<p>hi these are the probs from random variables and the binomial and geometric distributions.
im so confused. :((</p>
<ol>
<li>For college-bound high school seniors in 1996, the nationwide mean sat verval score was 505 with a standard deviation of about 110, and the mean sat math score was 508 with a standard deviation of about 110. students who do well on the verbal portion of the sat tend to do well on the math portion. if the two scores for each student are added, the mean of the combined scores is 1013. what is the standard deviation of the combined verbal and math scores ?</li>
</ol>
<p>The answer is (the standard devation can not be computed from the information given.)</p>
<ol>
<li>In a certain game, a fair die is rolled and a plalyer gains 20 points if the die shows a "6". if the die doesnt show a "6" the player loses 3 points. if the die were to be rolled 100 times, what would be the expected total gain or loss for the player?
(answer is a gain of about 83 points)</li>
</ol>
<p>3.If a customer rolls the dice and rents a second movie every thrusday for 30 consecutive weeks, what is the approximate probability that the total amound paid for these second movies will exceed $15.00?
a)0
b)0.09
c)0.14
d)0.91
i dont know the answer :(</p>
<ol>
<li><p>circuit boards are assembled by selecting 4 computer chips at random from a large batch of chips. in this batch of chips, 90 percent of the chips are acceptable. let x denote the number of acceptable chips out of a sample of 4 chips from this batch. what is the least probable value of x ?
a)0
b)1
c)2
d)4</p></li>
<li><p>supposed we select an SRS of size n=100 from a large population having proportion p of successes. let x be the number of successes in the sample. for which value of p would it be safe to assume the sampling distribution of x is approximately normal?
a) 0.01
b) 1/9
c) 0.975
d) 0.9999</p></li>
</ol>
<p>I’m in stats right now, hopefully I can help you a little bit. Please don’t be upset if they aren’t accurate, we’re doing inference right now and that’s confusing. But anyway here’s what I think:</p>
<p>1.standard deviation cannot be determined because it doesn’t give you the info to find it…pretty straightforward. for sample means std dev is: sigma/sqrt n and it doesn’t give you the sample size.</p>
<ol>
<li><p>so a distribution table would look like this:
roll: 1 2 3 4 5 6
x: -3 -3 -3 -3 -3 20
p(x): 1/6------------------------->
So expected value actually means mean of the data. so you do (x1<em>p1)+(x2</em>p2)…etc. leaving you with .83<em>100=83 (rounding can make you be off a bit)
*</em>*not 100% on this one</p></li>
<li><p>I think this one is missing some info? need a variance or something.</p></li>
<li><p>Well x is the least probable amount of working chips, which would mean a batch of 4 chips that are all defective, or 0. Not much stat involved there.</p></li>
</ol>
<p>5.p hat is probability of success/sample size, so x/100. One of the rules we learned is that our population of interest must be at least 10 times greater than our sample size. so which one would fit?
a. .01=x/100 x=1
b. 1/9=x/100 x=11.11
c. .975=x/100 x=97.5
d. .9999=x/100 x=99.5
The only one that fits is A.</p>
<p>For #1, the one assumption that people tend to make is that you can add the standard deviations together to get a standard deviation of 220. The problem is that when you add the data sets together, you can’t necessarily assume that the standard deviation would be the sum of the two standard deviations. You can try this with the data sets 1, 2, 3, 4, 5, 6 and 10, 10, 10, 20, 20, 20 to see what happens.</p>
<p>trav23’s answer to #2 is correct. Another way to calculate it would be to figure that in six rolls, you should average each result once. Achieveing each result once would leave you with a total of 20-3-3-3-3-3 = 5 points. You will go through this process 100/6 times, earning 5 points for each of those processes, for a total of 100/6*5 is approximately 83.</p>
<p>For #4, you could actually calculate the probabilities if you wanted to:
P(0 chips are effective) = 1<em>(.1)^4 = .0001
P(1 chip is effective) = 4</em>(.1)^3<em>(.9) = .0036
P(2 chips are effective) = 6</em>(.1)^2<em>(.9)^2 = .0486
P(4 chips are effective) = 1</em>(.9)^4 = .6561</p>
<p>thank you so muchhh!
im sorry i forgot put the actual problem for number 3, here the additional info.</p>
<p>Every thursday, matt and dave’s video venture has "roll-the-dice"day. A customer may choose to roll two fair dice and rent a second movie for an amount equal to the numbers uppermost on the dice witht the larger number first. for ex, if the customer rolls a two and a four, a second movie may be rented for $0.42. Let X represent the amount paid for a second movie on roll-the-dice day. the expected value of X Is 0.47$ and the standard deviation of X is 0.15$.</p>
<p>There are 21 possible outcomes: 15 outcomes where the dice have different numbers that can occur in two different but indistinguishable ways, and 6 outcomes where the dice have the same number that can occur in one way.</p>
<p>So, your data set in essence has 36 values:</p>