Ask your own question, for FREE!
Mathematics 18 Online
OpenStudy (anonymous):

Which method of gathering data is described by the statement, “You do manipulate variables”? Observed Study Experiment Simulation Indirect Study

jimthompson5910 (jim_thompson5910):

Hint: In an observed study, you are not affecting the outcome of any variable since you are merely observing (it's like you are invisible and not even there)

OpenStudy (anonymous):

Consider the experiment: A consumer advocacy group is curious about the effectiveness of pain medication in treating migraine headaches. They randomly give different doses of aspirin, Tylenol, and ibuprofen to migraine sufferers. They then measure the results and compare. What were the levels? The migraine patients The pain medication The aspirin, Tylenol, and ibuprofen The dosages of the drugs This is dosages of the drugs right? thanks

OpenStudy (anonymous):

@jim_thompson5910

jimthompson5910 (jim_thompson5910):

that is correct

OpenStudy (anonymous):

A mattress store boasts 4 out of 5 sleepers in a commercial that airs locally. I talked to the owner and found out he only talked to ten people. The situation listed above is an example of what? Regressive Stratification Variation Replication Blocking Would stratification be right?

OpenStudy (anonymous):

@jim_thompson5910

jimthompson5910 (jim_thompson5910):

not sure on this one, one sec

jimthompson5910 (jim_thompson5910):

you might be right, but I'm having a tough time verifying

OpenStudy (anonymous):

Its, i will just go with it. thanks !!

jimthompson5910 (jim_thompson5910):

np

OpenStudy (anonymous):

@jim_thompson5910 For this is .512 right its just .8 * .8 * .8 A couple plans to have three children. The chance they have a girl is 0.8. The gender of one child is independent of the gender of another child. What is the probability of having all girls when having three children? 0.008 0.096 0.384 0.512

jimthompson5910 (jim_thompson5910):

correct

OpenStudy (anonymous):

A family wants to use a simulation to find out the probability of having at least one boy in their first three children. The digit assignment is even for a girl and odd for a boy. The random digits are given as: 2731472867829684178912. What is the correct information about the fourth family in this simulation? The first born was a boy, followed by a girl, followed by a girl. The first born was a girl, followed by a boy, followed by a girl. The first born was a boy, followed by a boy, followed by a boy. The first born was a boy, followed by a girl, followed by a boy. Im very confused with this because it says even is girl, odd is boy. then the 4th family we are trying to find? how do we know which is the 4th family? if we just look at it the first number is even so its girl. so i went with the second answer

OpenStudy (anonymous):

@jim_thompson5910

jimthompson5910 (jim_thompson5910):

here's how I think of it you have a set of numbers: 2731472867829684178912 these represent whether you have a girl or boy (even or odd) if you break it up into groups of 3, you get: 273 147 286 782 968 417 891 2 the fourth family would be the block: 782 so that's odd, even, even, so you would get boy, girl, girl

jimthompson5910 (jim_thompson5910):

not sure if that's correct or not since it's hard to confirm something like this

OpenStudy (anonymous):

I think your right. i only have two more questions :) From the five number summary 7 13 17 21 43 found after the class has taken the exam, which statement below is true? The mean of the lowest 25% is 7. 25% of students earned 17 or fewer. 75% of students earned 17 or more. The highest score in the class was a 43. I got the third one 75 earned 17 or more is this correct?

OpenStudy (anonymous):

@jim_thompson5910

jimthompson5910 (jim_thompson5910):

17 is the median, which divides the data set into two halves (50% each)

jimthompson5910 (jim_thompson5910):

the 5 number summary goes like this min, Q1, median, Q3, max

OpenStudy (anonymous):

so if 50% divides each set. the mean of the lowest would be 7? so the first one

jimthompson5910 (jim_thompson5910):

since Q1 = 13 this means 75% of the class earned more than 13

jimthompson5910 (jim_thompson5910):

7 is the min, so that's incorrect

OpenStudy (anonymous):

so its not 75% earned more than 17. because if 75 earned more than 13 could we say 25% students earn 17 or fewer?

OpenStudy (anonymous):

@jim_thompson5910

jimthompson5910 (jim_thompson5910):

17 is the median anything above it is 50% anything below it is 50%

jimthompson5910 (jim_thompson5910):

so 50% of the students scored higher than 17 50% of the students scored lower than 17

jimthompson5910 (jim_thompson5910):

that's what it means to be the median

OpenStudy (anonymous):

so the highest score is 43 is basically the only thing we can chose?

jimthompson5910 (jim_thompson5910):

yes

jimthompson5910 (jim_thompson5910):

max = 43

OpenStudy (anonymous):

alright that was a little confusing for me aha. the last one i had was this

OpenStudy (anonymous):

On the math portion of the SAT, the mean score was 510 with a standard deviation of 95. On the verbal portion, the mean score was 530 with a standard deviation of 85. We now know the mean of the combined score is 1040 with a standard deviation of 127.48. Given that these are normal distributions, we can do normal calculations with the combined statistics. What score would you need in order to be in the top five percent? 1131.8 1197.7 1231.5 1249.7

OpenStudy (anonymous):

i dont really know what to do to find the top 5%

OpenStudy (anonymous):

@jim_thompson5910

jimthompson5910 (jim_thompson5910):

one sec

jimthompson5910 (jim_thompson5910):

unfortunately there's no other way but to use a calculator

jimthompson5910 (jim_thompson5910):

on that page you'll see that P(z>1249.69) = 0.05 so this means that if you score higher than 1249.69, you are in the top 5%

OpenStudy (anonymous):

Got it!! thanks so much for the help!!

jimthompson5910 (jim_thompson5910):

yw

Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!
Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!