(a) Find the mean u^p
11 answers
To find the mean \( \mu^p \) of the proportion of U.S. adults who drink coffee daily, we can use the following formula:
\[ \mu^p = p \]
where \( p \) is the population proportion. In this case, the National Coffee Association reports that 65% of U.S. adults drink coffee daily, which can be expressed as:
\[ p = 0.65 \]
Thus, the mean \( \mu^p \) is:
\[ \mu^p = 0.65 \]
So, the mean \( \mu^p \) rounded to four decimal places is:
\[ \mu^p = 0.6500 \]
To find the standard deviation \( \sigma^p \) of the sample proportion, we can use the following formula:
\[ \sigma^p = \sqrt{\frac{p(1 - p)}{n}} \]
where:
First, we calculate \( 1 - p \):
\[ 1 - p = 1 - 0.65 = 0.35 \]
Now we can substitute \( p \) and \( n \) into the formula:
\[ \sigma^p = \sqrt{\frac{0.65 \cdot 0.35}{225}} \]
Calculating the numerator:
\[ 0.65 \cdot 0.35 = 0.2275 \]
Now, the expression under the square root becomes:
\[ \frac{0.2275}{225} \approx 0.0010111111 \]
Now, take the square root:
\[ \sigma^p = \sqrt{0.0010111111} \approx 0.0317 \]
Thus, the standard deviation \( \sigma^p \) rounded to four decimal places is:
\[ \sigma^p \approx 0.0317 \]
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. This is valid because the sample size is sufficiently large.
From earlier calculations, we have: \[ \mu^p = p = 0.65 \] \[ \sigma^p \approx 0.0317 \]
We use the z-score formula to standardize the sample proportion:
\[ z = \frac{\hat{p} - \mu^p}{\sigma^p} \] Substituting in the values:
\[ z = \frac{0.69 - 0.65}{0.0317} \approx \frac{0.04}{0.0317} \approx 1.2627 \]
Now, we want to find the probability \( P(X > 0.69) \), which corresponds to \( P(Z > 1.2627) \).
Using standard normal distribution tables or a calculator, we can find the cumulative probability for \( z = 1.2627 \):
\[ P(Z < 1.2627) \approx 0.8962 \]
Therefore, the probability of \( P(Z > 1.2627) \) is:
\[ P(Z > 1.2627) = 1 - P(Z < 1.2627) = 1 - 0.8962 = 0.1038 \]
Thus, the probability that more than 69% of the sampled adults drink coffee daily is approximately:
\[ \boxed{0.1038} \]
To find the probability that the proportion of the sampled adults who drink coffee daily is between 0.64 and 0.70, we will again use the normal approximation to the binomial distribution. We will calculate the z-scores for both 0.64 and 0.70 using the mean and standard deviation calculated previously.
For \( \hat{p} = 0.64 \): \[ z_1 = \frac{0.64 - 0.65}{0.0317} \approx \frac{-0.01}{0.0317} \approx -0.3155 \]
For \( \hat{p} = 0.70 \): \[ z_2 = \frac{0.70 - 0.65}{0.0317} \approx \frac{0.05}{0.0317} \approx 1.5771 \]
Using the standard normal distribution table or a calculator:
For \( z_1 \approx -0.3155 \): \[ P(Z < -0.3155) \approx 0.3768 \]
For \( z_2 \approx 1.5771 \): \[ P(Z < 1.5771) \approx 0.9429 \]
Now, we need to find the probability that the proportion is between 0.64 and 0.70:
\[ P(0.64 < \hat{p} < 0.70) = P(Z < 1.5771) - P(Z < -0.3155) \]
\[ P(0.64 < \hat{p} < 0.70) \approx 0.9429 - 0.3768 = 0.5661 \]
Thus, the probability that the proportion of the sampled adults who drink coffee daily is between 0.64 and 0.70 is approximately:
\[ \boxed{0.5661} \]
To find the probability that less than 55% of the sampled adults drink coffee daily, we will again use the normal approximation to the binomial distribution.
\[ z = \frac{\hat{p} - \mu^p}{\sigma^p} \]
Substituting in the values:
\[ \mu^p = p = 0.65 \] \[ z = \frac{0.55 - 0.65}{0.0317} = \frac{-0.10}{0.0317} \approx -3.157 \]
Now we need to find \( P(Z < -3.157) \).
Using standard normal distribution tables or a calculator, we find that:
\[ P(Z < -3.157) \approx 0.0008 \]
Thus, the probability that less than 55% of the sampled adults drink coffee daily is approximately:
\[ \boxed{0.0008} \]
To determine whether it would be unusual for less than 55% of the sampled adults to drink coffee daily, we can refer to the probability we calculated in part (e).
The probability of finding less than 55% of the sampled adults drinking coffee daily was calculated to be:
\[ P(X < 0.55) \approx 0.0008 \]
In statistics, an event is often considered unusual if its probability is less than 5% (0.05). Since the probability we found (0.0008) is much less than 0.05, it can be classified as an unusual event.
Thus, it would be considered unusual for less than 55% of the sampled adults to drink coffee daily because the probability of this occurring is very low (approximately 0.08%). This suggests that a proportion of coffee drinkers that low is significantly below the expected average of 65%, indicating that something atypical may be influencing coffee consumption in this specific sample.
In summary, it is unusual if less than 55% of the sampled adults drink coffee daily due to the very low probability of this outcome occurring.
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