D=event that chip selected is defective
d=event that chip selected is NOT defective
Four possible scenarios for the first two selections:
P(DDD)=15/100*14/99*13/98=13/4620
P(DdD)=15/100*85/99*14/98=17/924
P(dDD)=85/100*15/99*14/99=17/924
P(ddD)=85/100*84/99*15/98=17/154
Probability of third selection being defective is the sum of all cases,
P(XXD)=P(DDD)+P(DdD)+P(dDD)+P(ddD)
=3/20
Answer:
Yes, result is significant ; PVALUE < α
Step-by-step explanation:
Given :
x = 536
n = sample size = 1012
Phat = x / n = 536 / 1012 = 0.5296 = 0.53
H0 : P0 = 0.5
H1 : P0 > 0.5
Test statistic :
(Phat - P0) ÷ sqrt[(P0(1 - P0)) / n]
1-P0 = 1 - 0.5 = 0.5
(0.53 - 0.5) ÷ sqrt[(0.5*0.5)/1012]
0.03 ÷ 0.0157173
= 1.9087
Pvalue :
Using the Pvalue from test statistic :
Pvalue = 0.02815
To test if result is significant :
α = 0.05
0.02815 < 0.05
Pvalue < α ; Hence, result is significant at α=0.05; Hence, we reject H0.
Answer:
it is 35
Step-by-step explanation:
Answer:
y
Step-by-step explanation:
Answer:
9 over 5
Step-by-step explanation:
