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Answer 1

What you want to do is find the value of \(k\) such that \(P(X\ge k)=0.10\), so that the area to the right of \(k\) in the distribution below is 10% of the total area under the curve:

To do this, you transform to the standard normal random variable:
\[Z=\frac{X-\mu}{\sigma}~~\iff~~X=\mu+Z\sigma\]
where \(\mu=35.0\) is the mean and \(\sigma=11.6\) is the standard deviation.
\[P(X\ge k)=P(35.0+11.6Z\ge k)=P\left(Z\ge\frac{k-35.0}{11.6}\right)=0.10\]
The \(z\) value that gives a right-tail probability of 10% (or equialently, a left-tail probability of 90%) is \(z\approx1.285\), so we have the equation
\[\frac{k-35.0}{11.6}=1.285\]
Solve for \(k\).