Question

1.0000 1.0000 0.2430 0.6035 0.8531 0.9601 0.9918 0.9987 0.9998 1.0000 1.0000 1.0000 1.0000 1.0000 ← -. 2-final 2016 istics 2.jpg 山 Academie Level: Thin Semester Fine 2016/2017 Ceerse Cede & Thle Statisties and Engineering Econony astracter Dr Abdulah El-Far Fieal Esam Time allewed:h Saturday 14Ja 201 STATISTICS AND INTEREST TABLES ARE ONLYALLOMp tions Question No,1115Marks] The percentage of students in four different faculty departments A·B, C and D are 40 30% 20% and 10% respectively. tryou know that tre ss%, s-andor, or each students have passed last esam, what is the probability that a randomly selected student has passed the exam. ISMarks) in a string of a different ways can 3 red, 4 yellow, and 2 blue bulbs be arranged wedding lights with ockets? Suppose that a random variable X has a probability density function given by 5Marks) 1.0000 elsewhere 1.0000 1.0000 1.0000 a- Find the value of k that makes this a probability function b- Find the Cumulative distribution function Fluj and use it to evaluate P(o.4sXs1) 15Marks) No. 216Marks The probability that a patient recovers from a rare blood disease is O.4. t 15 people are known to have contracted this disease, what is the probability that

Answer 1

ii.  We have 9 sockets, choose 3 for the 3 red lights, i.e. ⁹C₃ = 84 ways.

After screwing in those 3 red bulbs, there remain only 6 open sockets.
now we  have 6 sockets, choose 4 for the 4 yellow lights, i.e.  ⁶C₄ =15 ways.

After screwing in those 4 red bulbs, there remain only 2 open sockets.

now we have 2 sockets, choose 2 for the blue lights. That's ²C₂= 1 ways.

After screwing in those 2 bulbs, all the sockets have a bulb in them.

Total combination =  ⁹C_{3 *} ⁶C₄ * ²C₂ = 84 * 15 * 1 = 1260 ways

iii. a. We know

$\dpi{100} \fn_jvn \int_{0}^{1} f(x)dx = 1$

$\dpi{100} \fn_jvn \Rightarrow \int_{0}^{1} kx(x-1)dx = 1$

$\dpi{100} \fn_jvn \Rightarrow \int_{0}^{1} kx^{2}dx - \int_{0}^{1} kxdx = 1$

$\dpi{100} \fn_jvn \Rightarrow k\left [ \frac{1}{3} x^{3}\right ]^{1}_{0} - k\left [ \frac{1}{2} x^{2}\right ]^{1}_{0} = 1$

$\dpi{100} \fn_jvn \Rightarrow \frac{1}{3}k - \frac{1}{2}k = 1$

$\dpi{100} \fn_jvn \Rightarrow \frac{-1}{6}k = 1$

$\dpi{100} \fn_jvn k = -6$

b. We know cumulative distribution function is given by

$\dpi{100} \fn_jvn F_{x}(X)= \int f(x) dx$

  $\dpi{100} \fn_jvn =\int kx(x-1)dx$

$\dpi{100} \fn_jvn = k(\frac{1}{3}x^{3} - \frac{1}{2}x^{2})$

  $\dpi{100} \fn_jvn = -6(\frac{1}{3}x^{3} - \frac{1}{2}x^{2})$

  

now,

$\dpi{100} \fn_jvn P(0.4\leq X\leq 1) = F_{x}(1) - F_{x}(0.4)$

=$\dpi{100} \fn_jvn = 1 - 0.35202 = 0.64798$