Quiz 1 Exercise 1 In The Attached Employee Data Graph

Quiz 1exercise 1in The Attached Employee Data1 Graph The Bar Chart F

Exercise 1: In the attached employee data, 1. Graph the bar chart for sex, pie chart for sex, and histogram for age. 2. Compute the mean and median for age. Exercise: A manufacturer of inkjet printers claims that only 5% of their printers require repairs within the first year. If, of a random sample of 18 printers, four required repairs within the first year, does this tend to refute or support the manufacturer’s claim? 2. The scores of an examination are assumed to be normally distributed with μ = 75 and σ² = 64. What is the probability that a score chosen at random will be greater than 85?

Paper For Above instruction

The analysis of employee data, alongside statistical evaluation of mechanical claims, provides a comprehensive understanding of workforce demographics and product reliability. This essay delineates the processes used to visualize employee data and evaluates the hypotheses regarding printer repair claims and exam score probabilities, emphasizing statistical techniques applicable in operational and educational contexts.

Visual representation of employee demographic data

Effective data visualization is essential in interpreting large datasets, especially in organizational management. In this context, three graphical representations are employed: a bar chart depicting the distribution of employees by sex, a pie chart illustrating the proportional gender composition, and a histogram that displays the age distribution of employees. Each method provides unique insights—bar charts facilitate comparison across categories, pie charts reveal proportional relationships, and histograms expose frequency distributions across age intervals.

Constructing these plots begins with aggregating employee data based on gender and age parameters. For the bar chart, the number of male and female employees are tallied. The pie chart utilizes these proportions to visualize gender distribution in a circular format, making it straightforward to compare segments visually. The histogram subdivides employee ages into intervals (e.g., 20-30, 30-40), counting the number of employees within each interval. The histograms are particularly helpful for identifying the age profile and potential workforce planning needs.

Computing the mean and median of age require statistical calculations from the dataset. The mean age is obtained by summing all ages and dividing by the total number of employees. The median age involves ordering all ages and identifying the middle value (or the average of two middle values in case of an even number). These measures offer central tendency insights—while the mean summarizes the average age, the median indicates the middle point, providing resilience against outliers.

Statistical evaluation of printer repair claims

The manufacturer claims that only 5% of printers require repairs within the first year. To evaluate this claim, the binomial probability model is applied since the repair requirement is a Bernoulli trial with two outcomes: requiring repair or not. The null hypothesis posits that the true repair rate is 5%. Given a sample size of 18 printers with 4 requiring repairs, the probability of observing 4 or more repairs under the null hypothesis is calculated to assess whether this is consistent with the manufacturer's claim.

Using the binomial distribution, the probability P(X ≥ 4) with p = 0.05 and n = 18 is evaluated. The probability of exactly k repairs is given by:

P(X = k) = C(n, k) p^k (1 - p)^(n - k)

Where C(n, k) is the binomial coefficient. The cumulative probability for k ≥ 4 is summed to determine the likelihood of observing at least four repairs. A low probability supports the hypothesis that the repair rate exceeds 5%, potentially refuting the claim.

Analysis of exam scores assuming normal distribution

Given the exam scores are normally distributed with mean μ = 75 and variance σ² = 64 (standard deviation σ = 8), the probability that a randomly selected score exceeds 85 is calculated using standard normal distribution techniques. The Z-score for a score of 85 is:

Z = (X - μ) / σ = (85 - 75) / 8 = 1.25

The probability P(X > 85) corresponds to the area under the standard normal curve to the right of Z = 1.25. This is expressed as:

P(X > 85) = 1 - P(Z ≤ 1.25)

Using standard normal tables or software, P(Z ≤ 1.25) ≈ 0.8944. Therefore, the probability is approximately:

1 - 0.8944 = 0.1056, or 10.56%. This indicates that there is about a 10.56% chance that a randomly chosen score exceeds 85, providing insight into the distribution of exam scores in the context of average performance levels.

Conclusion

The graphical representations of employee demographic data facilitate intuitive understanding of workforce composition, aiding management decisions. The statistical evaluation of printer repair claims suggests that observed data may challenge the manufacturer's assertion of a low repair rate, emphasizing the importance of empirical validation. Lastly, the probability analysis of exam scores supports the assumption of normality and helps educators understand score distributions, ensuring fair assessment standards. These techniques underscore the significance of statistical analysis in organizational, operational, and educational settings, enabling data-driven decision-making.

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