Probabilistic System Analysis Homework from my College

Problem 1: Satisfaction Survey Analysis
Part (a): 95% Confidence Interval for Actual Proportion
Given:

Sample size: n = 119
Number satisfied: x = 99
Sample proportion: p^ = 99/119 = 0.8319

Solution:
For a 95% confidence interval, we use z(0.025) = 1.96
The confidence interval formula is:
p^ ± z × v[p^(1-p^)/n]
Calculating the standard error:
SE = v[0.8319 × 0.1681/119]
SE = v[0.001176]
SE = 0.0343
Margin of error:
ME = 1.96 × 0.0343 = 0.0672
95% Confidence Interval:
CI = 0.8319 ± 0.0672 = (0.7647, 0.8991)
Answer: The 95% confidence interval is (76.47%, 89.91%) or approximately (0.765, 0.899)
Part (b): Probability of Exactly 99 Satisfied (Given p = 0.89)
Given:

Population proportion (claimed): p = 0.89
n = 119, x = 99

Solution:
Since n is large and both np = 119(0.89) = 105.91 and n(1-p) = 119(0.11) = 13.09 are greater than 5, we can use the normal approximation to the binomial distribution.
Mean and Standard Deviation:
Mean = np = 119 × 0.89 = 105.91
Standard deviation = v[np(1-p)] = v[119 × 0.89 × 0.11] = v11.65 = 3.413
Using continuity correction for P(X = 99):
P(X = 99) ˜ P(98.5 < X < 99.5)
Standardizing:
Z1 = (98.5 - 105.91)/3.413 = -7.41/3.413 = -2.17
Z2 = (99.5 - 105.91)/3.413 = -6.41/3.413 = -1.88
Finding the probability:
P(X = 99) = F(-1.88) - F(-2.17)
= 0.0301 - 0.0150 = 0.0151
Answer: The probability is approximately 0.0151 or 1.51%
This low probability suggests that if the organizer's claim is correct, observing only 99 satisfied participants is quite unlikely, which supports the group's skepticism.

Problem 2: Sample Size for Election Prediction
Given:

Margin of error: E = 2.5% = 0.025
Confidence level: 95% (z = 1.96)

Solution:
The formula for minimum sample size when estimating a proportion is:
n = (z/E)² × p(1-p)
Since we don't know p (the true proportion), we use p = 0.5 to get the maximum (most conservative) sample size:
n = (1.96/0.025)² × 0.5 × 0.5
n = (78.4)² × 0.25
n = 6146.56 × 0.25
n = 1536.64
Rounding up: n = 1537 respondents
Answer: At least 1537 respondents must be surveyed to predict the election result within a 2.5% margin of error with 95% confidence.