Answer:
a) p-hat (sampling distribution of sample proportions)
b) Symmetric
c) σ=0.058
d) Standard error
e) If we increase the sample size from 40 to 90 students, the standard error becomes two thirds of the previous standard error (se=0.667).
Step-by-step explanation:
a) This distribution is called the <em>sampling distribution of sample proportions</em> <em>(p-hat)</em>.
b) The shape of this distribution is expected to somewhat normal, symmetrical and centered around 16%.
This happens because the expected sample proportion is 0.16. Some samples will have a proportion over 0.16 and others below, but the most of them will be around the population mean. In other words, the sample proportions is a non-biased estimator of the population proportion.
c) The variability of this distribution, represented by the standard error, is:
d) The formal name is Standard error.
e) If we divided the variability of the distribution with sample size n=90 to the variability of the distribution with sample size n=40, we have:

If we increase the sample size from 40 to 90 students, the standard error becomes two thirds of the previous standard error (se=0.667).
Answer: 30 - 4 = 26
Step-by-step explanation:
30 - 2x = 0
so after 2 days he will have done 4 hours so 30 - 4 = 26
If you add $15 + $27 because that's how much it costs for each student, you get $42. Then you divide $714 by $42, you get 17 and that's how many students are in the class this month.
Answer:
600 times .012=7.20
7.20 times 6= 43.20
43.20+ 600= 643.20
the correct answer is b- $643.20
Answer:
The expected value of Jordan gains is -1 dollar.
Step-by-step explanation:
Consider the following random variables. X := #of shots that Jordan makes. Then, we can can define the random variable Y of the earnings of Jordan in a game as follows
Y = 5 if X=2 (since he gets 10, but invested 5), Y=0 if X=1(since he gets the 5 back) and Y=-5 if X=0(since he doesn't get the money back). Then, in this case, we can define the probability as follows.
P(Y=5) = P(X=2), P(Y=0) = P(X=1), P(Y=-5)= P(X=0).
By definition, the expected value of Y is given by
. By the previous analysis, we have that
![E[Y] = 5\cdot P(X=2)-5P(X=0)](https://tex.z-dn.net/?f=E%5BY%5D%20%3D%205%5Ccdot%20P%28X%3D2%29-5P%28X%3D0%29)
We only need to calculate the probabilities for X. In this case, we can consider each shot independt from each other. Then, we can consider X to be distributed as a binomial random variable with n=2 trials and p=0.4 of success (since he has a 40% chance of winning).
Then, by definition

where 
Then,


Then,
![E[Y] = 5\cdot 0.16-5\cdot 0.36 = -1](https://tex.z-dn.net/?f=E%5BY%5D%20%3D%205%5Ccdot%200.16-5%5Ccdot%200.36%20%3D%20-1)