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The problem you're working on involves the normal distribution, where the mean and standard deviation of the battery life are provided. Here's a breakdown of the solution for each part:

Given:
- Mean (μ) = 100 hours
- Standard deviation (σ) = 20 hours
- Sample size (n) = 16 batteries

Since we're dealing with sample means, the standard deviation of the sample mean (also known as the standard error) is given by:
\[
\text{Standard Error (SE)} = \frac{\sigma}{\sqrt{n}} = \frac{20}{\sqrt{16}} = \frac{20}{4} = 5
\]

### Part (i) - Proportion of sample means greater than 90 hours:
We need to find the proportion of sample means that are greater than 90 hours. 

First, calculate the z-score for 90 hours:
\[
z = \frac{90 - 100}{5} = \frac{-10}{5} = -2
\]
Using the standard normal distribution table, the probability corresponding to a z-score of -2 is approximately 0.0228. This represents the proportion of sample means less than 90 hours.

Therefore, the proportion of sample means greater than 90 hours is:
\[
1 - 0.0228 = 0.9772
\]
So, approximately 97.72% of the sample means will be greater than 90 hours.

### Part (ii) - Proportion of sample means between 96 and 105 hours:
Next, we calculate the z-scores for 96 hours and 105 hours.

For 96 hours:
\[
z = \frac{96 - 100}{5} = \frac{-4}{5} = -0.8
\]
For 105 hours:
\[
z = \frac{105 - 100}{5} = \frac{5}{5} = 1
\]

From the z-table, the probability for \( z = -0.8 \) is approximately 0.2119, and for \( z = 1 \), it is 0.8413.

The proportion of sample means between 96 and 105 hours is:
\[
0.8413 - 0.2119 = 0.6294
\]
Thus, about 62.94% of the sample means will be between 96 and 105 hours.

### Part (iii) - Limits around the population mean that will contain 90% of sample means:
For this part, we need to find the z-scores that correspond to the middle 90% of the distribution. This leaves 5% in each tail of the distribution. From the z-table, the z-scores that correspond to 90% are approximately \( \pm 1.645 \).

The limits are given by:
\[
\mu \pm z \times \text{SE} = 100 \pm 1.645 \times 5
\]
Calculating the limits:
\[
100 \pm 8.225
\]
So, the limits are approximately \( 91.775 \) and \( 108.225 \).

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

---

Would you like more details on any part of this problem? Here are some related questions for further practice:

1. How does increasing the sample size affect the standard error of the sample mean?
2. What proportion of sample means would fall below 95 hours?
3. What is the probability of finding a sample mean above 110 hours?
4. How would the results change if the standard deviation were 30 hours instead of 20?
5. What is the interpretation of the z-score in the context of a normal distribution?

**Tip:** When working with normal distributions, always remember that standardizing values into z-scores is crucial for using standard normal tables or software tools.

The number of hours of life of a type of transistor battery is normally distributed with mean = 100 hours and standard deviation = 20 hours. If random samples of 16 batteries were selected, what proportion of the sample means would be (i) more than 90 hours, (ii) between 96 and 105 hours? Within what limits around the population mean will 90% of sample means lie?

Solve a problem involving the normal distribution of battery life with a mean of 100 hours and standard deviation of 20 hours. This problem focuses on calculating the proportion of sample means using z-scores and finding confidence intervals. Great for students learning about sampling distributions.

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