The National Coffee Association reported that 64% of U.S. adults drink coffee daily. A random sample of 225 U.S. adults is selected. Round your answers to at least four decimal places as needed.

Question

Use a TI-84 calculator and find the mean u^p

Answer 1

To find the mean \( \mu_p \) for the sample proportion of U.S. adults who drink coffee daily, we use the formula:

\[ \mu_p = p \]

where \( p \) is the population proportion. Given that 64% of U.S. adults drink coffee daily, we convert this percentage into a decimal:

\[ p = 0.64 \]

Thus, the mean \( \mu_p \) is:

\[ \mu_p = 0.64 \]

If you need to calculate the standard deviation \( \sigma_p \) of the sample proportion, you can use the formula:

\[ \sigma_p = \sqrt{\frac{p(1 - p)}{n}} \]

where:

• \( p = 0.64 \)
• \( n = 225 \)

Calculating \( \sigma_p \):

\[ \sigma_p = \sqrt{\frac{0.64(1 - 0.64)}{225}} = \sqrt{\frac{0.64 \times 0.36}{225}} = \sqrt{\frac{0.2304}{225}} \approx \sqrt{0.001024} \approx 0.0320 \]

To summarize:

• Mean \( \mu_p = 0.6400 \)
• Standard deviation \( \sigma_p \approx 0.0320 \)

Please let me know if you need further assistance!

Answer 2

(c) Find the probability that more than 69% of the sampled adults drink coffee daily.

Answer 3

To find the probability that more than 69% of the sampled adults drink coffee daily, we can use the normal approximation to the binomial distribution. First, we'll standardize the value using the mean \( \mu_p \) and standard deviation \( \sigma_p \) calculated previously.

Step 1: Calculate the mean and standard deviation

We already found:

• Mean \( \mu_p = 0.64 \)
• Standard deviation \( \sigma_p \approx 0.0320 \)

Step 2: Convert the sample proportion to a z-score

We want to find \( P(p > 0.69) \). We first convert \( p = 0.69 \) to a z-score using the formula:

\[ z = \frac{p - \mu_p}{\sigma_p} \]

Substituting in the values:

\[ z = \frac{0.69 - 0.64}{0.0320} \approx \frac{0.05}{0.0320} \approx 1.5625 \]

Step 3: Find the probability

Next, we will look up the z-score in the standard normal distribution table or use a calculator to find the probability.

Using a standard normal distribution table or calculator, we find:

\[ P(Z > 1.5625) = 1 - P(Z \leq 1.5625) \]

Using a calculator or a z-score table, we find:

\[ P(Z \leq 1.5625) \approx 0.9408 \quad (\text{this value may vary slightly based on the table or calculator used}) \]

So,

\[ P(Z > 1.5625) = 1 - 0.9408 \approx 0.0592 \]

Final Answer

Thus, the probability that more than 69% of the sampled adults drink coffee daily is approximately:

\[ P(p > 0.69) \approx 0.0592 \]

This means there's about a 5.92% chance that more than 69% of the sampled adults drink coffee daily.

Answer 4

(d) Find the probability that the proportion of the sampled adults who drink coffee daily is between 0.6 and 0.72.

Answer 5

To find the probability that the proportion of the sampled adults who drink coffee daily is between 0.60 and 0.72, we will again use the normal approximation to the binomial distribution.

Step 1: Calculate the z-scores for both proportions

We already calculated the mean \( \mu_p = 0.64 \) and standard deviation \( \sigma_p \approx 0.0320 \).

For \( p = 0.60 \):

\[ z_{0.60} = \frac{0.60 - 0.64}{0.0320} \approx \frac{-0.04}{0.0320} \approx -1.25 \]

For \( p = 0.72 \):

\[ z_{0.72} = \frac{0.72 - 0.64}{0.0320} \approx \frac{0.08}{0.0320} \approx 2.50 \]

Step 2: Find the probabilities for the z-scores

Next, we will find the probabilities corresponding to these z-scores using standard normal distribution tables or a calculator.

1. For \( z = -1.25 \):

Using a z-table or calculator:

\[ P(Z \leq -1.25) \approx 0.1056 \]

2. For \( z = 2.50 \):

Using a z-table or calculator:

\[ P(Z \leq 2.50) \approx 0.9938 \]

Step 3: Calculate the probability that the proportion is between 0.60 and 0.72

To find the probability that the sample proportion is between 0.60 and 0.72, we can subtract the two probabilities:

\[ P(0.60 < p < 0.72) = P(Z \leq 2.50) - P(Z \leq -1.25) \]

Substituting in the values we found:

\[ P(0.60 < p < 0.72) = 0.9938 - 0.1056 \approx 0.8882 \]

Final Answer

Thus, the probability that the proportion of the sampled adults who drink coffee daily is between 0.60 and 0.72 is approximately:

\[ P(0.60 < p < 0.72) \approx 0.8882 \]

This means there is about an 88.82% chance that the proportion of sampled adults drinking coffee daily falls between 0.60 and 0.72.

Answer 6

(e) Find the probability that less than 59% of sampled adults drink coffee daily.

Answer 7

To find the probability that less than 59% of the sampled adults drink coffee daily, we will use the normal approximation to the binomial distribution.

Step 1: Calculate the z-score for \( p = 0.59 \)

We previously calculated the mean and standard deviation:

• Mean \( \mu_p = 0.64 \)
• Standard deviation \( \sigma_p \approx 0.0320 \)

Now, we need to calculate the z-score for \( p = 0.59 \):

\[ z = \frac{p - \mu_p}{\sigma_p} \]

Substituting in the values:

\[ z = \frac{0.59 - 0.64}{0.0320} \approx \frac{-0.05}{0.0320} \approx -1.5625 \]

Step 2: Find the probability using the z-score

Next, we will find the probability corresponding to this z-score. We want to find:

\[ P(Z < -1.5625) \]

Using a standard normal distribution table or calculator, we can find this value.

1. Using a calculator or a z-table, we find:

\[ P(Z < -1.5625) \approx 0.0594 \]

Final Answer

Thus, the probability that less than 59% of the sampled adults drink coffee daily is approximately:

\[ P(p < 0.59) \approx 0.0594 \]

This means there is about a 5.94% chance that less than 59% of the sampled adults drink coffee daily.

Answer 8

(f) Would it be unusual if less than 60% of the sampled adults drink coffee daily?

Answer 9

To determine whether it would be unusual if less than 60% of the sampled adults drink coffee daily, we need to look at the probability associated with that event.

Step 1: Find the z-score for \( p = 0.60 \)

We already calculated the z-score for \( p = 0.60 \):

\[ z = \frac{0.60 - 0.64}{0.0320} \approx \frac{-0.04}{0.0320} \approx -1.25 \]

Step 2: Find the probability for \( z < -1.25 \)

Using a standard normal distribution table or calculator:

\[ P(Z < -1.25) \approx 0.1056 \]

Step 3: Interpretation

In general, we consider an event unusual if its probability is less than 5%. Since the probability of observing less than 60% of the sampled adults drinking coffee daily is approximately 10.56%, this probability is greater than 5%.

Conclusion

Therefore, it would not be considered unusual if less than 60% of the sampled adults drink coffee daily, as the probability of this happening is relatively high at about 10.56%.