A Survey Found That Women’s Height Is Normally Distributed

A Survey Found That Women Height Are Normally Distributed With A Mean

A survey found that women’s heights are normally distributed with a mean of 63.8 inches and a standard deviation of 2.8 inches. The survey also found that men’s heights are normally distributed with a mean of 68.1 inches and a standard deviation of 2.9 inches. The height requirements at an amusement park specify a minimum height of 4 feet (48 inches) and a maximum height of 6 feet 4 inches (76 inches).

Calculate the percentage of women who meet this height requirement. Similarly, determine the percentage of men who meet this requirement. Furthermore, if the height requirements are adjusted to exclude only the tallest 5% of men and the shortest 5% of women, find the new height requirements. Identify the corresponding heights for these adjusted limits.

Finally, analyze three randomly selected households with 1, 3, and 8 people respectively, to compute variances and probabilities related to sampling, as well as explore probabilities involving fruit fly populations and candy weights, using normal and binomial models as specified.

Paper For Above instruction

The analysis of height distributions among women and men at the amusement park involves understanding normal probability distributions and applying statistical concepts to determine proportions meeting specific height requirements. The heights of women are normally distributed with a mean (μ) of 63.8 inches and a standard deviation (σ) of 2.8 inches. Conversely, for men, the distribution has a mean of 68.1 inches and a standard deviation of 2.9 inches. To compute the proportion of women and men meeting the height requirement of between 48 inches and 76 inches, we utilize the properties of the normal distribution and the standard normal (z) scores.

Calculating Percentages Meeting Height Requirements

For women, the lower limit of 48 inches corresponds to the z-score:

z = (48 - 63.8) / 2.8 ≈ -5.64

Similarly, the upper limit of 76 inches corresponds to:

z = (76 - 63.8) / 2.8 ≈ 4.36

Given the symmetry and properties of the standard normal distribution, the probability that a woman's height falls between these z-scores is effectively 100%, as these z-scores are far in the tails of the normal distribution. However, the probability that a woman's height is between 48 inches and 76 inches can be calculated using a standard normal table or calculator, which yields approximately 0.9999 or 99.99%. This indicates that almost all women meet the height requirement.

For men, the z-scores are:

Lower: (48 - 68.1) / 2.9 ≈ -6.93

Upper: (76 - 68.1) / 2.9 ≈ 2.68

Similarly, due to the extremity of the lower z-score, the probability of a man being shorter than 48 inches is negligible. The probability that a man’s height is between 48 inches and 76 inches is approximately 0.9964 or 99.64%, indicating most men meet the requirement.

Adjusted Height Limits Excluding the Top and Bottom 5%

To exclude the shortest 5% of women, we find the z-score corresponding to the 5th percentile:

z = -1.645

Solving for the height:

height = μ + z × σ = 63.8 + (-1.645)(2.8) ≈ 63.8 - 4.606 ≈ 59.19 inches

To exclude the tallest 5% of men, z-score for the 95th percentile is 1.645:

height = 68.1 + 1.645×2.9 ≈ 68.1 + 4.76 ≈ 72.86 inches

Therefore, the new height requirements are at least approximately 59.19 inches for women and at most approximately 72.86 inches for men.

Sampling Variance and Probability Calculations

In the household survey, variances are calculated based on the sample sizes. For the samples of sizes 1, 3, and 8, with sampling done with replacement, the sample variance S² is computed based on the representative data. The probability of particular sample variances depends on the distribution and sample variance formulas, often involving chi-square or F-distributions.

Regarding the fruit flies, the probability of selecting two flies with certain gender compositions, and the sampling distribution of the proportion of females, can be modeled using binomial probabilities. The probability that exactly one female appears in a sample of size 2 with replacement, where the population has 2 females (Dana and Emily) and 2 males (Billy and Christen), is calculated using the binomial formula:

P = [Number of females / Total]^k × [Number of males / Total]^{n−k}

The probability distribution of the proportion of females is binomial with parameters n=2 and p=0.5 (assuming equal probability), with the mean of the distribution equal to p=0.5, and variance p(1−p)/n=0.25/2=0.125. The mean of the sampling distribution aligns with the population proportion, illustrating the unbiased nature of sample proportions in simple random sampling.

Candy Weight Analysis and Probabilities

The candy weights are normally distributed with a mean of 0.8588 grams and a standard deviation of 0.0523 grams. For a single candy, the probability that a randomly selected candy weighs more than 0.8542 grams can be found by converting to a z-score:

z = (0.8542 - 0.8588) / 0.0523 ≈ -0.0848

Referring to the standard normal table, the probability that a random candy weighs more than 0.8542 grams is approximately 0.5339.

For the entire sample of 452 candies, the sampling distribution of the mean weight will be approximately normal with mean 0.8588 grams and standard deviation (standard error):

SE = 0.0523 / √452 ≈ 0.00246

Calculating the probability that the sample mean exceeds 0.8542 grams involves finding the z-score:

z = (0.8542 - 0.8588) / SE ≈ -1.86

The probability that the mean weight of the 452 candies is at least 0.8542 grams is roughly 0.9683, indicating the company likely provides the claimed amount on the label, as the probability of sampling a mean below this is reasonably high.

Discrete Probability with Continuity Correction

For discrete variables like the number of passengers who do not show up, using the normal approximation with continuity correction enhances accuracy. The probability of exactly 4 no-shows is represented by the area to the right of 3.5, while the probability that the count is between 3.5 and 4.5 is the area between these bounds, computed through standard normal z-scores based on the binomial model.

Voter Turnout Probability

Assuming that 22% of eligible voters aged 18-24 voted in a previous study, and selecting 199 voters randomly, the probability that fewer than 48 voters voted is calculated using the normal approximation to the binomial distribution. The mean number of voters who voted is:

μ = np = 199 × 0.22 ≈ 43.78

The standard deviation is:

σ = √np(1−p) ≈ √199×0.22×0.78 ≈ 5.18

Applying the continuity correction, the z-score for fewer than 48 voters is:

z = (47.5 - 43.78) / 5.18 ≈ 0.72

From standard normal tables, the probability that fewer than 48 voters voted is approximately 0.7642, which provides insight into voting behavior patterns among young adults.

Conclusion

The various statistical analyses demonstrate applications of normal distribution, sampling, and probability calculations in real-world scenarios, ranging from height requirements and household surveys to quality control in manufacturing and election predictions. These methods underpin decision-making and policy formulation in health, industry, and civic engagement, showcasing the importance of understanding and applying statistical principles correctly.

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