Each video game that a video game shop sold last year was either for the PS4 or Xbox One. The shop sold new PS4 Games for USD 60 and new Xbox One Games for USD 30. The standard deviation for all of the video games sold last year was USD 16.
Quantity A: The percentage of PS4 video games sold last year
Quantity B: 50%
A Quantity A is greater.
B Quantity B is greater.
C The two quantities are equal.
D The relationship cannot be determined from the information given
On
This problem has actually no solution.
Denote $n_1$ the number of PS4 games sold and $n_2$ the number of XBox games sold, and $n=n_1+n_2$. If the proportion of PS4 games sold is $t$, this means $n_1=tn$ and $n_2=(1-t)n$.
There were $n$ items sold, and the variable we are interested in is the price, $x_i$, ($1\le i\le n$), with $x_i\in\{30,60\}$. Among these $n$ items, $n_1$ have price $60$ and $n_2$ items have price $30$.
The mean price is
$$\mu=\frac1n\sum_{i=1}^n x_i=\frac{60n_1+30n_2}{n}=60t+30(1-t)=30+30t$$
The variance of the prices is
$$\sigma^2=\frac1n\sum_{i=1}^n (x_i-\mu)^2=\frac{(60-\mu)^2n_1+(30-\mu)^2n_2}{n}\\=\frac{(30-30t)^2n_1+(30t)^2n_2}{n}=900[t(1-t)^2+t^2(1-t)]\\=900t(1-t)$$
And we are told the standard deviation is $16$, hence the variance is $256$.
However, $t(1-t)\le\frac{1}{4}$ for all real $t$, therefore $\sigma^2\le\frac{900}{4}=225$. There is no way it can reach $256$.
In case the inequality $t(1-t)\le\frac{1}{4}$ is not clear, you can write
$$\frac{1}{4}-t(1-t)=t^2-t+\frac14=(t-\frac12)^2\ge0$$
More generally, a sample from a distribution with two values $a$ and $b$ has variance $t(1-t)(a-b)^2$, $t\in[0,1]$ being the proportion of "$a$"s. Hence, depending on $t$, the standard deviation can be any value between $0$ and $\frac{|a-b|}{2}$. Here $\frac{|a-b|}{2}=15$.
We may attempt something to make sense of the question anyway: above I use the variance formula without bias correction, as it's a population variance (all sales). What happens if we apply the correction? The variance is increased, does it reach $256$?
Let's see.
$$\sigma^2=\frac{1}{n-1}\sum_{i=1}^n(x_i-\mu)^2=\frac{n}{n-1}\cdot900t(1-t)$$
Letting $t=1/2$ to get the maximum possible value, it's easy to check that the variance is decreasing as $n$ grows, and it's $<256$ when $n>8$. So it means $n$ must be between $2$ and $8$.
But we are not done yet. We must find a value of $n_1$ so that $\sigma^2$ is exactly $256$. There is a small number of candidates to check (and we can restrict this further to $n_1\le n$), and none leads to a variance exactly $256$.
Therefore, either way, the problem can't be solved.
The relationship cannot be determined. We have that, where $t$ is the percentage of PS4 games sold this year,
$$\sqrt{900 t^2 + (30 - 30 t)^2}/\sqrt 2 = 16$$
By the formula for standard deviation. so that $$t= 1/2 \pm \frac{\sqrt{31}}{30}$$ by the quadratic formula. Hence, it could have been the case that PS4 games were the majority or minority of games sold. D is correct.