Question

1 Let X denote the time in hours (rounded to nearest hour) for the quality control lab to provide results from a process sample. The probability mass function is given below. Determine the following values: 0 4 EK f(x) 0.25 P(X< 2 hours) P 1.75< X S4) PD( 군 1) F(X), the cumulative distribution E(X) and V(x) 0.15 0.45 0.05 0.1

Answer 1

Solution:

a.

$P(X<2)=P(X=0)+P(X=1)$

= 0.05 + 0.25

$=0.3$

b. $P(1.75<X\leq 4)=P(X=2)+P(X=3)+P(X=4)$

0.15 + 0.450.1

=0.7

c. $P(X\neq 1)=1-P(X=1)$

  

= 1-0.25

  

0.75

d. F(X), the cumulative distribution

x	f(x)	F(x) Cumulative distribution
0	0.05	0.05
1	0.25	0.3
2	0.15	0.45
3	0.45	0.9
4	0.1	1

e. E(X) and V(X)

E(X)-〉 ,.rf(z)

(0 × 0.05) (1 × 0.25) (2 × 0.15) (3 × 0.45) (4 × 0.1)

$=0+0.25+0.3+1.35+0.4$

$=2.3$

$V(X)=\sum (x-E(X))^{2}f(x)$

(x - Mean)'f(x) 0.2645 0.4225 0.0135 0.2205 0.289 Σ( y f(x) = 1.21 i(x) 0.05 0.25 0.15 0.45 0.1 2 4 x - Mean (x)-1.21