time exceeds 33 minutes = $\mathbf{\frac{1}{5}}$

F(x) = $\int\limits^x_{25} {\frac{1}{10}} \, dx$

F(x) = $^{x}_{25}|\frac{x}{10}|$

F(x) = $\frac{x}{10}\ -\ \frac{25}{10}$

=> F(x) = $\mathbf{\frac{(x - 25)}{10}}$

Answer 2

Given : The time (in minutes) for a lab assistant to prepare the equipment for a certain experiment is a random variable taking values between 25 and 35 minutes with p.d.f.

To Find : What is the probability that preparation time exceeds 33 minutes ?  c.d.f. of X.

Solution

probability that preparation time exceeds 33 minutes

$\int\limits^{35}_{33} {\frac{1}{10}} \, dx$

$=\left[\dfrac{x}{10}\right]\limits^{35}_{33}$

= 35/10 - 33/10

= 2/10

= 1/5

probability that preparation time exceeds 33 minutes  =  1/5

F(x) = $\int\limits^x_{25} {\frac{1}{10}} \, dx$

F(x) $=\left[\dfrac{x}{10}\right]\limits^{x}_{25}$

F(x) = x/10 - 25/10

=> F(x) = (x - 25)/10

probability that preparation time exceeds 33 minutes  =  1/5

F(x) = (x - 25)/10  

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