Question

math test normal distribution ?!?!

Answer 1

@iPwnBunnies

Answer 2

Sorry, I don't know about statistics.

Answer 3

do you know any one who is?

Answer 4

No, sorry.

Answer 5

@Jadeishere

Answer 6

which question do you need help with?

Answer 7

all of them like any you want to help me with

Answer 8

Okay

Answer 9

thank you

Answer 10

Okay i'm going from the bottom up. I'm not sure about 11.
10. I think that depends on what the person did to break the law. If they stole something, would you think it'd be fair to kill them? if the commited a mass genocide, and killed thousands of people, then yes, i do believe that they'd have the right. That does not mean they should kill. Period.
9. Not sure
8. 43%
7. you already know
6. 22%
5. 3
4. mean = 8, variance = 18.3 and the standard deviation is 4.3
3. 30th percentile is 113 and the 90th percentile is 200
2. it's the one that's crossed out in blue

Answer 11

for 11: the labels in order from left to right:
\(\mu-3\sigma,\,\,\, \mu-2\sigma,\,\,\, \mu-\sigma,\,\,\, \mu, \,\,\,\mu+\sigma, \,\,\,\mu+2\sigma,\,\,\, \mu+3\sigma\)

Answer 12

and just substitude \(\mu=50\) and \(\sigma=2\)

Answer 13

For 10, they are asking about BIAS, not whether the question is "right or wrong" (as @Jadeishere alluded to) . That means, will responders be more likely to answer in one way versus another because of how the question is asked.

Answer 14

For 9: This is the question about bias again. Here, you can say there would be a very likely potential bias of people responding to eating out more (since the interviewer is interviewing people at a restaurant...) It's a very selective sample and is not a good indicator of how a more general population (the sample is too homogeneous).

Answer 15

for 11 what steps would I show?

Answer 16

8: 93% ... Almost everyone is found within this range of rent. In fact, it looks like about a range of \(\pm \)2 standard deviations from the mean... which is about 95% of the data... so 93% is your closest bet

Answer 17

They already sketched the curve for you.. just write those labels by using \(\mu=50\) and \(\sigma=2\). For example, \(\mu-3\sigma=50-3(2) = 44\)

Answer 18

okay I think I understand

Answer 19

7: is correct.

6: You have that \(X\sim\text{Bin}(10, 0.61)\) where X = number of people who swallow a spider. The probability of success (swallowing at least 1 spider) is 0.61. So, you have either people swallowing at least 1 spider or not. The trials are independent and the success probability is the same on every trial, which is why we have a binomial distribution.

You want \(P(X=7)\), so
\[P(X=7)={10\choose7}(0.61)^7(1-0.61)^{10-7}=0.2237 \]

Answer 20

so it would be 22%

Answer 21

yes

Answer 22

okay thanks!!

Answer 23

The rest seems okay I think.. I didn't try entering all those data points for the questions about data sets but if you have access to a calculator/program that has stats functions, it should be very straightforward