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.
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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.
Corbin
4.6106 Answers
1. **Identify given information:**
- Mean (\(\mu\)):12
- Standard deviation (\(\sigma\)):3.25
- Sample size (\(n\)):20
- Sample mean (\(\bar{X}\)):13.5
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 approximatelyP(\bar{X} > 13.5) \approx 0.0194
- Mean (\(\mu\)):
- Standard deviation (\(\sigma\)):
- Sample size (\(n\)):
- Sample mean (\(\bar{X}\)):
2. **Calculate the standard error of the mean (SEM):**
3. **Convert the sample mean to a z-score:**
4. **Find the probability (P) corresponding to the z-score (using Z-tables or a calculator):**
Therefore, the probability that the average coffee consumption is more than 13.5 pounds is approximately
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