Question

assume the random variable ( x ) is normally distributed with mean ( mu = 50 ) and standard deviation ( sigma = 7 ). find the 77th percentile.
the 77th percentile is
(round to two decimal places as needed.)

Explanation:

Step1: Find the z - score corresponding to the 77th percentile

We know that if \(X\sim N(\mu,\sigma^{2})\), to find the \(p\)th percentile, we first find the \(z\) - score \(z_p\) such that \(P(Z\leq z_p)=p\%\) (where \(Z\) is the standard normal random variable \(Z=\frac{X - \mu}{\sigma}\)).
Using a standard normal table (or a calculator with a normal - distribution function, e.g., the invNorm function on a TI - 84: invNorm\((0.77,0,1)\)), we find that \(z\approx0.74\).

Step2: Use the formula for the \(x\) value in a normal distribution

The formula for \(X\) in terms of \(Z\) is \(X=\mu+z\sigma\).
Given \(\mu = 50\), \(\sigma=7\), and \(z = 0.74\), we substitute these values into the formula:
\(X=50+(0.74)\times7\).
First, calculate \((0.74)\times7 = 5.18\).
Then, \(X=50 + 5.18\).

Answer:

\(55.18\)