Question 1: The Following Shows The Temperatures High 839955

Question 1the Following Shows The Temperatures High Low And Weather

The following shows the temperatures (high, low) and weather conditions in a given Sunday for some selected world cities. For the weather conditions, the following notations are used: c = clear; cl = cloudy; sh = showers; pc = partly cloudy. City Hi Lo Condition Acapulco 99 77 pc Bangkok 92 78 pc Mexico City 77 57 sh Montreal 72 56 pc Paris 77 58 c Rome 88 68 cl

1. Is “Montreal” an element, variable, or observation?

2. Provide the observation for Rome.

3. Give an example of a categorical variable.

4. Provide the ranges for low and high temperatures.

Paper For Above instruction

In the context of the provided data, “Montreal” represents an element or unit of observation, specifically a city observed on a particular Sunday. It is a single data point within a dataset comprising various cities’ weather conditions. The observation for Rome encompasses the temperature highs and lows and the weather condition, which are 88°F high, 68°F low, and 'cl' indicating partly cloudy weather. An example of a categorical variable in this dataset is the weather condition, as it classifies the weather into categories such as clear, cloudy, showers, and partly cloudy. The temperature ranges for the cities are derived from the minimum and maximum observed values: the low temperatures range from 56°F (Montreal) to 78°F (Bangkok), while the high temperatures range from 72°F (Montreal) to 99°F (Acapulco). These ranges help understand the variability in temperature across different cities on the specified Sunday period.

Question 2: Student Grades Analysis

The student’s grades in 20 courses are: D, B, C, D, A, B, F, A, A, C, B, B, B, C, B, D, A, B, C, A. To analyze these grades, we first develop a frequency distribution:

• A: 4
• B: 7
• C: 3
• D: 3
• F: 1

This distribution indicates that B grades are most common, followed by A, C, D, and F.

For visualization, a bar chart titled “Grades Distribution” with each grade on the x-axis and frequency on the y-axis provides clear comparative insights. A pie chart illustrating the proportions of each grade, in percentage, complements this view by showing the relative distribution visually. These different charts serve unique purposes: while the bar chart effectively compares the frequencies directly, the pie chart conveys the proportion of each grade relative to the total.

The bar chart is preferable over the pie chart when demonstrating precise comparisons among categories, especially when the categories are numerous or differences are subtle, since bar charts allow easier reading of exact frequencies. Pie charts are better suited for showing proportionate parts of a whole when the focus is on the relative distribution rather than precise counts.

Question 3: Hours Worked Analysis

The number of hours worked per week by ten students is: 40, 35, 45, 20, 50, 38, 42, 40, 36, 44.

Calculations:

• Mean: Sum all hours and divide by 10: (40+35+45+20+50+38+42+40+36+44)/10 = 410/10 = 41 hours.
• Median: Arrange in order: 20, 35, 36, 38, 40, 40, 42, 44, 45, 50. The median is the average of the 5th and 6th values: (40+40)/2 = 40 hours.
• Mode: The most frequent value is 40 hours, appearing twice.

The mean provides an overall average but can be skewed by extreme values. The median reflects the middle value and is robust against outliers, making it a better measure of central tendency if the data is skewed. The mode indicates the most common value, which is useful for understanding the most typical workload.

The standard deviation, which measures dispersion around the mean, is calculated as approximately 7.4 hours. It indicates that most students’ hours are within roughly 7.4 hours of the mean, thus providing a sense of variability in work hours.

Question 4: Probabilities of Events A, B, C, D

Given probabilities: P(A) = 0.5, P(B) = 0.3, P(C) = 0.15, P(A U D) = 0.7, P(A ∩ C) = 0.05, P(A | B) = 0.22, P(A ∩ D) = 0.

To compute P(D):

P(A U D) = P(A) + P(D) - P(A ∩ D) → 0.7 = 0.5 + P(D) - 0, thus P(D) = 0.2.

To compute P(A ∩ B):

P(A | B) = P(A ∩ B) / P(B) → 0.22 = P(A ∩ B) / 0.3 → P(A ∩ B) = 0.066.

To compute P(A | C):

P(A | C) = P(A ∩ C) / P(C) → 0.05 / 0.15 ≈ 0.333.

The complement probability of C, P(C̄):

P(C̄) = 1 - P(C) = 0.85.

Mutually exclusive events cannot occur simultaneously; for example, if two events are mutually exclusive, P(A ∩ B) = 0.

Question 5: Binomial Probabilities for Machine Defects

The probability that an item is non-defective is 0.75, so defective probability is 0.25. For 8 items:

• Probability exactly one item is defective:

P(X=1) = C(8,1) (0.25)^1 (0.75)^7 ≈ 8 0.25 0.1335 ≈ 0.267.

• Probability exactly three are non-defective:

P(X=5 defective) = C(8,5) (0.25)^5 (0.75)^3 ≈ 56 0.00098 0.422 ≈ 0.023.

• Probability at least 6 are non-defective:

Sum of probabilities for X=6,7,8 non-defective:

P(X≥6) = P(X=6) + P(X=7) + P(X=8), which sums approximately to 0.213 + 0.057 + 0.01 ≈ 0.28.

Question 6: Salary Distribution

Assuming salaries are normally distributed with a mean of $55,000 and standard deviation of $4,000:

• The probability a graduate earns at least $52,700:

Z = (52,700 - 55,000) / 4,000 = -2,300 / 4,000 = -0.575. From Z-tables, P(Z ≥ -0.575) ≈ 0.717.

• Percentage earning less than $45,000:

Z = (45,000 - 55,000) / 4,000 = -10,000 / 4,000 = -2.5. P(Z ≤ -2.5) ≈ 0.0062, or 0.62%.

• Percent within one standard deviation of the mean:

Salaries within $55,000 ± $4,000: from $51,000 to $59,000. The probability is approximately 68% based on empirical rule.

• Range of salaries within one standard deviation:

$51,000 to $59,000.

Question 7: Programmer Income Probabilities

Data: Males ($250, $270, $255), Females ($285, $290), sum to six programmers.

Probability that a randomly selected salary exceeds $252:

Eligible salaries: $270, $285, $290, $255, $270. Total of 6; 4 exceed $252, so probability: 4/6 ≈ 0.667.

Probability that a randomly selected programmer is male:

Number of males = 3; total = 6; probability = 3/6 = 0.5.

Probability that a male programmer earns over $260:

Salaries over $260 among males: $270 only. Probability: 1/3 ≈ 0.333.

Question 8: Lunch Expenditure Analysis

Average expenditure: $6; standard deviation: $1.

Probability that the sample mean (e.g., of multiple lunches) ≥ $4.50:

Z = (4.50 - 6) / (1/√n). For the sample mean, assuming large enough sample, the standard error is used. If sample size n is 1, Z = -1.5. The probability that the mean ≤ $4.50 is about 0.067, so ≥ $4.50 corresponds to 1 - 0.067 ≈ 0.933.

Probability that the sample mean is exactly $7.90 is practically zero for continuous distributions.

Regarding Doria's lunch spending of $2.99, since this value is quite below the mean of $6 and assuming standard deviation of $1, the Z-score is (2.99 - 6)/1 = -3.01, indicating it is more than 3 standard deviations below the mean, an extremely rare event, thus not typical.

References

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