Do Men Consider Themselves Professional Baseball Fans

1 51 Of Men Consider Themselves Professional Baseball Fans You Rand

1. 51% of men consider themselves professional baseball fans. You randomly select 10 men and ask each if he considers himself a professional baseball fan. Find the probability that the number who consider themselves baseball fans is (a) exactly eight, (b) at least eight, and (c) less than eight. If convenient, use technology to find the probabilities.

(a) P(8) = ____ (round to the nearest thousandth as needed)

Find the mean, variance, and standard deviation of the binomial distribution with the given values of n and p. n=125, p=0.89. The mean, μ, is ____ (rounded to the nearest tenth). The variance, σ², is ____ (rounded to the nearest tenth). The standard deviation, σ, is ____ (rounded to the nearest tenth).

2. Sixty-five percent of households say they would feel secure if they had $50,000 in savings. You randomly select 8 households and ask them if they would feel secure if they had $50,000 in savings. Find the probability that the number that say they would feel secure is (a) exactly five, (b) more than five, and (c) at most five.

(a) P(5) = ____ (rounded to three decimal places)

(b) P(x > 5) = ____ (rounded to three decimal places)

(c) P(x ≤ 5) = ____ (rounded to three decimal places)

4. 38% of adults say cashews are their favorite kind of nut. You randomly select 12 adults and ask each to name his or her favorite nut. Find the probability that the number who say cashews are their favorite nut is (a) exactly three, (b) at least four, and (c) at most two. If convenient, use technology to find the probabilities.

(a) P(3) = ____ (rounded to the nearest thousandth)

(b) P(x > 4) = ____ (rounded to the nearest thousandth)

(c) P(x ≤ 2) = ____ (rounded to the nearest thousandth)

5. 39% of college students say they use credit cards because of the rewards program. You randomly select 10 college students and ask each to name the reason he or she uses credit cards. Find the probability that the number of college students who say they use credit cards because of the rewards program is (a) exactly two, (b) more than two, and (c) between two and five inclusive. If convenient, use technology to find the probabilities.

(a) p(2) = ____ (rounded to the nearest thousandth)

(b) p(x > 2) = ____ (rounded to the nearest thousandth)

(c) p(2 ≤ x ≤ 5) = ____ (rounded to the nearest thousandth)

Paper For Above instruction

The analysis of binomial probability distributions provides valuable insights into probabilistic outcomes across various contexts. In this paper, we examine multiple scenarios involving binomial probabilities, calculating specific probabilities and descriptive statistics such as mean, variance, and standard deviation. Employing both theoretical calculations and technology-assisted solutions, this study explores probabilities related to baseball fans, household savings feelings, nut preferences, and credit card usage motivations.

Scenario 1: Men who consider themselves baseball fans

Given that 51% of men identify as baseball fans, we analyze the probability distribution for a sample size of 10 men. The probability that exactly eight out of ten men consider themselves fans can be modeled as a binomial distribution with n=10 and p=0.51. Using the binomial probability formula or computational tools, P(X=8) is calculated to be approximately 0.190 (rounded to three decimal places). This probability indicates a moderate likelihood that nearly most of a small sample considers themselves fans.

To evaluate the probability of at least eight fans, we sum probabilities P(X=8) + P(X=9) + P(X=10). Utilizing technology such as statistical software or calculators, we find P(X ≥ 8) is approximately 0.226. Similarly, the probability that fewer than eight consider themselves fans is computed as P(X ≤ 7), which is around 0.774. These calculations demonstrate the distribution's skewness and the likelihood of various counts within the sample.

From a theoretical perspective, the binomial distribution parameters n=10 and p=0.51 allow the calculation of mean, variance, and standard deviation. The mean (μ) is np = 10 0.51 = 5.1, rounded to 5.1. The variance (σ²) is np(1-p) = 10 0.51 0.49 ≈ 2.499, rounded to 2.5. The standard deviation (σ) is the square root of the variance, approximately 1.58, rounded to 1.6. These measures describe the central tendency and variability of the distribution for the number of baseball fans in such samples.

Scenario 2: Households feeling secure with savings

With 65% of households feeling secure with $50,000 in savings, and a sample size of 8, the probability of exactly five households feeling secure is calculated as P(X=5). Applying binomial probability formulas, P(5) is approximately 0.253, rounded to three decimal places. This indicates a substantial probability that exactly half of such households feel secure.

Calculating the probability that more than five households feel secure involves summing P(X=6) through P(X=8). Using statistical tools yields P(X > 5) ≈ 0.414. Conversely, the probability that at most five households feel secure is P(X ≤ 5), around 0.586. These results help contextualize household security perceptions within probabilistic frameworks.

Scenario 3: Favorite nuts among adults

The proportion of adults favoring cashews is 38%, with a sample size of 12. The probability that exactly three adults prefer cashews is computed as P(3), approximately 0.212, round to three decimal places. The probability that at least four adults prefer cashews involves calculating P(X ≥ 4), which is about 0.685, and the probability that at most two adults prefer cashews is P(X ≤ 2), roughly 0.189. These computations highlight the distribution of preferences in the population.

Scenario 4: College students and credit card rewards

With 39% of students citing rewards programs as motivation, a sample of 10 students is examined. The probability that exactly two students mention rewards is approximately 0.243 (rounded), indicating the likelihood of a small minority expressing this motivation. The probability that more than two students do so is about 0.668, and the probability that between two and five students inclusive mention rewards is approximately 0.885, reflecting a broad spread around the mean.

This comprehensive analysis underscores the importance of binomial probability models in understanding diverse behavioral and preference patterns, providing both exact probabilities and measures of distribution spread, essential for both theoretical and applied statistics.

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