Question 1198869
**Formula for Standard Error of the Proportion**

The standard error of the proportion (SE) is given by:

SE = √[p * (1 - p) / n] 

where:

* p is the sample proportion
* n is the sample size

**Normality Assumption**

We can generally assume normality of the sampling distribution of the proportion if the following conditions are met:

* **n * p ≥ 10** 
* **n * (1 - p) ≥ 10**

**Now let's calculate the standard error for each case and check for normality:**

**a) n = 27, m = 0.24**

* SE = √[0.24 * (1 - 0.24) / 27] = √[0.24 * 0.76 / 27] ≈ 0.0758
* n * p = 27 * 0.24 = 6.48 
* n * (1 - p) = 27 * 0.76 = 20.52 
* **Normality may not be assumed** because n * p < 10.

**b) n = 56, m = 0.43**

* SE = √[0.43 * (1 - 0.43) / 56] = √[0.43 * 0.57 / 56] ≈ 0.0658
* n * p = 56 * 0.43 = 24.08
* n * (1 - p) = 56 * 0.57 = 31.92
* **Normality can be assumed** as both n * p and n * (1 - p) are greater than 10.

**c) n = 123, m = 0.47**

* SE = √[0.47 * (1 - 0.47) / 123] = √[0.47 * 0.53 / 123] ≈ 0.0448
* n * p = 123 * 0.47 = 57.81
* n * (1 - p) = 123 * 0.53 = 65.19
* **Normality can be assumed** as both n * p and n * (1 - p) are greater than 10.

**d) n = 578, m = 0.004**

* SE = √[0.004 * (1 - 0.004) / 578] = √[0.004 * 0.996 / 578] ≈ 0.0026
* n * p = 578 * 0.004 = 2.312
* n * (1 - p) = 578 * 0.996 = 575.688
* **Normality may not be assumed** because n * p < 10.

**Summary**

| Sample Size (n) | Sample Proportion (m) | Standard Error (SE) | Normality Assumed? |
|---|---|---|---|
| 27 | 0.24 | 0.0758 | No |
| 56 | 0.43 | 0.0658 | Yes |
| 123 | 0.47 | 0.0448 | Yes |
| 578 | 0.004 | 0.0026 | No |

I hope this helps! Let me know if you have any other questions.