Gabriela Garcia, a researcher at Honduras Coffee Corporation, is interested in determining the rate of coffee use per household in Honduras. She believes that annual household consumption has a normal distribution with a mean of 12 pounds and a standard deviation of 3.25 pounds. If Gabriela takes a sample of 20 households, what is the probability that Honduran households consume more than 13.5 pounds of coffee annually on average? Write your answer using 4 decimal places.

Name: Solved: Gabriela Garcia, a researcher at Honduras Coffee Corporation, is interested in determining the rate of coffee use per household in Honduras. She believes that annual household consumption has a normal distribution with a mean of 12 pounds and a standard deviation of 3.25 pounds. If Gabriela takes a sample of 20 households, what is the probability that Honduran households consume more than 13.5 pounds of coffee annually on average? Write your answer using 4 decimal places. - MathMaster
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Answer to a math question Gabriela Garcia, a researcher at Honduras Coffee Corporation, is interested in determining the rate of coffee use per household in Honduras. She believes that annual household consumption has a normal distribution with a mean of 12 pounds and a standard deviation of 3.25 pounds. If Gabriela takes a sample of 20 households, what is the probability that Honduran households consume more than 13.5 pounds of coffee annually on average? Write your answer using 4 decimal places.

$Expert avatar$

Corbin

4.6

107 Answers

1. **Identify given information:**
- Mean ($\mu$):

pounds
- Standard deviation ($\sigma$):

3.25

pounds
- Sample size ($n$):

households
- Sample mean ($\bar{X}$):

13.5

pounds

2. **Calculate the standard error of the mean (SEM):**

\text{SEM} = \frac{\sigma}{\sqrt{n}} = \frac{3.25}{\sqrt{20}} = \frac{3.25}{4.4721} \approx 0.7274

3. **Convert the sample mean to a z-score:**

z = \frac{\bar{X} - \mu}{\text{SEM}} = \frac{13.5 - 12}{0.7274} \approx \frac{1.5}{0.7274} \approx 2.0623

4. **Find the probability (P) corresponding to the z-score (using Z-tables or a calculator):**

P(Z > 2.0623) = 1 - P(Z \leq 2.0623) \approx 1 - 0.9806 = 0.0194

Therefore, the probability that the average coffee consumption is more than 13.5 pounds is approximately

P(\bar{X} > 13.5) \approx 0.0194

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