Can someone respond to my last question?
2 answers:
You might be interested in
P = 10% = 0.1
q = 1 - 0.1 = 0.9
P(at least one defective calculator) = P(1) + P(2) + P(3) + P(4) = 1 - P(0)
The brobability of a binomial distribution is given by
where: n = 4
Therefore,
P(at least one defective calculator) = 1 - 0.6561 = 0.3439
q = 1 - 0.1 = 0.9
P(at least one defective calculator) = P(1) + P(2) + P(3) + P(4) = 1 - P(0)
The brobability of a binomial distribution is given by
where: n = 4
Therefore,
P(at least one defective calculator) = 1 - 0.6561 = 0.3439
hello there
the answer is down below in a graph
hope it helps
Answer:
At 5% significance level, larger proportion of military personnel students protect their profiles on social-networking sites than young adults with profiles on social-networking sites.
Step-by-step explanation:
let p be the proportion of military personnel students who restrict access to their profiles. Then null and alternative hypotheses are:
: p=0.67 (67%)
: p>0.67
We need to calculate z-statistic of sample proportion:
z= where
- p(s) is the sample proportion of military students who restrict access to their profiles (
=0.78)
- p is the proportion assumed under null hypothesis. (0.67)
- N is the sample size (100)
Then z= ≈ 2.34
The corresponding p-value is 0.0096. Since 0.0096<0.05 (significance level) we can reject the null hypothesis and conclude at 5% significance level that larger proportion of military personnel students protect their profiles on social-networking sites than young adults with profiles on social-networking sites.