Question

If np ≥ 5 and nq ≥ 5, estimate P (fewer than 4) with n = 13 and p = 0.5 by using the normal distribution as an approximation to the binomial distribution; if np < 5 or nq <5, then state that the normal approximation is not suitable.
P (fewer than 4) =

Answer 1

P( X<4 ) = P( X <= 3)

Answer 2

does the directions state to use normal approximation continuity factor

Answer 3

0.019?

Answer 4

It doesn't say

Answer 5

I just can't grasp this one...

Answer 6

well i can tell you the exact answer, without approximating
binomcdf( 13, .5 , 3 ) = .046

Answer 7

the exact answer, without approximating binomcdf( 13, .5 , 3 ) = .0461425

Answer 8

It says u=np, standard deviation = npq , and z = x-u/standard deviation

Answer 9

the first thing you have to check is whether
np >= 5 and n(1-p) >= 5 , to use the normal curve as an approximation

Answer 10

so is that true here?

Answer 11

The approximation is suitable

Answer 12

right because they are both greater than 5

Answer 13

It states to round to four decimal places as needed.

Answer 14

I just don't know how to find the answer

Answer 15

I am so confused

Answer 16

how would you usually want to find the answer? what are your thoughts tell you?

Answer 17

Like I said, I am confused on this problem. I don't know how to find the answer. I have even tried the examples provided and just can't get it.

Answer 18

if given a normal distribution, with a mean, a standard deviation, and a value of x

what is your usual process to find a probability for the conditions of x? forget about the value sin the problem for a moment

Answer 19

**the values in the ...

Answer 20

does this sound familiar?

define a z score, and look up the proper tail area?

Answer 21

To calculate the np and nq

Answer 22

No, because I am so new at all of this

Answer 23

the only way they would ask you to do a normal approximation, is if youve done work with a normal distribution to begin with. You must have been exposed to it previously.

Answer 24

Yes, in the beginning but now it seems to be getting harder

Answer 25

spose i say:

given a mean of 0, and a standard deviation of 1, what is the probability of geting a value less then -2?

Answer 26

how do we work it?

Answer 27

how did you work it in the beginning?

Answer 28

0.8188

Answer 29

the final results is not important, and its wrong ...

HOW do we approach it? its the same approach we need to use now. with a slight modificaiton.

Answer 30

I have no idea I am sorry. This is so stressful for me

Answer 31

you mean to tell me you just tried to find the solution to what i just gave you, and you have no idea how you did it ..

that doesnt make any sense you know.

Answer 32

are you working with a stats calculator? or a table?

Answer 33

calculator

Answer 34

ti 83? 84? or other

Answer 35

83

Answer 36

did you use the normalCDF function?

Answer 37

I also have a statistics calculator app

Answer 38

Which I use more often

Answer 39

im more familiar with ti83, but in any case we want to use what is traditionally called a "normalCDF" function

Answer 40

how do you use your app? what do you input?

Answer 41

can you screenshot it?

Answer 42

One sec. What would the SD be?

Answer 43

the sd of a binomial distribution is sqrt(npq)

Answer 44

nm I have my ti83 open

Answer 45

so what is my first step?

Answer 46

you see the 2nd button top left? hit it

then hit the VARS button to open a menu

select the normalCDF function and get prepared to input values.

Answer 47

I see tcdf

Answer 48

nope, its a long word, normalCDF

Answer 49

scroll down if you have to, or back up ...

Answer 50

got it

Answer 51

hit enter to select it

Answer 52

k

Answer 53

it should say:

normalcdf(

and is waiting for inputs. you got that?

Answer 54

yes

Answer 55

we are looking for some left tail area, so we want an arbitrarily low value here ... i use -9999 then hit the comma button, should be above the 8 key

Answer 56

it was above the 7

Answer 57

got it

Answer 58

normalcdf(-9999,

the next input is the max of our interval, this is a modification. instead of 4, we want less than 4. we use 3.5

normalcdf(-9999, 3.5,

now we need to insert the mean, the mean of a binomial is np; in this case 13*.5, you can insert the calculation as is, itll work it for you

normalcdf(-9999, 3.5, 13*.5,

the last input is the standard deviation, let me know when you get this far

Answer 59

k got it

Answer 60

you are using the continuity correction factor, adding 0.5 to 3

Answer 61

Yes

Answer 62

the standard deviation of a binomial distribution is sqrt(npq) so thats our last input value

2nd, x^2, 13*.5*.95))

it should look like:
\[\text{normalcdf}(-9999,3.5,13*.5,\sqrt(13*.5*.95)~)\]
then hit enter

Answer 63

ugh, 13*.5*.5 read the p wring

Answer 64

how do i make that big symbol?

Answer 65

2nd, x^2

Answer 66

2nd, x^2, 13*.5*.5 ) )

Answer 67

its giving me an error

Answer 68

then try it again, make sure youve hit all the keys you need to hit

Answer 69

normalCDF( -9999, 3.5, 13*.5, sqrt(13*.5*.5) )

Answer 70

Ugh.... It is not working

Answer 71

[2nd]

[VARS]

[2]

-9999

[ , ]

3.5

[ , ]

13 * .5

[ , ]

[2nd]

[\(x^2\)]

13 * .5 * .5

[ ) ]

[enter]

Answer 72

I typed in everything you posted..

Answer 73

well, since youve been trying, this is what we get if we work the wolf

http://www.wolframalpha.com/input/?i=P%28z%3C%28+%283.5+-+.5%2813%29%29+%2F+sqrt%2813*.5*.5%29+%29%29

Answer 74

Did it again and it says 1: quit or 2: Goto

Answer 75

P(X<3.5) = .048046

rnd to your hearts content

Answer 76

2 for goto

Answer 77

imthinking you ar emissing a comma or a closing paranthesis, but thats just an idea

Answer 78

OMG it worked!!!!

Answer 79

Thank you!!!!!

Answer 80

yep, good luck :)

Answer 81

How did you end up putting that problem into wolfram???

Answer 82

Where did you get the 3.5?

Answer 83

the normal approximation of a binomial ... theres some adjustment to be made since one is discrete and the other continuous.