UBI Business To Hand In November 15, 2017 Status

UBI BUSINESS To Hand in 15th November th November 2017 STATISTICS-DISTRIBUTIONS HW GRADED 5% Binomial Distribution

A statistician working for major league baseball determined the probability that the hitter will be out on ground balls to be 0.80. In a game where there are 15 ground balls, find the probability that all of these were outs.

Flaws in a material tend to occur randomly and independently at a rate of one every 250 square feet. What is the probability that a carpet that is 8 x 10 feet contains no flaws?

The lifetimes of light bulbs that are advertised to last for 4000 hours are normally distributed with a mean of 4200 hours and a standard deviation of 175 hours. What is the probability that a bulb lasts longer than the advertised figure?

Use the t-table (p.288) to find the following values of t.

• a. If the significance level is 0.05 with 195 degrees of freedom, find t.
• b. If the significance level is 0.10 with 15 degrees of freedom, find t.
• c. If the significance level is 0.10 with 23 degrees of freedom, find t.
• d. If the significance level is 0.025 with 83 degrees of freedom, find t.

Paper For Above instruction

The following comprehensive analysis examines four fundamental statistical concepts—binomial distribution, Poisson distribution, normal distribution, and Student's t-test—applied to real-world scenarios. Each section provides detailed calculations and interpretations relevant to the respective problem, elucidating core statistical principles and their practical applications.

Binomial Distribution: Probability of All Ground Outs

The binomial distribution models the number of successes in a fixed number of independent Bernoulli trials, each with the same probability of success. In this context, success is defined as the hitter being out on a ground ball. The probability that the hitter will be out on a ground ball is given as p = 0.80, and there are n=15 ground balls in the game.

The probability that all 15 ground balls result in outs is calculated as:

P(all outs) = pⁿ = 0.80¹⁵

Calculating this value: 0.80¹⁵ ≈ 0.0352. Therefore, there is approximately a 3.52% chance that all ground balls are outs in this scenario.

This low probability underscores the rarity of such a perfect outcome, emphasizing the stochastic nature of baseball plays and the importance of probabilistic modeling in sports analytics.

Poisson Distribution: Probability of No Flaws in a Carpet

The Poisson distribution models the number of occurrences of a rare event within a fixed space or time frame, assuming these events occur independently at a constant rate. In this instance, flaws in a material occur at a rate λ, calculated based on the surface area.

The flaw rate is 1 flaw per 250 square feet. The carpet measures 8 feet by 10 feet, totaling an area of 80 square feet.

The expected number of flaws, λ, in this area is:

λ = (rate) × (area) = (1 / 250) × 80 = 0.32

The probability of observing zero flaws follows the Poisson probability mass function:

P(X=0) = e^-λ = e^-0.32

Calculating: e^-0.32 ≈ 0.726. This indicates approximately a 72.6% chance that the carpet contains no flaws, an insight critical for quality assurance in manufacturing processes.

Normal Distribution: Probability of Bulb Lasting Longer Than Advertised

Given the lifetime of light bulbs follows a normal distribution with mean μ = 4200 hours and standard deviation σ = 175 hours, the task is to compute the probability that a bulb lasts longer than the advertised mean of 4000 hours.

The z-score associated with 4000 hours is calculated as:

z = (X - μ) / σ = (4000 - 4200) / 175 ≈ -1.14

Since we are interested in the probability of lasting longer than 4000 hours, we need P(X > 4000), which is equivalent to P(z > -1.14).

From standard normal distribution tables, P(z -1.14) = 1 - 0.1271 = 0.8729.

Therefore, there is approximately an 87.29% chance that a bulb lasts longer than 4000 hours, reflecting the reliability of the product.

Student’s t-Distribution: Critical t-values for Various Significance Levels

Student's t-test is used to determine the significance of differences between means when the sample size is small or the population variance is unknown. The critical t-values depend on the degrees of freedom (df) and the significance level (α).

• a. For α = 0.05 and df = 195, the critical t-value is approximately 1.97.
• b. For α = 0.10 and df = 15, the critical t-value is approximately 1.34.
• c. For α = 0.10 and df = 23, the critical t-value is approximately 1.35.
• d. For α = 0.025 and df = 83, the critical t-value is approximately 2.01.

These t-values are obtained from t-distribution tables and are essential for hypothesis testing, allowing researchers to assess whether observed differences are statistically significant with specified confidence levels.

Conclusion

Through the analysis of these scenarios, we demonstrate the versatility and application of different statistical distributions and testing methods in real-world contexts. The binomial distribution effectively models binary outcomes in sports, the Poisson distribution characterizes rare events in manufacturing, the normal distribution assesses continuous variables like product lifespan, and the Student's t-test guides hypothesis testing in small samples. These tools are integral to data-driven decision-making across industries, emphasizing the importance of understanding statistical principles for accurate interpretation of data.

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