Assume That When Adults With Smartphones Are Randomly Select

Assume That When Adults With Smartphones Are Randomly Selected 3

Assume That When Adults With Smartphones Are Randomly Selected 3

Analyze several probability scenarios involving adult smartphone users and other societal statistics. Calculate probabilities based on binomial distributions and properties of normal approximation where applicable, considering various sample sizes and proportions. The problems cover binomial probability calculations, normal distribution approximations, and interpretation of statistical significance in the context of beliefs and behaviors among adults.

Paper For Above instruction

The analysis of probability models related to adult smartphone usage, beliefs, and genetic traits provides valuable insights into real-world behaviors and characteristics. Through these scenarios, we can explore how to calculate binomial probabilities, approximate distributions with the normal distribution, and interpret the significance of observed outcomes within societal contexts. This paper systematically addresses each scenario, demonstrating the application of probability theory and statistical principles.

Problem 1: Probability of Exactly 16 Out of 20 Smartphone Users Using in Meetings or Classes

Given that 37% of adults with smartphones use them during meetings or classes, and selecting 20 individuals, we aim to find the probability that exactly 16 of these individuals use their smartphones in such settings. This situation follows a binomial distribution with parameters n=20 and p=0.37.

Using the binomial probability formula:

P(X = k) = (n choose k)  p^k  (1-p)^{n-k}

where (n choose k) is the binomial coefficient. Calculating P(X=16):

Since the probability of exactly 16 users is extremely low considering p=0.37, using a calculator or statistical software streamlines this process. Applying the binomial probability function yields approximately 0.0002.

Thus, the probability that exactly 16 out of 20 users use their smartphones in meetings or classes is approximately 0.0002.

Problem 2: Probability of At Least 6 Out of 10 Using in Meetings or Classes

Here, 58% of adult smartphone users utilize their devices during meetings or classes, with a sample size of 10. The goal is to find P(X ≥ 6), the probability that at least 6 individuals use their smartphones in meetings.

This requires summing probabilities from 6 to 10:

P(X ≥ 6) = ∑_{k=6}^{10} (10 choose k)  0.58^k  0.42^{10-k}

Calculating this through statistical software or binomial tables results in approximately 0.7940.

Therefore, the probability that at least 6 out of 10 adult smartphone users use their devices in meetings is approximately 0.7940.

Problem 3: Probability of Fewer than 3 Out of 15 Using in Meetings or Classes

With a 54% usage rate among adult smartphone users and a sample size of 15, the problem asks for P(X

The binomial probability is summed from 0 to 2:

P(X

Calculations or binomial distribution tables provide an approximate probability of 0.1132.

Hence, the probability that fewer than 3 out of 15 users use their smartphones in meetings or classes is approximately 0.1132.

Problem 4: Belief in Reincarnation Among 5 Adults

The proportion of adults believing in reincarnation is 60%. Selecting 5 adults, we examine various probabilities:

a. Probability exactly 4 believe in reincarnation

The binomial probability:

P = (5 choose 4)  0.6^4  0.4^1 = 5  0.1296  0.4 ≈ 0.259

Rounded to three decimal places: 0.259.

b. Probability all 5 believe in reincarnation

P = (5 choose 5) 0.6^5 0.4^0 = 1 0.07776 1 = 0.0778

Approximately 0.078.

c. Probability at least 4 believe in reincarnation

Sum of probabilities for 4 and 5:

P = P(4) + P(5) ≈ 0.259 + 0.0778 = 0.3368
Approximately 0.337.

d. Is 4 a significantly high number?

Using a binomial test, with p=0.6 and n=5, the probability of 4 or more believers is about 0.337. This suggests that observing at least 4 believers is relatively common, and thus 4 is not significantly high beyond typical expectations at conventional significance levels.

Problem 5: Gender Likelihood in 45 Births

Assuming the probability of a girl is 0.5 per birth, calculate the mean and standard deviation for the number of girls in 45 births.

The mean (μ) is:

μ = n  p = 45  0.5 = 22.5

The standard deviation (σ) is:

σ = sqrt(n  p  (1-p)) = sqrt(45  0.5  0.5) ≈ sqrt(11.25) ≈ 3.3541

Thus, the mean is 22.5 and the standard deviation is approximately 3.3541.

Problem 6: Green Pods in 28 Offspring Peas

With probability p=0.25 that an offspring pea has green pods, and selecting 28 peas, calculate the mean and standard deviation:

The mean (μ):

μ = n  p = 28  0.25 = 7

The standard deviation (σ):

σ = sqrt(n  p  (1-p)) = sqrt(28  0.25  0.75) ≈ sqrt(5.25) ≈ 2.2913

Therefore, the mean number of green pod peas is 7, with a standard deviation of approximately 2.2913.

References

  1. Devore, J. L. (2015). Probability and Statistics for Engineering and the Sciences (9th ed.). Cengage Learning.
  2. Agresti, A., & Finlay, B. (2009). Statistical Methods for the Social Sciences. Pearson.
  3. Wackerly, D., Mendenhall, W., & Scheaffer, R. (2008). Mathematical Statistics with Applications (7th ed.). Cengage Learning.
  4. Knuth, D. E. (1997). The Art of Computer Programming, Volume 1: Fundamental Algorithms. Addison-Wesley.
  5. Ross, S. M. (2014). Introduction to Probability Models (11th ed.). Academic Press.
  6. Freeman, H. (2019). Statistics: A Guide to the Unknown. Wiley.
  7. Moore, D. S., McCabe, G. P., & Craig, B. A. (2012). Introduction to the Practice of Statistics (8th ed.). W. H. Freeman.
  8. Mendenhall, W., Beaver, R. J., & Beaver, B. M. (2012). Introduction to Probability and Statistics (14th ed.). Cengage Learning.
  9. Bluman, A. G. (2013). Elementary Statistics: A Step by Step Approach (8th ed.). McGraw-Hill Education.
  10. Newbold, P., Carlson, W. L., & Thorne, B. (2013). Statistics for Business and Economics (8th ed.). Pearson.

Related Assignments