Show All Work Chapter Five 1 If Light Bulbs Have Lives ✓ Solved

Show All Workchapter Five1 If Light Bulbs Have Lives That Are Normall

Analyze the problem of normally distributed light bulb lifespans and related probability questions with given means and standard deviations. Calculate percentages, percentiles, and probabilities based on standard normal distribution concepts. Address questions involving within-1-standard-deviation ranges, percentile calculations, and probability of exceeding or falling below specific values. Apply the empirical rule for normal distributions, and perform z-score calculations as needed to find probabilities and critical points, including using standard normal tables. Incorporate concepts of sampling distributions and the Central Limit Theorem for sample means, especially in questions involving annual precipitation and exam scores. Evaluate the significance of probabilities in hypothesis testing context, and compute simple probabilities from discrete sample spaces, such as dice rolls, coin flips, and drawing marbles. Use probability models to estimate expected values for biased coins, and interpret life expectancy data from actuarial tables to answer demographic questions involving mortality probabilities and life spans.

Sample Paper For Above instruction

In this paper, we address a series of statistical problems involving normal distributions, probability calculations, and hypothesis testing, demonstrating how to approach each with proper statistical techniques and reasoning.

Question 1: Probability of Light Bulb Lifespan Less Than Mean

The lifespans of light bulbs are normally distributed with a mean (μ) of 2500 hours and a standard deviation (σ) of 500 hours. To determine the percentage of bulbs that have a lifespan less than 2500 hours, we calculate the z-score:

z = (X - μ) / σ = (2500 - 2500) / 500 = 0

Using the standard normal distribution table, a z-score of 0 corresponds to a cumulative probability of 0.5 or 50%. Thus, 50% of light bulbs have a lifespan less than 2500 hours.

Question 2: Percentage Within One Standard Deviation

For bulbs with a mean of 370 hours and a standard deviation of 5 hours, the percentage of bulbs with lifetimes within one standard deviation of the mean—that is, between 365 and 375 hours—is determined by the empirical rule. Approximately 68% of data in a normal distribution lies within ±1σ of the mean. Therefore, 68% of the bulbs have lifespans between 365 and 375 hours.

Question 3: Phone Bills—Interval Covering 68%

Her monthly phone bills are normally distributed with a mean of $60 and a standard deviation of $12. Applying the empirical rule, approximately 68% of her bills fall within one standard deviation of the mean:

Lower bound: $60 - $12 = $48

Upper bound: $60 + $12 = $72

Hence, 68% of her bills are between $48 and $72.

Question 4: 25th Percentile of Phone Bills

For bills with mean $50 and standard deviation $10, to find the 25th percentile, we find the z-score corresponding to the 25th percentile, which is approximately -0.674 (from z-tables). The raw score (X) is then:

X = μ + zσ = 50 + (-0.674)(10) = 50 - 6.74 = $43.26

Therefore, the 25th percentile is approximately $43.26.

Question 5: Diameter of Bolts Greater Than 0.32 Inches

The diameters are normally distributed with μ = 0.30 inches and σ = 0.01 inches. Calculating the z-score for 0.32 inches:

z = (0.32 - 0.30) / 0.01 = 2

The probability of a z-score greater than 2 is 1 - P(z ≤ 2). From standard normal tables, P(z ≤ 2) ≈ 0.9772, so the probability that a bolt exceeds 0.32 inches is:

1 - 0.9772 = 0.0228 or 2.28%.

Question 6: Precipitation Less Than 90.8 Inches Over 25 Years

The mean annual precipitation is 88 inches with a standard deviation of 10 inches. For the mean of 25 years, the standard error of the mean is:

σₓ̄ = σ / √n = 10 / √25 = 2

The z-score for 90.8 inches is:

z = (90.8 - 88) / 2 = 1.4

P( x̄

Question 7: Sample Mean Less Than 70 in a Test

Exam scores are normally distributed with μ=73 and σ=7.73. For a sample of 24 students, the standard error of the mean is:

SE = 7.73 / √24 ≈ 1.577

Calculating z for a mean score of 70:

z = (70 - 73) / 1.577 ≈ -1.91

Using standard normal tables, P(z

Question 8: Scores Between –1 and +1 Standard Deviations

Standardized scores within ±1σ of the mean (score = 50, SD = 10):

Range: 50 ± 10, i.e., 40 to 60

b. The percentage of all scores falling between –1 and +1 SD is approximately 68%, by empirical rule.

c. A score at +2 SD is:

50 + 2×10 = 70

d. The percentage of scores between +1 and +2 SD is approximately 13.6%, since about 34% fall between 1 and 2 SD, with the remaining 34% between 0 and 1 SD and the total within 2 SDs being 95%. The difference gives the percentage between +1 and +2 SD.

