Question

Algebra 2 Help Please! Binomial Distribution.

Answer 1

According to a national survey, 16% of U.S. adults use the internet to make telephone calls. In a random survey of 100 adults what is the probability that at most 20 of them use the internet to make telephone calls?

Answer 2

@jim_thompson5910 @Miracrown

Answer 3

do you have a calculator like a TI83 or TI84?

Answer 4

N=100
P=0.16

Np=100*0.16 = 16 > 5
N(1-p)=100*0.84=84 > 5

Mean u = np = 16
Standard deviation o = sqrt (np(1-p)) = sqrt(100(0.16)(0.84)) = 4

Z= 20-16/4 = 4/4 = 1

P(x<_20) = *Not sure how to answer this part*

Answer 5

oh that works too

Answer 6

No I do not! @jim_thompson5910

Answer 7

let me look over your work

Answer 8

sqrt(100(0.16)(0.84)) is not equal to 4

Answer 9

It was about equal to but the sign would not work

Answer 10

sqrt(100(0.16)(0.84)) = 3.67 roughly

Answer 11

since we are approximating the binomial distribution with a normal distribution, we need to use the continuity correction factor

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

Answer 12

Using the continuity correction factor, we get
\[\Large P(X \le 20) \to P(X < 20+0.5)\]
\[\Large P(X \le 20) \to P(X < 20.5)\]

Answer 13

x = 20.5
mu = 16
sigma = 3.67


z = (x - mu)/sigma
z = (20.5 - 16)/3.67
z = 1.23


P( X < 20.5) = P(Z < 1.23)

You will then use a table like this one

https://www.stat.tamu.edu/~lzhou/stat302/standardnormaltable.pdf

to compute P(Z < 1.23)

according to that table, P(Z < 1.23) is approximately 0.89065

So P(X < 20.5) = 0.89065 approximately