DS301 Fall 2018 Homework 7 Page 1

Ds301 Fall2018 Hw7 Page 1ds 30101 Fall 2018 Homework 7 Q1 The W

Identify and analyze the probability and expected value related to traffic signal waiting times and student phone call distributions using concepts of probability, expected value, and the standard normal distribution.

Paper For Above instruction

This paper addresses a set of probability problems related to wait times at a traffic signal and the distribution of student phone calls. These problems are based on the application of continuous probability distributions, specifically the uniform distribution, and the standard normal distribution (Z-scores). Excelling in understanding and calculating these probabilities and expected values is essential for analyzing real-world scenarios such as traffic delays and behavioral patterns among students.

Part 1: Expected Waiting Time and Probabilities at the Traffic Signal

The first problem involves analyzing the waiting time at a traffic signal where the wait time is uniformly distributed between zero and five minutes. The key is understanding the properties of the continuous uniform distribution, which is characterized by a constant probability density over its interval. The uniform distribution over the interval [a, b], in this case [0, 5], has an expected value (mean) calculated as (a + b)/2, and its probabilities can be derived directly from the uniform density function.

The expected waiting time for the traffic signal can be computed as:

Expected value (E[X]) = (a + b)/2 = (0 + 5)/2 = 2.5 minutes

This means that, on average, a vehicle would wait 2.5 minutes at the signal.

The probability that the wait time is less than 1.5 minutes is the probability that a uniformly distributed random variable takes values in [0, 1.5]. Since the uniform distribution is flat, the probability is proportional to the interval length:

P(wait

Thus, there is a 30% chance that a vehicle will pass in less than 1.5 minutes.

Next, the probability that waiting exceeds 2 minutes is the complement of waiting 2 minutes or less:

P(wait > 2) = 1 - P(wait ≤ 2) = 1 - (2 / 5) = 1 - 0.4 = 0.6

Therefore, there is a 60% probability that a vehicle gets stuck for more than two minutes at the signal.

Part 2: Z-Values for the Standard Normal Distribution

The second problem deals with finding z-values for specified probabilities in the standard normal distribution. Given the symmetry and the properties of the normal distribution, z-scores corresponding to certain cumulative probabilities can be found using Z-tables or statistical software.

For the given probabilities:

  • Pr(Z ≤ z) = p
  • Pr(Z > z) = 1 - p
  • Pr(−z₁ ≤ Z ≤ z₂) = p
  • Pr(−0 ≤ Z ≤ z) = 0.3315

Using standard Z-tables or cumulative distribution functions:

  • Pr(Z ≤ z) = 0.5 implies z = 0, since the standard normal is symmetric about zero.
  • Pr(Z > z) = 0.1 corresponds to z ≈ 1.28, since 10% of the distribution lies above z=1.28.
  • Pr(−z₁ ≤ Z ≤ z₂) = 0.25 could correspond to z ≈ ±0.67, assuming equal tails.
  • Pr(−0 ≤ Z ≤ z) = 0.3315 implies z ≈ 0.43, based on the cumulative probability from zero upwards.

Part 3: Student Phone Calls Distribution Analysis

The third problem examines the number of phone calls made by students, which follows a normal distribution with a mean of 11 calls per day and a standard deviation of 4. Using the properties of the normal distribution, probabilities for specific call counts can be calculated via Z-scores.

1. Probability that a student takes between 3 and 17 calls per day:

The Z-scores for 3 and 17 are:

Z for 3: Z = (3 - 11) / 4 = -8 / 4 = -2

Z for 17: Z = (17 - 11) / 4 = 6 / 4 = 1.5

The probability between these Z-values corresponds to:

P(3

Using standard Z-tables:

Pr(Z

Pr(Z

Therefore, P(3

This indicates there is approximately a 91.04% chance that a student takes between 3 and 17 calls per day.

2. Probability that a student takes more than 7 calls per day:

Z for 7: Z = (7 - 11) / 4 = -1

From Z-tables, Pr(Z > -1) ≈ 0.8413, so the probability of more than 7 calls is approximately 84.13%.

3. Probability that a student takes more than 19 calls per day:

Z for 19: Z = (19 - 11) / 4 = 8 / 4 = 2

Pr(Z > 2) ≈ 0.0228, indicating about a 2.28% chance of taking more than 19 calls per day.

Conclusion

Applying uniform and normal probability distributions allows for comprehensive insights into real-world events such as traffic delays and student behaviors. Accurate calculations of expected values and tail probabilities are crucial for planning and decision-making. As demonstrated, these distributions provide valuable tools to forecast and understand variability in everyday phenomena, supporting both academic research and practical problem-solving.

References

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  • Evans, M., Hastings, N., & Peacock, B. (2010). Statistical Distributions. Wiley.
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  • Miller, R. W., & Freund, J. E. (2009). Probability and Statistics for Engineering and the Sciences. Pearson.
  • Ross, S. M. (2014). Introduction to Probability Models. Academic Press.
  • Walpole, R. E., Myers, R. H., Myers, S. L., & Ye, K. (2012). Probability and Statistics for Engineers and Scientists. Pearson.
  • Siegel, S., & Castellan, N. J. (1988). Nonparametric Statistics for the Behavioral Sciences. McGraw-Hill.
  • Knüppel, P. (2011). Applied Statistics in Business and Economics. Springer.
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  • Moore, D. S., Notz, W. I., & Fligner, M. A. (2013). The Basic Practice of Statistics. W. H. Freeman.

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