 Consider a normal population distribution with the value of 𝜎 known.
(a)
What is the confidence level for the interval
x ± 2.88𝜎⁄
	n
?
(Round your answer to one decimal place.)
(b)
What is the confidence level for the interval
x ± 1.41𝜎⁄
	n
?
(Round your answer to one decimal place.)
(c)
What value of z𝛼⁄2 in the CI formula below results in a confidence level of 99.7%? (Round your answer to two decimal places.)
 
x − z𝛼⁄2 · 
𝜎
	n
, x + z𝛼⁄2 · 
𝜎
	n
(d)
Answer the question posed in part (c) for a confidence level of 70%. (Round your answer to two decimal places.) 

Let's work through each part of the question methodically.

### (a) Confidence level for the interval \( x \pm \frac{2.88\cdot\sigma}{\sqrt{n}} \)

The general form for a confidence interval is:

\[
x \pm z_{\alpha/2} \cdot \frac{\sigma}{\sqrt{n}}
\]

Where:
- \( z_{\alpha/2} \) is the critical value from the standard normal distribution that corresponds to the desired confidence level.
- \( \sigma \) is the known population standard deviation.
- \( n \) is the sample size.

In this case, we are given \( z_{\alpha/2} = 2.88 \).

To find the confidence level, we can use the relationship between \( z_{\alpha/2} \) and the confidence level. The area between \( -z_{\alpha/2} \) and \( +z_{\alpha/2} \) represents the proportion of the standard normal distribution covered by the confidence interval.

Using a standard normal distribution table or calculator, the area to the left of \( z = 2.88 \) is approximately 0.9980. The total area covered by the confidence interval is therefore:

\[
2 \times (0.9980 - 0.5) = 0.996
\]

So, the confidence level is approximately 99.6%.

**Answer**: The confidence level is **99.6%**.

### (b) Confidence level for the interval \( x \pm \frac{1.41\cdot\sigma}{\sqrt{n}} \)

Given \( z_{\alpha/2} = 1.41 \), we follow the same procedure as in part (a).

The area to the left of \( z = 1.41 \) is approximately 0.9207. The total area covered by the confidence interval is:

\[
2 \times (0.9207 - 0.5) = 0.8414
\]

So, the confidence level is approximately 84.1%.

**Answer**: The confidence level is **84.1%**.

### (c) Value of \( z_{\alpha/2} \) for a 99.7% confidence level

For a 99.7% confidence level, the area in the tails is:

\[
1 - 0.997 = 0.003
\]

Dividing this evenly between the two tails gives:

\[
\alpha/2 = 0.0015
\]

Looking up the corresponding \( z_{\alpha/2} \) value for \( \alpha/2 = 0.0015 \) in the standard normal distribution table gives \( z_{\alpha/2} = 3.00 \).

**Answer**: The value of \( z_{\alpha/2} \) for a 99.7% confidence level is **3.00**.

### (d) Value of \( z_{\alpha/2} \) for a 70% confidence level

For a 70% confidence level, the area in the tails is:

\[
1 - 0.70 = 0.30
\]

Dividing this evenly between the two tails gives:

\[
\alpha/2 = 0.15
\]

Looking up the corresponding \( z_{\alpha/2} \) value for \( \alpha/2 = 0.15 \) in the standard normal distribution table gives \( z_{\alpha/2} = 1.04 \).

**Answer**: The value of \( z_{\alpha/2} \) for a 70% confidence level is **1.04**.

---

Let me know if you want further clarification! Here are some relative questions for practice:

1. How is the confidence interval affected if the sample size increases?
2. What happens to the width of the confidence interval if the confidence level increases?
3. How do you interpret a 95% confidence interval in practice?
4. Why is the standard normal distribution used for constructing confidence intervals?
5. How does the z-value change for a 90% confidence level?

**Tip**: As the confidence level increases, the corresponding z-value also increases, making the confidence interval wider.

Consider a normal population distribution with the value of 𝜎 known.
(a) What is the confidence level for the interval x ± 2.88𝜎⁄n? (Round your answer to one decimal place.)
(b) What is the confidence level for the interval x ± 1.41𝜎⁄n? (Round your answer to one decimal place.)
(c) What value of z𝛼⁄2 in the CI formula below results in a confidence level of 99.7%? (Round your answer to two decimal places.) x − z𝛼⁄2 · 𝜎⁄n, x + z𝛼⁄2 · 𝜎⁄n
(d) Answer the question posed in part (c) for a confidence level of 70%. (Round your answer to two decimal places.)

Explore how to calculate confidence levels for intervals using Z-scores in a normal population distribution. This problem includes key formulas like x ± z_{α/2} * (σ/√n) and covers confidence levels from 70% to 99.7%. Suitable for college-level students.

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