Question

Some people believe in the force (yes, like in Star Wars). To test if young Anakin Skywalker has the force with him, he is told that some cards that Yoda can see but Anakin cannot contain a picture of a planet, a lightsaber, a laser rifle, or a spaceship. As Yoda looks at 20 such cards in turn, Anakin tries to guess what is on the card Yoda is looking at. Of course, Anakin has a 25% chance of simply guessing correctly.

a) Verify that the count of correct guesses in 20 cards follows a binomial distribution and write the notation.
b) What is the mean number of correct guesses?

Answer 1

c) What is the probability that Anakin guesses all 20 cards correctly?
d) Suppose Anakin guesses correctly on 10 of the cards. What is the probability of him doing this well or better by chance? Do you think he has the force with him?

Answer 2

@jim_thompson5910

Answer 3

a) for this part, you just need to verify the conditions needed to use the binomial distribution

hint: one of those conditions is that you have independent events (one event doesn't affect the other)

Answer 4

b)

mean = n*p

n = number of trials
p = probability of success

Answer 5

Ok I got that part as X= number of correct guesses
P = probability of correct answer ----> 0.25
Q = probability of wrong answer ----> 0.75
Parameter is:
---------> (20,0.25)?

Answer 6

good

Answer 7

you might also see this notation

B(n,p) = B(20,0.25)

Answer 8

you can also say for part a) is that you have two outcomes (he gets the guess right or he gets the guess wrong)

so that adds more support for a binomial distribution

Answer 9

ok cool

Answer 10

Now (b)? @jim_thompson5910

Answer 11

mean = n*p

n = number of trials
p = probability of success

Answer 12

Ok

Answer 13

20*25?

Answer 14

0.25 but yes

Answer 15

5

Answer 16

you got it, mean = 5

Answer 17

Ok ! how do i find the sd

Answer 18

sd = sqrt(n*p*(1-p))

Answer 19

how do i find that

Answer 20

n = 20
p = 0.25

plug them in

Answer 21

sd = sqrt(n*p*(1-p))

sd = sqrt(20*0.25*(1-0.25))

sd = ???

Answer 22

sqrt(20*0.25*(1-20)

Answer 23

yep

Answer 24

then use a calc to evaluate

Answer 25

wait u did sqrt(20*0.25*(1-0.25))

Answer 26

is it yours? or mine

Answer 27

sqrt(5 * (.75)

Answer 28

Is it 3.75 @jim_thompson5910 ?

Answer 29

no that's too high

Answer 30

oh i didn't see yours fully, you made a typo

it *should* be this

sd = sqrt(20*0.25*(1-0.25))

Answer 31

Oh okz

Answer 32

i did that, and got 3.75...

Answer 33

try it again

Answer 34

i did lol

Answer 35

it's NOT 3.75

Answer 36

1.94?

Answer 37

ok here's what the google calculator is saying (you can use google as a calculator)

http://www.google.com/search?hl=&q=sqrt%2820*0.25*%281-0.25%29%29&sourceid=navclient-ff&rlz=1B3GGLL_enUS420US420&ie=UTF-8

Answer 38

ok much better, the sd is roughly 1.9364916731037

Answer 39

Coool ,i got that!!

Answer 40

if you round that to two places, you get 1.94

Answer 41

kk! now c?

Answer 42

wasn't that c?

Answer 43

b was the mean

Answer 44

No

Answer 45

btw i only see a and b

Answer 46

and b makes no mention of the sd

Answer 47

ohhh okay heres c and d

Answer 48

c) What is the probability that Anakin guesses all 20 cards correctly?
d) Suppose Anakin guesses correctly on 10 of the cards. What is the probability of him doing this well or better by chance? Do you think he has the force with him?

Answer 49

yeah unfortunately i don't have the force like anakin does lol

Answer 50

haha

Answer 51

how do i find c

Answer 52

type this into your calculator

(0.25)^20

to get the answer for c

Answer 53

i got 9.094947e-13

Answer 54

yeah it's very small

Answer 55

so either write that or say 0.0000000000009094947

Answer 56

kk

Answer 57

d?

Answer 58

P(X = k) = (n C k)*(p)^(k)*(1-p)^(n-k)

P(X = 10) = (20 C 10)*(0.25)^(10)*(1-0.25)^(20-10)

P(X = 10) = (20 C 10)*(0.25)^(10)*(0.75)^(20-10)

P(X = 10) = (184756)*(0.25)^(10)*(0.75)^10

P(X = 10) = ???

Answer 59

0.00992227527

Answer 60

Can u look at this? A new medical test provides a false positive result for Hepatitis 2% of the time. That is, a perfectly healthy subject being tested for Hepatitis will test as being infected 2% of the time. In research, the test is given to 30 healthy (not having Hepatitis) subjects. Let X be the number of subjects who test positive for the disease.

a) What is the probability that all 30 subjects will appropriately test as not being infected?
b) What are the mean and standard deviation of X?
c) To what extent do you think this is a viable test to use in the field of medicine?

I only need C.

Answer 61

you got it, congrats

and now that I look back up, i see parts c and d in the post below a and b...lol i feel stupid

Answer 62

c) To what extent do you think this is a viable test to use in the field of medicine?

Answer 63

I found a and b..

a) P(r=30)
= 1*1*0.5455
= .5455 or .55 <---

b) Mean = np
= 30*0.98
= 29.4 <---

Standard deviation = sqrt(npq)
= sqrt (30*0.98*0.02)
= 0.7668 <---

c)

Answer 64

well you'd need it to be very accurate, the question is: how accurate?

Answer 65

yea..

Answer 66

hmm let me think

Answer 67

kk

Answer 68

did u get it?

Answer 69

does it mention anywhere you need to make a bunch of distribution tables?

Answer 70

no

Answer 71

ok hold on

Answer 72

basically the value of p is fixed, but you can change the value of n

the question is: when does this test become unreliable when you have n reach a certain point?

Answer 73

yep

Answer 74

what should i say

Answer 75

well let's say you want the average of false positives to be 10 or less (that's your limit and what you would consider accurate...if you get more than 10 on average, then it's inaccurate)

So

mean = np

10 = np

10 = n*0.02

10/0.02 = n

500 = n

n = 500

This means that if you test 500 people, then you'll have on average, 10 false positives.

So the max number of people to be tested is 500 people. Anything more than this will make the test unreliable according to your definition above

Answer 76

that's one way to do it I think

Answer 77

Ok cool! so c should be : So the max number of people to be tested is 500 people. Anything more than this will make the test unreliable according to your definition above

Answer 78

Yeah your test is only viable for up to 500 people max

Answer 79

ok thanks. and my a and b above are correct?

Answer 80

according to your error set up (ie how much error you're willing to tolerate)

Answer 81

huh

Answer 82

where are you lost

Answer 83

are they

Answer 84

are my answers right

Answer 85

for a and b

Answer 86

oh a is correct, b is not

Answer 87

what is b ?

Answer 88

well the second part of b is correct, so that's good

Answer 89

but the mean you got is incorrect

Answer 90

ok@ thanks!!

Answer 91

you use 0.02 NOT 0.98