Question

problem 4
a. x is a normally distributed random variable with a standard deviation of 11.72. find the
mean of x when 99.84% of the area lies to the left of 42.61.

Explanation:

Step1: Find the z - score

We know that if \(P(X < x)=0.9984\), we look up the z - score in the standard normal distribution table. The z - score \(z\) corresponding to an area of \(0.9984\) is \(z = 3\) (from the standard normal table: \(P(Z<3)=0.9987\approx0.9984\) considering rounding in table values).

Step2: Use the z - score formula

The z - score formula is \(z=\frac{x-\mu}{\sigma}\). We are given \(x = 42.61\), \(\sigma=11.72\), and \(z = 3\).
Substitute the values into the formula: \(3=\frac{42.61-\mu}{11.72}\).
Multiply both sides by \(11.72\): \(3\times11.72=42.61-\mu\).
\(35.16 = 42.61-\mu\).
Solve for \(\mu\): \(\mu=42.61 - 35.16\).

Answer:

\(\mu = 7.45\)