To find the mean ($\mu$) and standard deviation ($\sigma$) of the normal distribution, we can use the fact that in a normal distribution, the percentages provided correspond to certain z-scores.

Given:

• 31% of the items are under 35.
• 12% of the items are over 62.

Steps:

1. Identify the z-scores associated with the given percentages:

   • For $31\%$ of the items under 35, this implies $P(X \leq 35) = 0.31$. From the z-table (or using a calculator), the corresponding z-score is approximately $-0.50$.
   • For $12\%$ of the items over 62, this implies $P(X \geq 62) = 0.12$. This means $P(X \leq 62) = 1 - 0.12 = 0.88$. From the z-table, the corresponding z-score is approximately $1.18$.

2. Set up the z-score formulas: The z-score formula is: $z = \frac{X - \mu}{\sigma}$ where $X$ is the value, $\mu$ is the mean, and $\sigma$ is the standard deviation.

   Using this formula for both given points:

   • For $X = 35$, $z = -0.50$: $-0.50 = \frac{35 - \mu}{\sigma}$
   • For $X = 62$, $z = 1.18$: $1.18 = \frac{62 - \mu}{\sigma}$

3. Solve the system of equations:

   • From the first equation: $-0.50 = \frac{35 - \mu}{\sigma} \implies 35 - \mu = -0.50\sigma$ \mu = 35 + 0.50\sigma \tag{1}
   • From the second equation: $1.18 = \frac{62 - \mu}{\sigma} \implies 62 - \mu = 1.18\sigma$ \mu = 62 - 1.18\sigma \tag{2}

4. Substitute and solve for $\sigma$: Equating equations (1) and (2): $35 + 0.50\sigma = 62 - 1.18\sigma$ Combine like terms: $1.68\sigma = 27$ Solve for $\sigma$: $\sigma = \frac{27}{1.68} \approx 16.07$

5. Find $\mu$: Substitute $\sigma = 16.07$ into equation (1): $\mu = 35 + 0.50(16.07) \approx 35 + 8.04 = 43.04$

Final Answer:

• The mean $\mu \approx 43.04$
• The standard deviation $\sigma \approx 16.07$

Let me know if you'd like more details or have any questions.

Here are 5 related questions to expand your understanding:

1. How would the values change if 50% of the items were under 35?
2. What is the probability that a randomly selected item falls between 35 and 62?
3. If the standard deviation was smaller, say 10, how would that impact the distribution?
4. How do z-scores help in comparing different normal distributions?
5. What would be the z-score if the value is 50?

Tip: Z-scores are a great tool for understanding how far a value is from the mean in terms of standard deviations!