Question

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Answer 1

Communications 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?

this is either a binomial distribution or a poisson distribution.

my bet is binomial

Answer 2

do you have a ti83? or do you want to count these by hand? its the sum of 20 terms

Answer 3

is exactness required or can we approximate it with a normal distribution?

Answer 4

what is the mean of a binomial distribution? do you recall?

Answer 5

that doesnt look familiar to me

(.16 + .84) ^ 100 = 1, which is the binomial distribution for this

the first 20 terms of the expansion would determine the probability

100/0 (.16)^0 (.84)^100
+ 100/1 (.16)^1 (.84)^99
+ 100/2 (.16)^2 (.84)^98
+ 100/3 (.16)^3 (.84)^97
.....
+ 100/20 (.16)^20 (.84)^80

Answer 6

the sum of a distribution has a probability of 1, right?

Answer 7

we have .16 given, .16+q = 1 what is q?

Answer 8

no

.16 + q = 1

q = 1 - .16 = .84

Answer 9

the binomial distribution can be formualted as:

(1)^100 = 1

(.16 + .84)^100 = 1

we agree so far?

Answer 10

now we dont want the whole thing, we just want part of it: 0 to 20, out of 100, to be exact

Answer 11

at most 20 means: 0, or1, or2, or3, or ... or20 and no more

Answer 12

correct

now, the binomial distribution can be approximated if you dont feel like adding up all 20 terms

Answer 13

do you recall something like this?
\[z_n=\frac{x-\mu}{\sigma}\]

Answer 14

no, but we are about to fill it in :)

if 16% is considered average, then what is the average (the mean) out of 100 people?

Answer 15

umm, no

16% of 100 is 16

Answer 16

mean is stated in the problem: 16% of yada yada .... 16% of 100 is just 16

the variance of a binomial distibuion is npq:

npq = 100(.16)(.84)

\[\sigma=\sqrt{npq}=sqrt(100(.16)(.84))\]

one more thing to adjust and that is the value of x