Question

About 55% of emergency room visits are unnecessary. Assuming each emergency room visit is independent. (Hints: first define the random variable X. Then write the objective of each question in terms of X.)
a) If you randomly select 15 patients from those visiting the emergency department, what is the chance that all of them need to be in the emergency room?

b) What is the expected number from the 15 randomly selected patients, who don’t need to be in the emergency room and what is the standard deviation of that number? @theEric

Answer 1

This is clearly a binomial distribution. Think of each patient as a Bernoulli trial. Each visit (trial) is independent of the others.

Answer 2

wow so i do that 15 times? @wio

Answer 3

how do i go about b..

Answer 4

You don't have to do it 15 times... you only have to do it for \(X=15\)

Answer 5

The expected value for a binomial distribution is \(n\times p\) where \(n\) is the total trials, \(p\) is the probability of each trial

Answer 6

so will the n be wat i got from a? or n = 15

Answer 7

The standard distribution of binomial distribution is \(\sqrt{np(1-p)}\)

Answer 8

The results of a are not needed to do b.

Answer 9

so the b ask of those who do not need to be in the emergency.will it be equal to n* p= where n= 15 and p =.55? thus 15* .55

Answer 10

and do i subtract the answer i get from that from 1? to get those who dont need to?

Answer 11

You don't need to subtract it from 1

Answer 12

Oh, and on second look, it looks like for a you should find \(X=0\) actually.

Answer 13

for b?

Answer 14

@wio

Answer 15

for a

Answer 16

so i only work for 0??

Answer 17

im getting quite confused here,,,if u can help me here.will be very glad

Answer 18

15 are needed means 0 are unnecessary.

Answer 19

so that is for b then or..because b is the one asking for those who dont need to br in the emergency

Answer 20

or i find both 0 and 15 for a

Answer 21

Just find \(\Pr(X=0)\) for a.

Answer 22

\[
X\sim B(n,p)
\]\[\Pr(X=k) = \binom k n p^k(1-p)^{n-k}
\]

Answer 23

I mean \[
\Pr(X=k) = \binom n k p^k(1-p)^{n-k}
\]

Answer 24

ok.so here i use 0 ryt..

Answer 25

We use \(k=0\)

Answer 26

so the answer i get become my final answer?

Answer 27

Is is the probability that all visits of the 15 are necessary.

Answer 28

i had 0.174

Answer 29

that becomes my anser ryt?

Answer 30

wat of b ?

Answer 31

Think long and hard. I have to go.

Answer 32

awww

Answer 33

jux let me knw how im supposed to go about..like the figures for n times p.

Answer 34

@wio