A list of twelve numbers have a range of eight, a mean, median & mode of five, if you picked one at random you had a fifty percent chance of a prime number, & there are as many even numbers as odd. List all twelve members.

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Solution to follow …

We have twelve numbers, so $n=12$ & the mean is
$$\stackrel{\_}{x}=\frac{\Sigma {x}_{i}}{12}=5$$
$$\Sigma {x}_{i}=60$$
The median is five too but as there are an even number of elements the mode is the mean of the sixth & seventh element when listed in order, so
$$\mathrm{median}=\frac{{x}_{6}+{x}_{7}}{2}=5$$
$${x}_{6}+{x}_{7}=10$$

Furthermore, the mode is also five, so both ${x}_{6}$ and ${x}_{7}$ are five, perhaps more.
The range is eight, so

$${x}_{\mathrm{max}}-{x}_{\mathrm{min}}=8$$
If the numbers are ordered this becomes

$${x}_{12}-{x}_{1}=8$$
Knowing the
$P(X=\mathrm{prime})=\mathrm{0.5}$
so there must be six prime numbers (some are repeated, eg. five)

$${x}_{1}+{x}_{2}+{x}_{3}+{x}_{4}+{x}_{5}+5+5+{x}_{8}+{x}_{9}+{x}_{10}+{x}_{11}+{x}_{1}+8=60$$
$$2{x}_{1}+{x}_{2}+{x}_{3}+{x}_{4}+{x}_{5}+{x}_{8}+{x}_{9}+{x}_{10}+{x}_{11}=42$$
-oOo-