Below, n is the sample size, p is the population proportion, and ^p is the sample proportion. First, check if the assumptions are satisfied to use the normal distribution for probabilities. If appropriate, use the Central Limit Theorem to find the indicated probability.

Question

n= 200
p= 0.13
It [is or isnt] appropriate to use the normal distribution for probabilities.
p(0.12 < ^p < 0.16) = ?

Answer 1

To determine whether it is appropriate to use the normal distribution for probabilities, we can check the conditions of the Central Limit Theorem (CLT). According to the CLT, the sample proportion \(\hat{p}\) can be approximated by a normal distribution if the following conditions are satisfied:

1. \(np \geq 10\)
2. \(n(1-p) \geq 10\)

where:

• \(n\) is the sample size,
• \(p\) is the population proportion, and
• \(1-p\) is the proportion of failures.

Given:

• \(n = 200\)
• \(p = 0.13\)

Let's check the two conditions:

1. Calculate \(np\): \[ np = 200 \times 0.13 = 26 \] This satisfies \(np \geq 10\).

2. Calculate \(n(1-p)\): \[ 1 - p = 1 - 0.13 = 0.87 \] Now, calculate \(n(1-p)\): \[ n(1-p) = 200 \times 0.87 = 174 \] This also satisfies \(n(1-p) \geq 10\).

Since both conditions are satisfied, it is appropriate to use the normal distribution for probabilities.

Now, we can use the normal approximation to find the probability \(P(0.12 < \hat{p} < 0.16)\).

First, we need to calculate the mean and standard deviation of the sample proportion \(\hat{p}\):

Step 1: Calculate the Mean \(\mu_{\hat{p}}\) and Standard Deviation \(\sigma_{\hat{p}}\)

• The mean \(\mu_{\hat{p}} = p\): \[ \mu_{\hat{p}} = 0.13 \]

• The standard deviation \(\sigma_{\hat{p}} = \sqrt{\frac{p(1-p)}{n}}\): \[ \sigma_{\hat{p}} = \sqrt{\frac{0.13 \times 0.87}{200}} \approx \sqrt{\frac{0.1131}{200}} \approx \sqrt{0.0005655} \approx 0.0238 \]

Step 2: Calculate the Z-scores for 0.12 and 0.16

Now we'll calculate the Z-scores to standardize the proportions.

• For \( \hat{p} = 0.12 \): \[ Z_{0.12} = \frac{0.12 - 0.13}{0.0238} \approx \frac{-0.01}{0.0238} \approx -0.4202 \]

• For \( \hat{p} = 0.16 \): \[ Z_{0.16} = \frac{0.16 - 0.13}{0.0238} \approx \frac{0.03}{0.0238} \approx 1.2627 \]

Step 3: Find the Probabilities

Using a standard normal distribution table (or a calculator):

• \(P(Z < -0.4202) \approx 0.3377\)
• \(P(Z < 1.2627) \approx 0.8962\)

Step 4: Calculate the Probability

Now, we can find the probability: \[ P(0.12 < \hat{p} < 0.16) = P(Z < 1.2627) - P(Z < -0.4202) \] \[ P(0.12 < \hat{p} < 0.16) \approx 0.8962 - 0.3377 \approx 0.5585 \]

Conclusion

Thus, the probability that the sample proportion is between 0.12 and 0.16 is approximately \(0.5585\).

In conclusion, it is appropriate to use the normal distribution for probabilities, and \(P(0.12 < \hat{p} < 0.16) \approx 0.5585\).