Question

Confused on both, studying for an exam. Please help.

27. The following is an inaccurate probability dis- tribution for a given random variable, x. State two errors with this probability distribution. 2 40 60 px) 12 60 30 60 18 60 28. The following is a hypothetical probability dis- tribution of the number of dreams recalled (per night) among students during a final exam week. How many dreams should we expect a student to recall during final exam week? Hint: Compute the expected value of this probability distribution. Number of Dreams Recalled 0 4 p(x) )22 11 24 31 .12

Answer 1

Let us get started with discussing probability :

$Probability = \frac{Number Of Favorable Outcome}{Total Possible Outcome}$

(27) Here, we have to state two errors with this probability distribution:

Here, in our first random variable "0" it's respective probability is given   $-\frac{12}{60}$ that can not be possible

because Probabilities always must be lie between $0$ and $1$. But here it is a negative fraction that lies outside of this range.

Next, error is that the Sum of Total Probabilities must be equal to 1

while here, if we add all the given Probabilities :$-\frac{12}{60} + \frac{30}{60} + \frac{40}{60} + \frac{18}{60}$   $= -\frac{38}{30} = -1.266$   

that means it doesn't follow the laws of probabilities .

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(28) Here , the next question we have to find the "expectation"

$E[X] = \sum_{i=1}^{n}x_ip_i$ , where $x_i$ is the random variable in the data

and $p_i$ is the probabilities of their corresponding random variables .

$\implies$$E[X]$  $=$$0 * (.22) + 1 * (.11) + 2*(.24) + 3*(.31) + 4 * (.12)$

$\implies$   $E[X]$  $=$$0 + .11 + .48 + .93 + .48$

$\implies$$E[X]$$=$$2.0$

Ans : We should expect 2 dreams a student to recall during final exam week .