Answer:
a. answer to a can be found in the attached file
b. Pr[survival] = Pr[good&survive]+Pr[medium&survive]+Pr[low&survive]=
0.24+0.06+0.05 = 0.35
c. Assume that the seed has a 0.2 chance of dying before it lands in a habitat. What is its overall probability of survival?
Pr[survival] = Pr[survival|lands] * Pr[lands] = 0.35 * 0.2 = 0.07
Step-by-step explanation:
"A seed randomly blows around a complex habitat. It may land on any of three different soil types: a high-quality soil that gives a 0.8 chance of seed survival, a medium-quality soil that gives a 0.3 chance of survival, and a low-quality soil that gives only a 0.1 chance of survival. These three soil types (high, medium, and low) are present in the habitat in proportions of 30:20:50, respectively. The probability that a seed lands on a particular soil type is proportional to the frequency of that type in the habitat. a. Draw a probability tree to determine the probabilities of survival under all possible circumstances. b. What is the probability of survival of the seed, assuming that it lands"c. Assume that the seed has a 0.2 chance of dying before it lands in a habitat. What is its overall probability of survival?
a. Find the probability tree as attached below
b. Pr[survival] = Pr[good&survive]+Pr[medium&survive]+Pr[low&survive]=
0.24+0.06+0.05 = 0.35
c. Assume that the seed has a 0.2 chance of dying before it lands in a habitat. What is its overall probability of survival?
Pr[survival] = Pr[survival|lands] * Pr[lands] = 0.35 * 0.2 = 0.07
The sum means adding, so your answer is f + 6
Answer:
The answer is: B) Yes, this is a valid inference because it took a random sample from the neighborhood.
And apart from the random sample, 40% of families agreed with her to attend the parade at least.
Step-by-step explanation:
Answer: 0.0930
Step-by-step explanation:
As per given , we have
H0: μcoffee = 6
Ha: μcoffee < 6
She finds z = −1.68 with one-sided P-value P = 0.0465.
The P-value for two-tailed test is calculated by :
For z= -1.68 , we have
![=2(1-P(z\leq1.68))\ \ [\because\ P(Z>z)=1-P(Z\leq z)]](https://tex.z-dn.net/?f=%3D2%281-P%28z%5Cleq1.68%29%29%5C%20%5C%20%5B%5Cbecause%5C%20P%28Z%3Ez%29%3D1-P%28Z%5Cleq%20z%29%5D)
![=2(1-0.9535213)\ \ \ [\text{ NORM.S.DIST(z, TRUE}]\\\\=2( 0.0464787)=0.0929573\approx0.0930](https://tex.z-dn.net/?f=%3D2%281-0.9535213%29%5C%20%5C%20%5C%20%5B%5Ctext%7B%20NORM.S.DIST%28z%2C%20TRUE%7D%5D%5C%5C%5C%5C%3D2%28%200.0464787%29%3D0.0929573%5Capprox0.0930)
Hence, the correct two-sided P-value for z = −1.68 is 0.0930 .
