http://www.purplemath.com/modules/specfact.htm
Considering that the subjects are chosen without replacement, they are not independent, and the probability cannot be found using the binomial distribution.
The binomial distribution and the hypergeometric distribution are quite similar, as:
- They find the probability of exactly x successes on n repeated trials.
- For each trial, there are only two possible outcomes.
- The difference is that the binomial distribution is for independent trials, that is, in each trial, the probability of success is the same, while the hypergeometric distribution is for dependent trials.
- If the sample is without replacement, the trials are not independent, thus the hypergeometric distribution is used, not the binomial.
A similar problem is given at brainly.com/question/21772486
Find the midpoint:
m= x1+x2/2; y1+y2/2
m= 9+-1/2; 8+-2/2
m= 8/2; 6/2
m= (4,3)
(4,3) is your answer.
I hope this helps!
~kaikers
Answer:
X = 2
Step-by-step explanation:
6(2) = 12
6 X 2 = 12
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.
