Question

X is a normally distributed random variable with a mean of 11.0 and a standard deviation of 3.50. Find the value x such that P(X < x) is equal to 0.86

Answer 1

Let \(Y\) denote the standardized normal distribution, which has \(Z=\dfrac{X-\mu}{\sigma}\).

So, you have
\[P(X<x)=0.86=P(\mu+\sigma Z<x)=P(11+3.5^2Z<x)\]
Using a table for \(z\)-values, you can find that the appropriate \(z\) that gives you 0.86, so you would just have to solve for \(x\):
\[11+3.5^2z=x\]

Answer 2

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