question_answer

Let X denotes the number of hours, you study during a randomly selected school days. The probability that X can take the values x has the following form, where k is any unknown constant,  \[P(x)=\left\{ \begin{matrix}    0.1 & \text{if}\,\,x=0  \\    k\,x, & \text{if}\,\,x=1\,\,or\,\,2  \\    k(5-x), & \text{if}\,\,x=3\,\,or\,\,4  \\    0, & \text{otherwise}  \\ \end{matrix} \right.\]           (i) Find the value of k.               (ii) What is the probability that you study (a) atleast 2 h?               (b) exactly 2 h?   Why early morning is considered as the best time to study, explain?

Answer:

Given, the random variable X denotes the number of hours, you study during a randomly selected school days and X has a probability distribution P(X) of the form \[P(X)=\left\{ \begin{matrix}    0,\,1, & \text{if}\,\,x=0  \\    kx, & \text{if}\,\,x=1\,\,or\,\,2  \\    k\,(5-x) & \text{if}\,\,x=3\,\,or\,\,4  \\    0, & \text{otherwise}  \\ \end{matrix} \right.\] It can be written as

X	0	1	2	3	4
P(X)	0.1	k	2k	2k	k

(i) Here,\[P(0)+P(1)+P(2)+P(3)+P(4)=1\]      [\[\because \] sum of probabilities in a probability distribution is one] \[\Rightarrow \]   \[0.1+k+2k+k=1\] \[\Rightarrow \]   \[0.1+6k=1\] \[\Rightarrow \]   \[6k=1-0.1\] \[\Rightarrow \]   \[6k=1-\frac{1}{10}=\frac{10-1}{10}\] \[\Rightarrow \]   \[k=\frac{9}{10}\times \frac{1}{6}\] \[\therefore \]      \[k=\frac{3}{20}\] (ii) (a) P(study at least 2 h) \[=P(X\ge 2)\]  \[=P(2)+P(3)+P(4)\]             \[=2k+2k+k=5k=5\times \frac{3}{20}=\frac{3}{4}\]                (b) P(study exactly 2 h) = P(X = 2) = P(2)             \[=2k=2\times \frac{3}{20}=\frac{3}{10}\] Early morning is considered as the best time for study because after a good night sleep, we have fresh and untidy mind. Also, there is no external disturbance in morning time.