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 = 111
p= 0.59
It [is or isnt?] appropriate to use the normal distribution for probabilities.
p(0.56< ^p <0.60) = [?]

Answer 1

To determine if it is appropriate to use the normal distribution for calculating probabilities related to the sample proportion \(^p\), we need to check if the sample size \(n\) is large enough based on the population proportion \(p\).

The conditions for using the normal approximation for the sample proportion are:

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

Let's calculate these:

1. \(np = 111 \times 0.59 = 65.49\)
2. \(n(1 - p) = 111 \times (1 - 0.59) = 111 \times 0.41 = 45.51\)

Both \(np\) and \(n(1 - p)\) must be greater than or equal to 10. Here,

• \(65.49 \geq 10\) (satisfied)
• \(45.51 \geq 10\) (satisfied)

Since both conditions are satisfied, it is appropriate to use the normal distribution to approximate probabilities related to the sample proportion.

Next, we will calculate the probability \(P(0.56 < \hat{p} < 0.60)\) using the Central Limit Theorem.

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

1. Mean of \(\hat{p} = p = 0.59\)

2. Standard deviation of \(\hat{p} = \sqrt{\frac{p(1 - p)}{n}} = \sqrt{\frac{0.59(1 - 0.59)}{111}} = \sqrt{\frac{0.59 \times 0.41}{111}}\)

Let's calculate this standard deviation:

\[ \text{Standard deviation} = \sqrt{\frac{0.59 \times 0.41}{111}} \approx \sqrt{\frac{0.2419}{111}} \approx \sqrt{0.0021788} \approx 0.0467 \]

Now we can standardize the values for \(\hat{p}\):

• For \( \hat{p} = 0.56 \): \[ z_1 = \frac{0.56 - 0.59}{0.0467} \approx \frac{-0.03}{0.0467} \approx -0.6420 \]

• For \( \hat{p} = 0.60 \): \[ z_2 = \frac{0.60 - 0.59}{0.0467} \approx \frac{0.01}{0.0467} \approx 0.2144 \]

Now we can use a standard normal distribution table or calculator to find the probabilities associated with these z-scores:

• \(P(Z < z_1) \approx P(Z < -0.6420)\)
• \(P(Z < z_2) \approx P(Z < 0.2144)\)

Using the standard normal distribution tables:

1. For \(z_1 \approx -0.6420\): \(P(Z < -0.6420) \approx 0.2611\)

2. For \(z_2 \approx 0.2144\): \(P(Z < 0.2144) \approx 0.5859\)

Now, we calculate \(P(0.56 < \hat{p} < 0.60)\):

\[ P(0.56 < \hat{p} < 0.60) = P(Z < 0.2144) - P(Z < -0.6420) \approx 0.5859 - 0.2611 = 0.3248 \]

Thus, the final answer is:

• It is appropriate to use the normal distribution for probabilities.
• \(P(0.56 < \hat{p} < 0.60) \approx 0.3248\)