I have been trying to solve this math problem, but I end up with a different answer, I suspect the publisher probably used a different approach but I am not sure which one.
Here is my attempt:
My cumulative distribution function, $F(x) = P(X \leq x)$ is same as what they have.
To find the median I used this definition: The median is the value of $X$ for which $P(X \leq x)$ is greater than or equal to 0.5 and $P(X \geq x)$ is greater than or equal to 0.5.
So the median is 2 because $P(X \leq 2) = 0.7$ and $P(X \geq 2) = 0.7$
I will appreciate any help.
I agree that the median $x_m$ is $2$ and must be an integer. The random variable is distributed as $X\sim Bin(4,0.5)$
Then we have to find the integer of $X$ which satisfies the following two conditions at the same time:
$P(X\leq x_m)\geq 0.5$
$P(X\geq x_m)\geq 0.5$
For $x_m=1$ we get:
$\sum_{k=0}^1 \binom{4}{k} \cdot 0.5^k\cdot 0.5^{4-k}=\frac{1}{16}+\frac{4}{16}=\frac{5}{16} \not \geq 0.5 \quad \quad \large{\color{red}{\times}}$
$\sum_{k=1}^4 \binom{4}{k} \cdot 0.5^k\cdot 0.5^{4-k}=\frac{4}{16}+\frac{6}{16}+\frac{4}{16}+\frac{1}{16}=\frac{15}{16} \geq 0.5 \quad \quad \large{\color{green}{\checkmark}}$
We have to check the next higher integer since the first condition is not fulfilled: $x_m=2$
$\sum_{k=0}^2 \binom{4}{k} \cdot 0.5^k\cdot 0.5^{4-k}=\frac{1}{16}+\frac{4}{16}+\frac{6}{16}=\frac{11}{16} \geq 0.5 \quad \quad \large{\color{green}{\checkmark}}$
$\sum_{k=2}^4 \binom{4}{k} \cdot 0.5^k\cdot 0.5^{4-k}=\frac{6}{16}+\frac{4}{16}+\frac{1}{16}=\frac{11}{16} \geq 0.5 \quad \quad \large{\color{green}{\checkmark}}$
Both conditions are fulfilled. I´ve never got the number $0.7$ but the median is indeed $2$.