Let "X" be a random variable such that "X(p)=\\text{height of $p$}" for every pupil "p" in the room.
We have "P(X\\in[150,160])=\\frac{4}{14}=\\frac{2}{7}", "P(X\\in[160,170])=\\frac{4}{14}=\\frac{2}{7}", "P(X\\in[170,180])=\\frac{6}{14}=\\frac{3}{7}"
Alos suppose that there is an uniform distribution on "[150,160]", "[160,170]" and "[170,180]".
So "\\rho_X(x)=\\begin{cases}\n0,&\\text{if $x\\not\\in[150,180]$}\\\\\n\\frac{1}{35},&\\text{if $x\\in[150,160)$}\\\\\n\\frac{1}{35},&\\text{if $x\\in[160,170)$}\\\\\n\\frac{3}{70},&\\text{if $x\\in[170,180]$}\n\\end{cases}="
"=\\begin{cases}\n0,&\\text{if $x\\not\\in[150,180]$}\\\\\n\\frac{1}{35},&\\text{if $x\\in[150,170)$}\\\\\n\\frac{3}{70},&\\text{if $x\\in[170,180]$}\n\\end{cases}", where "\\rho_X" is probability density function of "X".
So average height is "EX=\\int\\limits_{\\mathbb R}x \\rho_X(x)dx=\\int\\limits_{150}^{170}\\frac{1}{35}xdx+\\int\\limits_{170}^{180}\\frac{3}{70}xdx="
"=\\frac{1}{70}x^2\\bigl|_{150}^{170}+\\frac{3}{140}x^2\\bigl|_{170}^{180}=\\frac{1165}{7}"
Answer: "\\frac{1165}{7}" cm
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