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Lowest score earned to be qualified for an interview is 92.82
X : Marks scored in interview
X ~ N(80,100)
Here we need to find x such that P(X > x) = 0.10
using standard normal table,
we get P( z > 1.282) = 0.10
It's always easy to use Standard normal distribution to solve questions of normal distribution because standard normal is special case of normal distribution with mean 0 and standard deviation 1.
To convert x (which follows Normal distribution) to standard normal use:
so,
<em>hence,</em>
<em> </em>
Hence Lowest score earned to be qualified for an interview is 92.82
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