Respuesta :
Answer:
Probability of winning first situation: 0.0793
Probability of winning second situation: 0.0146
You should go with the first option
Step-by-step explanation:
See attached photo for answers. For this situation you need to identify p-hat, p, q and n.
P-hat, p, and q all stay the same for both situations, only n changes.
p-hat is the proportion of green balls we want, which is 0.7
p = 98/159 = 0.62
q = 0.38 because q = 1 - p, which in this case is q = 1 - 0.62 = 0.38
n = 73 for the first situation, and n = 175 for the second situation
Using the normal probability distribution and the central limit theorem, it is found that:
a)
With 73 draws, there is a 0.0808 = 8.08% probability of getting more than 70% green balls.
With 175 draws, 0.015 = 1.5% probability of getting more than 70% green balls.
b)
Due to the higher probability of getting more than 70% green balls, 73 draws should be chosen.
Normal Probability Distribution
In a normal distribution with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the z-score of a measure X is given by:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
- It measures how many standard deviations the measure is from the mean.
- After finding the z-score, we look at the z-score table and find the p-value associated with this z-score, which is the percentile of X.
Central Limit Theorem
- It states that the sampling distribution of the sample means with size n can be approximated to a normal distribution.
- For a proportion p in a sample of size n, the sampling distribution of the sample proportion will be approximately normal with mean [tex]\mu = p[/tex] and standard deviation [tex]s = \sqrt{\frac{p(1-p)}{n}}[/tex]
Item a:
- 98 green balls out of 158, thus [tex]p = \frac{98}{158} = 0.6203[/tex]
Out of 73 draws:
[tex]s = \sqrt{\frac{p(1-p)}{n}} = \sqrt{\frac{0.6203(0.3797)}{73}} = 0.0568[/tex]
The probability of more than 70% green balls is 1 subtracted by the p-value of Z when X = 0.7, thus:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
By the Central Limit Theorem
[tex]Z = \frac{X - \mu}{s}[/tex]
[tex]Z = \frac{0.7 - 0.6203}{0.0568}[/tex]
[tex]Z = 1.4[/tex]
[tex]Z = 1.4[/tex] has a p-value of 0.9192.
1 - 0.9192 = 0.0808
With 73 draws, there is a 0.0808 = 8.08% probability of getting more than 70% green balls.
Out of 175 draws:
[tex]s = \sqrt{\frac{p(1-p)}{n}} = \sqrt{\frac{0.6203(0.3797)}{175}} = 0.0367[/tex]
Then:
[tex]Z = \frac{X - \mu}{s}[/tex]
[tex]Z = \frac{0.7 - 0.6203}{0.0367}[/tex]
[tex]Z = 2.17[/tex]
[tex]Z = 2.17[/tex] has a p-value of 0.985.
1 - 0.985 = 0.015.
With 175 draws, 0.015 = 1.5% probability of getting more than 70% green balls.
Item b:
Due to the higher probability of getting more than 70% green balls, 73 draws should be chosen.
A similar problem is given at https://brainly.com/question/24663213
