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Let's break down and solve the problem step by step:

### Given:
- The life of the transistor battery is normally distributed with:
  - **Mean** (\(\mu\)) = 100 hours
  - **Standard deviation** (\(\sigma\)) = 20 hours
- Sample size (\(n\)) = 16 batteries.

### To find:
1. **Proportion of sample means** more than 90 hours.
2. **Proportion of sample means** between 96 and 105 hours.
3. **Limits around the population mean** within which 90% of sample means will lie.

#### Solution Approach:
1. **Calculate the standard error of the mean (SEM)**:
   The standard error is given by:
   \[
   \text{SEM} = \frac{\sigma}{\sqrt{n}} = \frac{20}{\sqrt{16}} = \frac{20}{4} = 5
   \]
   This SEM will be used in the next calculations, as we are dealing with sample means.

2. **Standardization (Z-scores)**:
   To find the required proportions, we convert the values to Z-scores using:
   \[
   Z = \frac{X - \mu}{\text{SEM}}
   \]
   where \(X\) is the specific value of the sample mean.

---

### 1. Proportion of sample means more than 90 hours:
   We need to find \(P(\bar{X} > 90)\).

   - For \(X = 90\):
     \[
     Z = \frac{90 - 100}{5} = \frac{-10}{5} = -2
     \]
   Using the standard normal distribution, the probability corresponding to \(Z = -2\) is approximately 0.0228 (from Z-tables). This is the area to the left of \(Z = -2\).

   Therefore, the proportion of sample means more than 90 is:
   \[
   P(\bar{X} > 90) = 1 - 0.0228 = 0.9772
   \]
   So, approximately **97.72%** of sample means will be more than 90 hours.

---

### 2. Proportion of sample means between 96 and 105 hours:
   We need to find \(P(96 < \bar{X} < 105)\).

   - For \(X = 96\):
     \[
     Z = \frac{96 - 100}{5} = \frac{-4}{5} = -0.8
     \]
   - For \(X = 105\):
     \[
     Z = \frac{105 - 100}{5} = \frac{5}{5} = 1
     \]

   From the Z-table:
   - The probability for \(Z = -0.8\) is approximately 0.2119.
   - The probability for \(Z = 1\) is approximately 0.8413.

   The proportion between 96 and 105 hours is:
   \[
   P(96 < \bar{X} < 105) = 0.8413 - 0.2119 = 0.6294
   \]
   So, approximately **62.94%** of sample means will lie between 96 and 105 hours.

---

### 3. Limits around the population mean for 90% of sample means:
   For this, we need the Z-value that corresponds to a central 90% interval in the standard normal distribution.

   The Z-values that correspond to the middle 90% are approximately \(\pm 1.645\). We will now convert these Z-values back to the sample mean limits.

   Using the formula:
   \[
   X = \mu + Z \times \text{SEM}
   \]
   - For the lower limit:
     \[
     X_{\text{lower}} = 100 - 1.645 \times 5 = 100 - 8.225 = 91.775
     \]
   - For the upper limit:
     \[
     X_{\text{upper}} = 100 + 1.645 \times 5 = 100 + 8.225 = 108.225
     \]

   Therefore, **90% of the sample means** will lie between **91.78 hours and 108.23 hours**.

---

Would you like further clarification or more details on any part of the solution?

### Further Related Questions:
1. What is the probability of the sample mean being less than 96 hours?
2. How would the results change if the standard deviation were 25 hours instead of 20?
3. How can we compute the proportion of sample means between 95 and 105 hours?
4. If the sample size increased to 25, what would be the new standard error of the mean?
5. How would the proportion of sample means more than 110 hours be calculated?

### Tip:
Always ensure to adjust for the sample size when calculating probabilities involving sample means by using the standard error instead of the population standard deviation.

The number of hours of life of a type of transistor battery is normally distributed with a mean of 100 hours and a standard deviation of 20 hours. If random samples of 16 batteries were selected, what proportion of the sample means would be: 1) More than 90 hours? 2) Between 96 and 105 hours? Also, within what limits around the population mean will 90% of sample means lie?

Solve this statistics problem involving the normal distribution of transistor battery life. Calculate the proportion of sample means more than 90 hours and between 96 and 105 hours. Also, find the confidence interval for 90% of the sample means. Suitable for undergraduate statistics courses.

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