Question

assume the random variable x is normally distributed with mean μ = 50 and standard deviation σ = 7. find the 97th percentile. the 97th percentile is (round to two decimal places as needed.)

Explanation:

Step1: Find the z - score for the 97th percentile

The 97th percentile means that \(P(Z\leq z)=0.97\). Using a standard normal table or a calculator with a normal - distribution function (such as the invNorm function on a TI - 84: invNorm\((0.97,0,1)\)), we find that \(z\approx1.88\).

Step2: Use the z - score formula \(z=\frac{x-\mu}{\sigma}\) to solve for \(x\)

We know that \(\mu = 50\), \(\sigma=7\), and \(z = 1.88\).
Substitute the values into the formula: \(1.88=\frac{x - 50}{7}\).
Multiply both sides by 7: \(1.88\times7=x - 50\).
\(13.16=x - 50\).
Add 50 to both sides: \(x=50 + 13.16\).

Answer:

\(63.16\)