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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.

twnly
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Meghan C
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1 Answers1

3

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:

  1. $P(X\leq x_m)\geq 0.5$

  2. $P(X\geq x_m)\geq 0.5$

For $x_m=1$ we get:

  1. $\sum_{k=0}^1 \binom{4}{k} \cdot 0.5^k\cdot 0.5^{4-k}=\frac{1}{16}+\frac{4}{16}=\frac{5}{16} \not \geq 0.5 \quad \quad \large{\color{red}{\times}}$

  2. $\sum_{k=1}^4 \binom{4}{k} \cdot 0.5^k\cdot 0.5^{4-k}=\frac{4}{16}+\frac{6}{16}+\frac{4}{16}+\frac{1}{16}=\frac{15}{16} \geq 0.5 \quad \quad \large{\color{green}{\checkmark}}$

We have to check the next higher integer since the first condition is not fulfilled: $x_m=2$

  1. $\sum_{k=0}^2 \binom{4}{k} \cdot 0.5^k\cdot 0.5^{4-k}=\frac{1}{16}+\frac{4}{16}+\frac{6}{16}=\frac{11}{16} \geq 0.5 \quad \quad \large{\color{green}{\checkmark}}$

  2. $\sum_{k=2}^4 \binom{4}{k} \cdot 0.5^k\cdot 0.5^{4-k}=\frac{6}{16}+\frac{4}{16}+\frac{1}{16}=\frac{11}{16} \geq 0.5 \quad \quad \large{\color{green}{\checkmark}}$

Both conditions are fulfilled. I´ve never got the number $0.7$ but the median is indeed $2$.

callculus42
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