1. CIVL 181 Tutorial 3 PMF to CDF PDF to CDF Mean and s.d., c.o.v. Formulae for reference

2. From PMF to CDF 3.1Two constructions A, B have the PMF below: Assume activity A and B are s.i., draw PMF and CDF of the total no. of days required. 345 t activity A P (T A = t) 0.3 0.5 0.2 4 5 6 t activity B P (T B = t) 0.2 0.6

3. From PMF to CDF 1. Observe the possible values (7 ≤ X ≤ 11) 2. Compute the associated probabilities of each value by theorem of total probability

4. From PMF to CDF P (X 7 ) = P(A 3 ) P(B 4 ) = 0.3 x 0.2 = 0.06 P (X 8 ) = P(A 3 )P(B 5 ) + P(A 4 )P(B 4 ) P (X 9 ) = P(A 3 )P(B 6 ) + P(A 4 )P(B 5 ) + P(A 5 )P(B 4 ) …… P(X 7 ) = 0.06; P(X 8 ) = 0.28; P(X 9 ) = 0.4; P(X 10 ) = 0.22; P(X 11 ) = 0.04; P(X 7 ) + P(X 8 ) + P(X 9 ) + P(X 10 ) + P(X 11 ) = ?

5. PMF CDF as a practice 7 t P (X) 891011 0.06 0.28 0.4 0.04 0.22

6. From PDF to CDF Suppose a PDF follows a square relationship with its variable x, x ranges from 3 to 6. Draw its PDF and CDF.

7. From PDF to CDF f(x) = x 2, 3 ≤ x ≤ 6? The first thing to do is CHECK! Should setup f(x) as f(x) = kx 2 and find k.

8. k?k? Area property: the area of PDF is 1 x 36 f(x) 9/63 36/63

9. CDF Wrong! Because F(1) = 1 + 9/63 There are integration constants Use boundary condition: C = - 9 / 63

10. Remember! CDF always starts from 0 and end at 1 Integration const may exist, especially when x are not start from 0. Use boundary condition to eliminate x 36 F(x) 1

11. Mean and s.d., c.o.v. Q1: If daily temperature is 20 o c, is that suitable for human to leave? The place may be a desert with 40 o c in daytime and 0 0 c at night Q2: For two r.v. X and Y, if X has 1000 times larger s.d. than Y, is that X more dispersed? Think of expressing concerte strength in terms of MPa and kPa (and compare the numerical value)

12. Formulae for reference Discrete: Continuous: s.d. = Median: Skewness, Kurtosis very seldom used.