Question

assume the random variable x is normally distributed with mean \\( \mu = 50 \\) and standard deviation \\( \sigma = 7 \\). find the 67th percentile. the 67th percentile is (round to two decimal places as needed)

Explanation:

Step1: Find the z - score

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 \(Z=\frac{X - \mu}{\sigma}\)).
For the \(67\)th percentile, \(p = 0.67\). Using a standard normal table (or a calculator with a normal - distribution function, such as the inverse - norm function on a TI - 84: invNorm(0.67,0,1)), we find that \(z_{0.67}\approx0.44\).

Step2: Use the formula for the \(x\) value

The formula for converting a \(z\) - score to an \(x\) value in a normal distribution is \(x=\mu+z\sigma\).
Given \(\mu = 50\), \(\sigma=7\), and \(z = 0.44\).
Substitute the values into the formula: \(x=50+(0.44)\times7\).

$$x=50 + 3.08$$

Answer:

\(53.08\)