Students pass a test if they score 50% or more. The marks of a large number of students were sampled and the mean and standard deviation were calculated as 42% and 8% respectively. Assuming this data is normally distributed, what percentage of student pass the test?
let X be the sore random variable.
"\\begin{aligned}\n\n&E(X)=42\n\n\\%=0.42 \\\\\n\n&S\n\nD(X)=8 \\%=0.08 \\\\\n\n&P(X>0.50)=?\n\n\\end{aligned}"
We know that if "X \\sim N(\\mu, \\sigma)" then,
"\\begin{aligned}\n\nz&=\\frac{X-\\mu}{\\sigma} \\sim N(0,1) \\\\\n&\\text{So,}\\ P(\\frac{X-0.42}{0.08}>\\frac{0.50-0.42}{0.08})\\\\\n\n&=P(z>1)\n\n\\end{aligned}"
From z-table calculator.
"P(X>0.50)=P(z>1)=0.15866"
So "15.866 \\%" students will pass the test.
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