BAE 242 W16 Assignment 2

Question 1

1. 0.178 or 17.8%2. 0.5 + 0178 – (0.5*0.178) = 0.589 or 58.9%3. 1.28 * 2 + 7 = 9.56ms

Question 2

6,000 8,000 10,000 12,000 14,000 16,000 18,000 20,000

-2.5

-2

-1.5

-1

-0.5

0

0.5

1

1.5

2

2.5

Quantiles

Quantiles

The scatterplots form a relatively straight line. Therefore I can conclude with reasonable confidence that the attendance is an approximately normal distributed variable.

Question 3

20/sqrt(25) = 4 ohms. Z value for 95 = -2.5Φ (95-100/4) → NORMSDIST(-1.25) = 0.1056 or 10.56%

Question 4

STDEV(of data) = 9.315ẋ (mean) = 40.41n (sample size) = 32t* for 95% confidence = 2.2622.262 * (9.315/sqrt(32)) = ±3.7248

Therefore the 95% confidence interval for the mean number of touchdowns is:40.41 ±3.7248 or 44.1348 & 36.6852