The probability that A wakes up before the alarm rings is 0.4. Then, the mean and variance of the number of times A wakes up before the alarm rings in the next 7 days, respectively are:

Question

Self-study helps students to build confidence in learning. It boosts the self-esteem of the learners. Recent surveys suggested that close to 50% of learners were self-taught using internet resources and upskilled themselves.

A student may spend 1 hour to 6 hours in a day in upskilling self. The probability distribution of the number of hours spent by a student is given below:

\[ P(X = x) = \begin{cases} kx^2, & \text{for } x = 1, 2, 3, \\ 2kx, & \text{for } x = 4, 5, 6, \\ 0, & \text{otherwise.} \end{cases} \]

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)(a) Find the mean number of hours spent by the student.
(iii)(b) Find \( P(1 < X < 6) \).

Answer 1

Step 1: Recognise the binomial setup.
Waking up before the alarm is a success with probability \(p=0.4\), repeated over \(n=7\) independent days, so the count of successes follows \(X\sim B(7,0.4)\).

Step 2: Use the direct formulas for mean and variance of a binomial variable.
\[ \text{Mean} = np, \qquad \text{Variance} = np(1-p) \]

Step 3: Substitute the numbers.
\[ \text{Mean} = 7(0.4) = 2.8 \]
\[ \text{Variance} = 7(0.4)(0.6) = 2.8\times0.6 = 1.68 \]

Step 4: State the pair of values.
\[ \boxed{2.8,\ 1.68} \]

The probability that A wakes up before the alarm rings is 0.4. Then, the mean and variance of the number of times A wakes up before the alarm rings in the next 7 days, respectively are:

Show Hint

The Correct Option is C

Solution and Explanation

Top Questions on Probability Distribution

Questions Asked in AP EAPCET exam