CBSE 2024 · Region 3 · Set 1 · Q36 · 4 marks
Self-study helps students to build confidence in learning. It boosts the self-esteem of the learners. Recent surveys suggested that close to $\displaystyle 50 \%$ learners were self-taught using internet resources and upskilled themselves.
A student may spend $\displaystyle 1$ hour to $\displaystyle 6$ hours in a day in upskilling self. The probability distribution of the number of hours spent by a student is given below : \[\mathrm{P}(\mathrm{X}=\mathrm{x})= \begin{cases}\mathrm{kx}^{2}, & \text { for } \mathrm{x}=1,2,3 \\ 2 \mathrm{kx}, & \text { for } \mathrm{x}=4,5,6 \\ 0, & \text { otherwise }\end{cases} \] where x denotes the number of hours. Based on the above information, answer the following questions :(i)Express the probability distribution given above in the form of a probability distribution table.(ii)Find the value of k .(iii)Find the mean number of hours spent by the student.Find $\displaystyle \mathrm{P}(1<\mathrm{X}<6)$. Case Study - $\displaystyle 2$
Self-study helps students to build confidence in learning. It boosts the self-esteem of the learners. Recent surveys suggested that close to $\displaystyle 50 \%$ learners were self-taught using internet resources and upskilled themselves.
A student may spend $\displaystyle 1$ hour to $\displaystyle 6$ hours in a day in upskilling self. The probability distribution of the number of hours spent by a student is given below : \[\mathrm{P}(\mathrm{X}=\mathrm{x})= \begin{cases}\mathrm{kx}^{2}, & \text { for } \mathrm{x}=1,2,3 \\ 2 \mathrm{kx}, & \text { for } \mathrm{x}=4,5,6 \\ 0, & \text { otherwise }\end{cases} \] where x denotes the number of hours. Based on the above information, answer the following questions :
(i)
Express the probability distribution given above in the form of a probability distribution table.
(ii)
Find the value of k .
(iii)
Find the mean number of hours spent by the student.
Find $\displaystyle \mathrm{P}(1<\mathrm{X}<6)$. Case Study - $\displaystyle 2$
Marking-scheme solution
X & $\displaystyle 1$ & $\displaystyle 2$ & $\displaystyle 3$ & $\displaystyle 4$ & $\displaystyle 5$ & $\displaystyle 6$
$\displaystyle \mathrm{P}(\mathrm{X})$ & k & $\displaystyle 4$ k & $\displaystyle 9$ k & $\displaystyle 8$ k & $\displaystyle 10$ k & $\displaystyle 12$ k
\begin{aligned}
& \mathrm{k}+4 \mathrm{k}+9 \mathrm{k}+8 \mathrm{k}+10 \mathrm{k}+12 \mathrm{k}=1
& \Rightarrow \mathrm{k}=\frac{1}{44}
\end{aligned}
$$
ProbabilityRandom Variable and its Probability DistributionApplycase_studymedium
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CBSE Class 12 Mathematics past-paper question from the 2024board exam, with the answer as CBSE’s own marking scheme gives it. Where our answers come from.