No reason, or no evidence to support your claim/proof
Answer:
The sample size needed is 2401.
Step-by-step explanation:
In a sample with a number n of people surveyed with a probability of a success of
, and a confidence level of
, we have the following confidence interval of proportions.
![\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}](https://tex.z-dn.net/?f=%5Cpi%20%5Cpm%20z%5Csqrt%7B%5Cfrac%7B%5Cpi%281-%5Cpi%29%7D%7Bn%7D%7D)
In which
z is the zscore that has a pvalue of
.
The margin of error of the interval is:
![M = z\sqrt{\frac{\pi(1-\pi)}{n}}](https://tex.z-dn.net/?f=M%20%3D%20z%5Csqrt%7B%5Cfrac%7B%5Cpi%281-%5Cpi%29%7D%7Bn%7D%7D)
95% confidence level
So
, z is the value of Z that has a pvalue of
, so
.
Estimate the sample size needed if no estimate of p is available so that the confidence interval will have a margin of error of 0.02.
We have to find n, for which M = 0.02.
There is no estimate of p available, so we use ![\pi = 0.5](https://tex.z-dn.net/?f=%5Cpi%20%3D%200.5)
![M = z\sqrt{\frac{\pi(1-\pi)}{n}}](https://tex.z-dn.net/?f=M%20%3D%20z%5Csqrt%7B%5Cfrac%7B%5Cpi%281-%5Cpi%29%7D%7Bn%7D%7D)
![0.02 = 1.96\sqrt{\frac{0.5*0.5}{n}}](https://tex.z-dn.net/?f=0.02%20%3D%201.96%5Csqrt%7B%5Cfrac%7B0.5%2A0.5%7D%7Bn%7D%7D)
![0.02\sqrt{n} = 1.96*0.5](https://tex.z-dn.net/?f=0.02%5Csqrt%7Bn%7D%20%3D%201.96%2A0.5)
![\sqrt{n} = \frac{1.96*0.5}{0.02}](https://tex.z-dn.net/?f=%5Csqrt%7Bn%7D%20%3D%20%5Cfrac%7B1.96%2A0.5%7D%7B0.02%7D)
![(\sqrt{n})^{2} = (\frac{1.96*0.5}{0.02})^{2}](https://tex.z-dn.net/?f=%28%5Csqrt%7Bn%7D%29%5E%7B2%7D%20%3D%20%28%5Cfrac%7B1.96%2A0.5%7D%7B0.02%7D%29%5E%7B2%7D)
![n = 2401](https://tex.z-dn.net/?f=n%20%3D%202401)
The sample size needed is 2401.
Answer:
The probability that the new baby just born is a boy is 0.6
Step-by-step explanation:
- B=Event where the nurse picked up a boy
- G=Event where the nurse picked up a girl
Applying Bayes's Theorem:
P(A|B)=P(A) P(B|A) / [ P(A) P(B|A)+ P(C) P(B|C) ]
Where:
- P(A|B)=probability that the new baby just born is a boy given that the nurse picked up a boy.
- P(A)=Probability that the woman had a boy= 1/2
- P(C)=Probability that the woman had a girl=1/2
The probability for P(A) and P(C) is 1/2 becuase there are only two options. Girl or Boy
n= number of girls born
3+n= Total number of children in the nursery
Then,
P(A|B)=![\frac{\frac{1}{2} \frac{3}{3+n} }{\frac{1}{2} \frac{3}{3+n} + \frac{1}{2} \frac{2}{3+n}}](https://tex.z-dn.net/?f=%5Cfrac%7B%5Cfrac%7B1%7D%7B2%7D%20%20%5Cfrac%7B3%7D%7B3%2Bn%7D%20%7D%7B%5Cfrac%7B1%7D%7B2%7D%20%20%5Cfrac%7B3%7D%7B3%2Bn%7D%20%2B%20%5Cfrac%7B1%7D%7B2%7D%20%20%5Cfrac%7B2%7D%7B3%2Bn%7D%7D)
P(A|B)=3/5
P(A|B)=0.6
Answer:
2.34 ounces per bag
Step-by-step explanation:
Marcos is making bags of trail mix for hocking club he will use this amount makes 25 bags of trail mix.
The total ounces of nuts:
22 ounces of walnuts + 10.1 ounces of almonds + 26.4 ounces of cashews
= 58.5 ounces of nuts
25 bags = 58.5 ounces
1 bag = x ounces
x = 58.5 ounces/25 bags
x = 2.34 ounces per bag