Question

machine fires balls a distance that is approximately normally distributed. the mean distance, $\mu$, is unknown and the
ation is 1.2 feet. if 5% of balls go farther than 70 feet, find $\mu$.
feet (round to 3 decimal places.)

Explanation:

Step1: Use the standard normal distribution property

Let \(X\) be the distance the ball is fired. \(X\sim N(\mu,\sigma^{2})\), where \(\sigma = 1.2\). We know that \(P(X>70)=0.05\). Then \(P(X\leq70) = 1 - 0.05=0.95\).
The \(z\) - score formula is \(z=\frac{x-\mu}{\sigma}\). From the standard normal distribution table (\(Z\sim N(0,1)\)), the \(z\) - value corresponding to a cumulative probability of \(0.95\) is \(z = 1.645\) (approximate value from the standard normal table).

Step2: Substitute into the \(z\) - score formula

We have \(z=\frac{70-\mu}{1.2}\), and since \(z = 1.645\), we can solve the equation \(1.645=\frac{70-\mu}{1.2}\) for \(\mu\).
Multiply both sides of the equation by \(1.2\): \(1.645\times1.2=70 - \mu\).
\(1.974=70-\mu\).
Then \(\mu=70 - 1.974\).

Answer:

\(68.026\)