Question 1 Continued: Significance Testing

a. The probability of rolling a sum of 12 with two dice is 1/36 ≈ 0.0278, which is less than 0.05, so it is considered significant. In hypothesis testing language, such a rare event would be significant at α=0.05.

b. With 500 school bus routes, 480 arrived on time, so the probability that a bus arrives late is (20/500) = 0.04. Since 0.04

Question 2: Probability of At Least One Head in 3 Coin Flips

The probability of no heads (all tails): (1/2)^3 = 1/8 = 0.125. Therefore, the probability of at least one head is:

1 - 0.125 = 0.875 or 87.5%.

Question 3: Probability of Equally Likely Events

With 64 equally likely events, each event's probability is:

1/64 ≈ 0.015625.

Question 4: Probability of Selecting a Blue Marble

Total marbles = 4 + 3 + 7 = 14

Probability of selecting a blue marble:

3/14 ≈ 0.2143 or 21.43%.

Question 5: Income at Least $98,000

Suppose the incomes are: $112,000, $90,000, $94,000, $98,000, $82,000, $94,000, $98,000, $78,000. The probability that a randomly selected member earns at least $98,000 is the number of members earning ≥$98,000 divided by total members:

Members earning ≥$98,000: 2 (the two with $98,000) out of 8 total members, so:

2/8 = 0.25 or 25%.

Question 6: Expected Number of Heads with Biased Coin

The coin's probability of heads is p=¼. Expectation in 32 flips: E = n × p = 32 × ¼ = 8 heads.

Question 7: Mortality and Life Expectancy from Table

To compute expected remaining lifespan for females aged 60:

Sum over the death probabilities weighted by age. Approximate calculations from the table can be conducted, assuming data are available. The same applies to males aged 70.

Number of females surviving to age 61: about 100,000 × (1 - probability of death at age 60). The exact calculations depend on the data, but typically, a significant proportion remain alive past age 60 and 70.

This allows estimation of average life expectancy and survival counts, based on the given probabilities and initial populations.

Conclusion

Through these problems, it is evident that understanding the properties of the normal distribution and associated probability calculations is essential in various real-world contexts—ranging from product quality control and demographic analysis to hypothesis testing in statistics. Properly applying z-scores, empirical rules, and probability rules enables practitioners to make informed decisions and predictions based on data distributions.

References

• Devore, J. L. (2015). Probability and Statistics for Engineering and the Sciences (8th ed.). Cengage Learning.
• Moore, D. S., McCabe, G. P., & Craig, B. A. (2017). Introduction to the Practice of Statistics (9th ed.). W. H. Freeman.
• Ross, S. M. (2014). Introduction to Probability and Statistics (11th ed.). Academic Press.
• Wackerly, D. D., Mendenhall, W., & Scheaffer, R. L. (2014). Mathematical Statistics with Applications (7th ed.). Cengage Learning.
• Lind, D. A., Marchal, W. G., & Wathen, S. A. (2014). Statistical Techniques in Business & Economics (16th ed.). McGraw-Hill Education.
• Freund, J. E. (2010). Modern Elementary Statistics (12th ed.). Pearson.
• Ghasemi, A., & Zahediasl, S. (2012). Normality tests for statistical analysis: a guide for non-statisticians. International Journal of Endocrinology and Metabolism, 10(2), 486-489.
• Wilcox, R. R. (2012). Introduction to Robust Estimation and Hypothesis Testing. Academic Press.
• Siegel, S., & Castellan, N. J. (1988). Nonparametric Statistics for the Behavioral Sciences. McGraw-Hill.
• Ott, R. L., & Longnecker, M. (2010). An Introduction to Statistical Methods and Data Analysis. Brooks/Cole.