Missed Questions Week 2 Quiz 8. Personality Test Score 0-2 ✓ Solved

Missed Questions Week 2 Quiz #8. Personality test scored 0-2

Missed Questions Week 2 Quiz #8. Personality test scored 0-200 has mean 95 and standard deviation 24. a. Using Chebyshev's theorem, what minimum percentage of scores lie between 59 and 131? b. Using Chebyshev's theorem, what minimum percentage of scores lie between 47 and 143? c. If the distribution is bell-shaped, using the empirical rule, what percentage of scores lie between 47 and 143? d. If bell-shaped, according to the empirical rule, approximately 68% of scores lie between __ and __.

Week 2 Homework #1. Personality test scored 0-200 has mean 106 and standard deviation 19. e. Using Chebyshev's theorem, what minimum percentage of scores lie between 68 and 144? f. Using Chebyshev's theorem, what minimum percentage of scores lie between 77.5 and 134.5? g. If the distribution is bell-shaped, using the empirical rule, what percentage of scores lie between 68 and 144? h. If bell-shaped, according to the empirical rule, approximately 99.7% of scores lie between __ and __.

Problem 2. A cab company has mean fare $28.57 and standard deviation $3.21 from O'Hare to the Drake Hotel. a. Using Chebyshev's theorem, at least 8/99 (about 89%) of fares lie between what dollars and what dollars? (Round to 2 decimals.) b. Using Chebyshev's theorem, what minimum percentage of fares lie between $22.15 and $34.99? c. If bell-shaped, using the empirical rule, what percentage of fares lie between $22.15 and $34.99? d. If bell-shaped, approximately 68% of fares lie between __ and __.

Problem 8. A fair coin is tossed 3 times. Let R be the number of tails. Define X = R^2 - R - 1. List all eight outcomes and compute the value of X for each.

Problem 17. A surgery is successful 85% of the time. For a random sample of 10 surgeries, what is the probability that more than 8 are successful? Carry intermediate computations to at least four decimal places and round the final answer to two decimal places.

Week 1 Homework #14. Red blood cell counts (in 10^6 cells/µL) measured on 6 days: 52, 48, 49, 50, 55, 52. Find the sample standard deviation. Round to two decimal places.

Problem 16. An investment company reports last year its clients on average made a profit of 10% (mean). Which claim must be true? Choose from: a. Two years ago, some clients made at least 10%. b. Last year all clients made less than 17%. c. At least one client made exactly 10%. d. Fewer than half made 10% or less. e. At least one client made at least 10%. f. None of the above.

Paper For Above Instructions

Introduction

This paper answers the statistical questions above using Chebyshev's theorem, the empirical (68-95-99.7) rule, binomial probability, and sample standard deviation computations. Each subproblem is solved step-by-step with concise justification and numerical results. The methods used are standard in elementary probability and statistics (Wackerly et al., 2008; Ross, 2014).

1. Quiz #8: Personality test (mean=95, sd=24)

a. Interval 59 to 131. Distance from mean = 36, so k = 36/24 = 1.5. Chebyshev's theorem guarantees at least 1 - 1/k^2 = 1 - 1/2.25 = 0.555555... = 55.56% of scores lie in [59,131] (Chebyshev bound) (Wackerly et al., 2008).

b. Interval 47 to 143. Distance = 48, k = 48/24 = 2. Chebyshev gives at least 1 - 1/4 = 0.75 = 75% of scores lie in [47,143].

c. If the distribution is bell-shaped (approximately normal), 47 and 143 are 2 standard deviations from the mean (mean ± 2σ). By the empirical rule, about 95% of scores lie between 47 and 143 (Moore et al., 2017).

d. The empirical rule states ~68% of data lie within ±1σ. With mean 95 and σ = 24, 68% lie between 95 - 24 = 71 and 95 + 24 = 119.

2. Week 2 Homework #1: Personality test (mean=106, sd=19)

e. Interval 68 to 144: distance = 38, k = 38/19 = 2. Chebyshev: at least 1 - 1/4 = 75% of scores in [68,144].

f. Interval 77.5 to 134.5: distance = 28.5, k = 28.5/19 = 1.5. Chebyshev: at least 1 - 1/(1.5^2) = 55.56% lie in [77.5,134.5].

g. If bell-shaped, 68 to 144 is ±2σ from the mean (2*19=38), so empirical rule gives approx 95% of scores in that interval.

h. For 99.7% (empirical rule), we use ±3σ: 106 ± 3(19) = 106 ± 57, so approximately between 49 and 163 cover 99.7% of scores in a bell-shaped distribution.

