Question 1168822
Let's solve this problem step-by-step.

**1. Define the Parameters**

* Population proportion (p) = 0.30
* Sample size (n) = 97
* Sample proportion (p̂)
* We want to find P(p̂ > 0.2)

**2. Check the Conditions for a Normal Approximation**

To use the normal approximation for the sampling distribution of p̂, we need to check if:

* np ≥ 10
* n(1 - p) ≥ 10

Let's check:

* np = 97 * 0.30 = 29.1 ≥ 10 (condition met)
* n(1 - p) = 97 * 0.70 = 67.9 ≥ 10 (condition met)

Since both conditions are met, we can use the normal approximation.

**3. Calculate the Mean and Standard Deviation of the Sampling Distribution**

* Mean of p̂ (μ_p̂) = p = 0.30
* Standard deviation of p̂ (σ_p̂) = √[p(1 - p) / n]
    * σ_p̂ = √[(0.30 * 0.70) / 97]
    * σ_p̂ = √(0.21 / 97)
    * σ_p̂ = √0.00216494845
    * σ_p̂ ≈ 0.04652

**4. Calculate the Z-score**

We want to find P(p̂ > 0.2). To do this, we need to convert 0.2 to a z-score:

* z = (p̂ - μ_p̂) / σ_p̂
* z = (0.2 - 0.30) / 0.04652
* z = -0.10 / 0.04652
* z ≈ -2.15

**5. Find the Probability**

We want to find P(p̂ > 0.2), which is equivalent to P(Z > -2.15).

* Using a standard normal distribution table or calculator, we find:
    * P(Z < -2.15) ≈ 0.0158
* Since P(Z > -2.15) = 1-P(Z < -2.15)
    * P(Z > -2.15) = 1-0.0158 = 0.9842

**6. Final Answer**

The probability that the sample proportion is greater than 0.2 is approximately 0.9842.

Answer = 0.9842