Question

I need some help
A student answers all 48 questions on a multiple-choice test by guessing. Each question has four possible answers, only one of which is correct. Find the probability that the student gets exactly 15 correct answers. Use the normal distribution to aproximate the binomial distribution.

I used BinomCdf(48,.25,15) and get 0.0767 but that isn't one of the choices. The choices are
0.7967
0.0606
0.0823
0.8577

Answer 1

you use the binomPdf function since you want exactly one probability. You aren't summing a bunch of binomial probabilities

Answer 2

if it asked something like "what is the probability the student gets at most 15 correct", then you'd be computing P(X <= 15) which would lead you to using binomCdf

Answer 3

I'm sorry I was using PDF, CDF got me 0.8768. I accidentally typed it in wrong

Answer 4

oh and it says "Use the normal distribution to aproximate the binomial distribution. "

I missed that part

Answer 5

mean = n*p = 48*0.25 = 12

standard deviation = sqrt(n*p*(1-p)) = sqrt(48*0.25*(1-0.25)) = 3

Answer 6

so you'll use the normal distribution with mean mu = 12 and std dev sigma = 3 to approximate this binomial distribution

Answer 7

So I would use normalpdf?

Answer 8

That's getting me 0.0807

Answer 9

normal cdf actually

look at this page

https://people.richland.edu/james/lecture/m170/ch07-bin.html

don't forget about the `Continuity Correction Factor`

Answer 10

What would I make my boundaries? I know about the add.5 and subtract .5, I'm not given much to work with

Answer 11

you'd have the lower boundary be 14.5
the upper boundary would be 15.5

Answer 12

I'm getting 0.0807 when I compute the area under the curve from 14.5 to 15.5

Answer 13

That's what I'm getting also. It's not a choice though :-\ see where I'm getting confused?

Answer 14

the more accurate value of P(X = 15) is 0.0767

I'm using the geogebra web app (the probability subsection) to compute these probabilities

https://web.geogebra.org/#probability

Answer 15

strange how none of the listed choice match

Answer 16

That's really what threw me off. I've tried just about everything and nothing is matching out.

Answer 17

maybe they want you to use a table instead of a calc?

Answer 18

No our teacher hates the table, any time they ask us to use it she makes us use the calculator. We didn't really learn how to do anything with the tables.

Answer 19

hmm let me try something

x = 14.5
z = (x-mu)/sigma
z = (14.5-12)/3
z = 0.83333
z = 0.83


x = 15.5
z = (x-mu)/sigma
z = (15.5-12)/3
z = 1.1666
z = 1.17


use a table like this one
http://www.stat.ufl.edu/~athienit/Tables/Ztable.pdf

to find that

P(z < 0.83) = 0.7967
P(z < 1.17) = 0.9525

So
P(0.83 < z < 1.17) = P(z < 1.17) - P(z < 0.83)
P(0.83 < z < 1.17) = 0.8790 - 0.7967
P(0.83 < z < 1.17) = 0.0823

Answer 20

that's odd how I was able to land on 0.0823 but your teacher didn't show you how to use the table or that she hates the tables. I don't blame her. I prefer the calculator too

Answer 21

sorry I made a typo, I meant to say

P(z < 1.17) = 0.8790

Answer 22

Maybe she's not the one who wrote out the questions? Idk but Thank you so much!!!!! I've been stressing over this for hours.

Answer 23

hours!

Answer 24

That happens a lot. Often the teacher uses questions found online somewhere or they were written by the department chair (those questions are then shared amongst the whole department to make things easier)

Answer 25

well I went back and finished all the others. We were assigned 400 problems to. (I would hope she didn't write all of them)

Answer 26

the ppl who make tests can kiss butt