3. Problem 2: Cab fares (mean=$28.57, sd=$3.21)

a. The prompt intends the Chebyshev bound with k = 3 (1 - 1/9 = 8/9 ≈ 88.89%). Using k = 3, the interval is mean ± 3σ = 28.57 ± 3(3.21) = 28.57 ± 9.63, giving [18.94, 38.20]. Rounded to two decimals: $18.94 and $38.20 (Wackerly et al., 2008).

b. Interval $22.15 to $34.99: distance = 28.57 - 22.15 = 6.42, k = 6.42/3.21 = 2. Chebyshev guarantees at least 75% of fares lie in [$22.15,$34.99].

c. If fares are approximately normal, the interval corresponds to ±2σ so empirical rule gives ≈95% of fares in [$22.15,$34.99] (Moore et al., 2017).

d. Empirical rule for 68%: mean ± 1σ = 28.57 ± 3.21 yields [$25.36,$31.78].

4. Problem 8: Coin tossed three times, X = R^2 - R - 1

List of outcomes (R = number of tails) and X values:

  • HHH: R=0 → X = 0^2 - 0 - 1 = -1
  • HHT: R=1 → X = 1 - 1 - 1 = -1
  • HTH: R=1 → X = -1
  • THH: R=1 → X = -1
  • HTT: R=2 → X = 4 - 2 - 1 = 1
  • THT: R=2 → X = 1
  • TTH: R=2 → X = 1
  • TTT: R=3 → X = 9 - 3 - 1 = 5

This exhausts all eight equiprobable outcomes; X takes values -1, 1, and 5 according to R.

5. Problem 17: Binomial probability (n=10, p=0.85)

We want P(X > 8) = P(X ≥ 9) = P(9) + P(10). Using the binomial formula P(k) = C(10,k) p^k (1-p)^(10-k):

P(9) = C(10,9) (0.85)^9 (0.15)^1 = 10 0.85^9 0.15.

Compute 0.85^9 = 0.2316 (approx to 4 decimals 0.2316) and 0.85^10 = 0.1969 (0.1969). Thus

P(9) ≈ 10 0.2316169 0.15 = 0.3474 (intermediate 0.3474).

P(10) = 0.85^10 ≈ 0.1969.

Total P(X ≥ 9) ≈ 0.3474 + 0.1969 = 0.5443. Rounded to two decimals: 0.54 (Ross, 2014).

6. Week 1 Homework #14: Sample standard deviation

Data: 52, 48, 49, 50, 55, 52. Mean = (52+48+49+50+55+52)/6 = 306/6 = 51.

Deviations: 1, -3, -2, -1, 4, 1. Sum of squared deviations = 1+9+4+1+16+1 = 32.

Sample variance = 32/(n-1) = 32/5 = 6.4. Sample standard deviation = sqrt(6.4) ≈ 2.5298 → rounded to two decimals: 2.53 (OpenStax, 2016).

7. Problem 16: Mean profit claim logic

When a finite collection of client profits has arithmetic mean 10%, there must be at least one client whose profit is greater than or equal to the mean; otherwise, if all values were strictly less than 10% the mean would be

Conclusion

All subproblems have been solved using standard theorems: Chebyshev for distribution-free guaranteed bounds, the empirical rule for approximately normal data, binomial probability for independent Bernoulli trials, and the sample standard deviation formula for sample variability. Results were rounded as requested, and intermediate values were provided where indicated.

References

  • DeGroot, M. H., & Schervish, M. J. (2012). Probability and Statistics (4th ed.). Pearson.
  • Moore, D. S., McCabe, G. P., & Craig, B. A. (2017). Introduction to the Practice of Statistics (9th ed.). W. H. Freeman.
  • Wackerly, D., Mendenhall, W., & Scheaffer, R. (2008). Mathematical Statistics with Applications (7th ed.). Cengage Learning.
  • Ross, S. M. (2014). A First Course in Probability (9th ed.). Pearson.
  • OpenStax. (2016). Introductory Statistics. OpenStax CNX. https://openstax.org
  • NIST/SEMATECH. (2012). e-Handbook of Statistical Methods. https://www.itl.nist.gov/div898/handbook/
  • StatTrek. (n.d.). Chebyshev Theorem. https://stattrek.com/
  • Khan Academy. (n.d.). Empirical Rule and Standard Deviation. https://www.khanacademy.org/
  • Agresti, A., & Franklin, C. (2017). Statistics: The Art and Science of Learning from Data (4th ed.). Pearson.
  • Rice, J. A. (2006). Mathematical Statistics and Data Analysis (3rd ed.). Cengage Learning.

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