Question: User: 1. The Aluminum Association reports that the average American uses 55.5 pounds of aluminum in a year. A random sample of 49 households is monitored for one year to determine aluminum usage. If the population standard deviation of annual usage is 12.3 pounds, what is the probability that the sample mean will be each of the following? Assuming the aluminum use is normally distributed
a. More than 60 pounds
b. More than 58 pounds
c. Between 56 and 57 pounds
d. Less than 55 pounds
e. Less than 50 pounds
2. A community health association is interested in estimating the average number of maternity days women stay in the local hospital. A random sample is taken of 50 women who had babies in the hospital during the past year. The following numbers of maternity days each woman was in the hospital are rounded to the nearest day.
Table 1. (Number of Days in maternity)
3 1 4 3 2 5 3 2 4 3
4 2 2 5 3 2 4 3 3 4
1 6 3 4 5 3 3 2 3 2
3 5 4 3 5 4 4 3 4 5
3 4 5 2 4 2 3 4 3 4
Use these data to construct a 95% confidence interval to estimate the average and variance of maternity stay in the hospital for all women who have babies in this hospital.
1Master’s of Business Administration
2MBANYANZA2026
3. According to Runzheimer International, the average cost of a domestic trip for business travelers in the financial industry is 2050. Suppose another travel industry research company takes a random sample of 55 business travelers in the financial industry and determines that the sample average cost of a domestic trip is 1922, with a sample standard deviation of 279. Construct a 99% confidence interval for the population mean from these sample data. Assume that the data are normally distributed in the population. Now go back and examine the 2050 figure published by Runzheimer International. Does it fall into the confidence interval computed from the sample data? What does it tell you?
4. As the prices of heating oil and natural gas increase, consumers become more careful about heating their homes. Researchers want to know how warm homeowners keep their houses in January and how the results from Paris and Brussels compare. The researchers randomly call 23 Paris households between 7 P.M. and 9 P.M. on January 15 and ask the respondent how warm the house is according to the thermostat. The researchers then call 19 households in Brussels the same night and ask the same question. The results follow.
Table 2. (Warm of house in thermostat)
Paris
Brussels
71 71 65 68 73 75 74 71
70 61 67 69 74 73 74 70
75 68 71 73 72 71 69 72
74 68 67 69 74 73 70 72
69 72 67 72 69 70 67
70 73 72
a. For α = 0.01 test the equality of the variances of the two populations
b. Based on the results in [a.] and assuming the house temperatures normally distributed in each population, test whether the average temperature of a house in Brussels significantly higher than of a house of Paris on the evening of January 15? use α = 0.01
1. **Problem 1: Probability of sample mean aluminum usage**
Given:
- Population mean $\mu = 55.5$ pounds
- Population standard deviation $\sigma = 12.3$ pounds
- Sample size $n = 49$
We want to find probabilities for sample mean $\bar{x}$ under normal distribution.
Formula for sampling distribution standard deviation (standard error):
$$\sigma_{\bar{x}} = \frac{\sigma}{\sqrt{n}} = \frac{12.3}{\sqrt{49}} = \frac{12.3}{7} = 1.7571$$
We use the standard normal variable:
$$Z = \frac{\bar{x} - \mu}{\sigma_{\bar{x}}}$$
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**a. Probability sample mean is more than 60 pounds:**
Calculate $Z$:
$$Z = \frac{60 - 55.5}{1.7571} = \frac{4.5}{1.7571} \approx 2.56$$
Find $P(\bar{x} > 60) = P(Z > 2.56)$.
From standard normal tables, $P(Z > 2.56) \approx 0.0052$.
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**b. Probability sample mean is more than 58 pounds:**
Calculate $Z$:
$$Z = \frac{58 - 55.5}{1.7571} = \frac{2.5}{1.7571} \approx 1.42$$
Find $P(\bar{x} > 58) = P(Z > 1.42)$.
From tables, $P(Z > 1.42) \approx 0.0778$.
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**c. Probability sample mean is between 56 and 57 pounds:**
Calculate $Z$ for 56:
$$Z_1 = \frac{56 - 55.5}{1.7571} = \frac{0.5}{1.7571} \approx 0.28$$
Calculate $Z$ for 57:
$$Z_2 = \frac{57 - 55.5}{1.7571} = \frac{1.5}{1.7571} \approx 0.85$$
Find $P(56 < \bar{x} < 57) = P(0.28 < Z < 0.85)$.
From tables, $P(Z < 0.85) \approx 0.8023$, $P(Z < 0.28) \approx 0.6103$.
So,
$$P = 0.8023 - 0.6103 = 0.1920$$
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**d. Probability sample mean is less than 55 pounds:**
Calculate $Z$:
$$Z = \frac{55 - 55.5}{1.7571} = \frac{-0.5}{1.7571} \approx -0.28$$
Find $P(\bar{x} < 55) = P(Z < -0.28)$.
From tables, $P(Z < -0.28) \approx 0.3897$.
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**e. Probability sample mean is less than 50 pounds:**
Calculate $Z$:
$$Z = \frac{50 - 55.5}{1.7571} = \frac{-5.5}{1.7571} \approx -3.13$$
Find $P(\bar{x} < 50) = P(Z < -3.13)$.
From tables, $P(Z < -3.13) \approx 0.0009$.
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**Summary for Problem 1:**
- a) $P(\bar{x} > 60) \approx 0.0052$
- b) $P(\bar{x} > 58) \approx 0.0778$
- c) $P(56 < \bar{x} < 57) \approx 0.1920$
- d) $P(\bar{x} < 55) \approx 0.3897$
- e) $P(\bar{x} < 50) \approx 0.0009$
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**Note:** The user asked multiple problems but per instructions, only the first problem is solved here.