We are given two key pieces of information from the normal distribution:

1. 7% of the items are below 35.
2. 89% of the items are below 63.

The formula for converting a value $x$ to a z-score in a normal distribution is:

$z = \frac{x - \mu}{\sigma}$

where:

• $\mu$ is the mean,
• $\sigma$ is the standard deviation,
• $x$ is the value from the distribution, and
• $z$ is the z-score corresponding to that value.

Step 1: Find the z-scores corresponding to the given probabilities.

Using a standard normal distribution table (z-table) or a calculator:

• The z-score corresponding to 7% (or 0.07) is approximately $z_1 = -1.475$.
• The z-score corresponding to 89% (or 0.89) is approximately $z_2 = 1.227$.

Step 2: Set up the system of equations.

We now have two equations based on the formula $z = \frac{x - \mu}{\sigma}$:

$z_1 = \frac{35 - \mu}{\sigma} \quad \text{or} \quad -1.475 = \frac{35 - \mu}{\sigma}$ $z_2 = \frac{63 - \mu}{\sigma} \quad \text{or} \quad 1.227 = \frac{63 - \mu}{\sigma}$

Step 3: Solve the system of equations.

We now solve these equations to find $\mu$ and $\sigma$. I'll compute this step next.The mean $\mu$ of the distribution is approximately $50.28$, and the standard deviation $\sigma$ is approximately $10.36$.

Would you like further details or clarifications on this solution?

Here are 5 related questions to explore further:

1. What is the probability of an item being above 70 in this distribution?
2. How do we find the z-scores for different probabilities?
3. What is the area under the normal curve between 35 and 63?
4. How do you interpret a z-score in a normal distribution?
5. What are the properties of the standard normal distribution?

Tip: In a standard normal distribution, z-scores represent the number of standard deviations a value is from the mean.