Are The Summary Statistics For The Weekly Payroll Of A Small

Are The Summary Statistics For The Weekly Payroll Of A Smallc

Here are the summary statistics for the weekly payroll of a small company: lowest salary equals $300, mean salary $500, median $500, range $1400, IQR $700, first quartile $400, standard deviation $350. Based on this information, analyze the distribution of salaries and determine whether it is symmetric, skewed to the left, or skewed to the right, providing a clear explanation.

Using the empirical rule, consider a problem involving GPA distribution. At a college, GPA scores are normally distributed with a mean of 3.0 and a standard deviation of 0.6. Calculate the percentage of students with GPA between 2.4 and 3.6. Additionally, evaluate a scenario involving mortgage interest rates in Massachusetts, where rates follow a normal distribution with a mean of 5.8% and a standard deviation of 0.15%. Find the proportion of rates exceeding 6.0%. For the standard normal distribution, determine the area to the right of a Z-score of 5.1, identify the Z-score with 15% area to the right, and calculate the fraction of major league baseball games lasting less than two hours, given a mean of 170 minutes and a standard deviation of 15 minutes.

Further, interpret the Z table to find the cumulative probability for Z

In a music library with 4929 songs averaging 243.5 seconds duration (SD 114.78 seconds), find the Z-score for a 359-second song. For the distribution of game lengths in baseball, where mean is 170 minutes and SD is 15 minutes, calculate the maximum game duration longer than which only 10% of games last. Lastly, evaluate the area under the standard normal curve to the right of Z = -0.24.

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The analysis of the distribution of salaries based on the provided summary statistics reveals insights into the nature of the payroll data for the small company. The key metrics include the mean salary of $500, a median equal to the mean, a range of $1400, and an interquartile range (IQR) of $700. The standard deviation is notably high at $350, which suggests significant variability in salaries. The distribution's skewness can be inferred considering the relationship between these statistics, especially the mean, median, and the spread measures.

In statistical analysis, a symmetric distribution typically exhibits the mean equal to the median, with the measures of spread not indicating significant skewness. Given that the mean and median are both $500, and the data's range and IQR indicate a broad spread, the distribution appears centered around these central measures. Nonetheless, the large standard deviation relative to the mean hints at potential skewness, especially considering the lowest salary is $300, which is below the mean and median, with a high maximum implied by the range ($1700). This suggests that the tail of the salary distribution might be longer on the higher end, implying a right skew.

Supporting this reasoning, the skewness can be subtly assessed through comparison of mean and median with the spread. Since the mean equals the median, a common assumption is that the distribution is symmetric; however, the high standard deviation and the explicitly high maximum salaries (implied from the range) could skew the distribution to the right. Typically, salary distributions tend to be right-skewed because higher salaries are less frequent but significantly larger, pulling the mean upwards relative to the median.

Moving to the application of the empirical rule, which states that for a normal distribution approximately 68% of data falls within one standard deviation of the mean, 95% within two standard deviations, and 99.7% within three, these principles aid in estimating the spread of data. In the GPA scenario, where the mean is 3.0 and SD is 0.6, the interval from 2.4 to 3.6 spans from (mean - SD) to (mean + SD), capturing approximately 68% of students' GPAs, consistent with the empirical rule (Everitt & Skrondal, 2010). Similarly, for mortgage interest rates in Massachusetts, with a mean of 5.8% and SD of 0.15%, rates exceeding 6.0% correspond to a Z-score of approximately 4.67, indicating a very small proportion of rates above this level, roughly 0.003%, effectively negligible in practical terms (Zweig & Campbell, 1993).

Computing probabilities and areas under normal curves are fundamental in statistical inference. For example, the area to the right of Z = 5.1 is virtually zero, indicating the probability of a value exceeding this Z-score is negligible. Conversely, identifying the Z-score with an area of 0.15 to the right involves finding the inverse of the cumulative distribution function (z-table), which yields approximately Z = 1.04 (Miller, 2020). These calculations are integral in assessing extreme values and understanding distributions.

Applying these principles to baseball game durations, with a mean of 170 minutes and SD of 15 minutes, to find the duration longer than which only 10% of games last, involves calculating the Z-score corresponding to the 90th percentile. Using standard normal tables, the Z-score is about 1.28, translating to a game length of approximately 182 minutes (170 + 1.28 * 15).

Understanding areas under the normal curve also aids in interpreting Z-table values such as the probability that Z is less than 1.34, which is approximately 0.9099, or about 90.99%. For Jen's phone bills, with a mean of $55 and SD of $12, the percentage between $19 and $91 can be estimated using the empirical rule. The interval from $19 to $91 encompasses approximately 99.7% of bills, considering these bounds are roughly ±3 SD from the mean (Everitt & Skrondal, 2010). This demonstrates the extensive coverage of normal distribution within three standard deviations.

Further, calculating the area to the left of Z = 1.11 involves cumulative probability, approximately 0.8665, indicating that about 86.65% of values are below this Z-score. For the Z-score corresponding to 75% area to the right, or equivalently 25% to the left, the Z-value is approximately -0.67. Also, given that the area to the right of Z = 0.8 is 0.2, the Z-score from the standard normal distribution is about 0.84.

Regarding heart rates of patients with heart disease, the fact that the central 95% lies between 73 bpm and 121 bpm allow us to estimate the mean and SD. Assuming the middle 95% corresponds to approximately ±2 SDs, the mean is roughly at the midpoint (97 bpm), and SD can be approximated as (121 - 73) / 4 or about 12 bpm, aligning with the empirical rule. Similarly, for systolic blood pressure in young women, the proportion with BP between 96 mmHg and 144 mmHg corresponds to within ±2 SDs from the mean of 120 mmHg, indicating approximately 95% coverage (Zweig & Campbell, 1993).

In the case of the song durations, with a mean of 243.5 seconds and a SD of 114.78 seconds, the z-score for a 359-second song is calculated as (359 - 243.5)/114.78 ≈ 1.01, showing it is over one standard deviation above the mean. For baseball game durations, finding the cutoff where only 10% of games last longer involves z = 1.28, and calculating the corresponding time: 170 + (1.28 * 15) = approximately 182 minutes.

The area to the right of Z = -0.24 can be calculated as 1 - P(Z

References

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