Question

Write a problem for which the following would be used in finding the solution.a) Binomialpdf(20,0.25,12)b) Binomialcdf(20,0.25,12)

Answer 1

@jim_thompson5910

Answer 2

Binomialpdf notation is

Binomialpdf(n,p,k)

n = # of trials
p = probability of success on 1 trial
k = number of exact successes you want

Answer 3

so Binomialpdf(20,0.25,12) means

n = 20
p = 0.25
k = 12

Answer 4

effectively allowing you to find the probability of getting exactly 12 successes (out of 20 trials total)

Answer 5

so what ideas come to mind?

Answer 6

uh maybe we can say for like an AP test? 20 tests, 25% success rate and we want 12 people to pass to make the school look good?

Answer 7

@jim_thompson5910

Answer 8

that works

Answer 9

okay what about for b?

Answer 10

the 'cdf' is very similar but it's a cumulative probability

cdf = cumulative density function

Answer 11

basically it's a bunch of pdfs added up

Answer 12

example

Binomialcdf(20,0.25,2) = Binomialpdf(20,0.25,0)+Binomialpdf(20,0.25,1)+Binomialpdf(20,0.25,2)

Answer 13

the Binomialcdf allows you to answer questions like "what is the probability of getting at least X or at most X" stuff like that

So again, this is where you're adding up a bunch of individual pdfs

Answer 14

so could we do the scores for AP tests over the years?

Answer 15

or you can ask something like "what is the probability of at most 12 succeeding"

that's the same as saying

"probability of 0 succeeding + probability of 1 succeeding + ... + probability of 12 succeeding"

Answer 16

oh okay!

Answer 17

can you help with one more?

Answer 18

sure

Answer 19

A quarterback for the Seattle Seahawks completes 54% of his passes. Let the random variable X be the number of passes completed in 20 attempts. Conduct a simulation of the 20 attempts using the following random digits. Be sure to state how you assign your digits.

98726 10983 56239 42042 76520 68276 58239 48729 84912 87491

a) What proportion of passes were completed?
b) How does this compare with what "should have happened" theoretically?

Answer 20

"A quarterback for the Seattle Seahawks completes 54% of his passes"

so p = 0.54

Answer 21

54% = 54/100

so if he threw 100 passes, then we expect him to make 54 passes (average)

Answer 22

okay

Answer 23

how do we use "98726 10983 56239 42042 76520 68276 58239 48729 84912 87491" ?

well we can make this rule

break up the sequence into number pairs. So the first number is 98, the second is 72, third is 61, etc etc

Rule: if you get a number from 1 to 54 (inclusive), he completes the pass. If you get a number from 55 to 100 (inclusive), then he fails to complete the pass

Answer 24

okay

Answer 25

So based on that rule, the first three numbers: 98, 72, 61 means he fails his first three passes

but the next number 09 tells us that on the 4th pass, he actually completes it (his first completion)

make sense?

Answer 26

okay give me a sec to get A

Answer 27

i got 18/25 @jim_thompson5910

Answer 28

how did you get 25? he only did 20 attempts (ie 20 trials)

Answer 29

actually it would be 7/25

Answer 30

98726 10983 56239 42042 76520 68276 58239 48729 84912 87491


98
72
61
09
83
56
23
94
20
42
76
52
06
82
76
58
23
94
87
29

84
91
28
74
91

--------------------------------------------------------------------------

Rule: if you get a number from 1 to 54 (inclusive), he completes the pass. If you get a number from 55 to 100 (inclusive), then he fails to complete the pass


98 - fail
72 - fail
61 - fail
09 - successful pass
83 - fail
56 - fail
23 - successful pass
94 - fail
20 - successful pass
42 - successful pass
76 - fail
52 - successful pass
06 - successful pass
82 - fail
76 - fail
58 - fail
23 - successful pass
94 - fail
87 - fail
29 - successful pass


There are 8 successful passes out of 20 total, so the probability for part a) is 8/20 = 2/5

Answer 31

oh i iused all the numbers i dint stop after 20 my bad

Answer 32

@jim_thompson5910

Answer 33

what's your question

Answer 34

is that it for A?

Answer 35

yes 2/5 = 0.4

Answer 36

and for B it would be this is less then what "should've happened" theorectially

Answer 37

yes

empirically we got 40%

theoretically it should be 54%

Answer 38

the more trials you do, the closer you should get to the theoretical probability (assuming the theoretical probability is accurate)

Answer 39

okay than you again! can you help with one more question or do you not have time?

Answer 40

post it and show me what you got so I can check

Answer 41

What are the conditions necessary for using the formula for standard deviation (sigma = sqrt(p(1-p)/n)
when examining a population from a sampling distribution?

Answer 42

check out this page

http://stattrek.com/estimation/confidence-interval-proportion.aspx

Answer 43

are these the two conditions?
the sampling method is simple random sampling.
The sample is sufficiently large. As a rule of thumb, a sample is considered "sufficiently large" if it includes at least 10 successes and 10 failures.

Answer 44

look just above the "alert" box

Answer 45

When the population size at least 10 times larger than the sample size?

Answer 46

yes

Answer 47

N > 10n

N = population size
n = sample size

Answer 48

okay thank you so much! you're the best!

Answer 49

If \(\Large N \ge 10n\) is false, then \(\Large \sqrt{\frac{N-n}{N-1}}\) needs to be tacked on (as shown on the page).