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) Suppose the distribution of unnecessary emergency room visits can be approximated by a normal distribution. What is the probability that AT LEAST two of them need to be in the emergency room? @dumbcow

Answer 1

http://en.wikipedia.org/wiki/Binomial_distribution#Normal_approximation

what is sample size?

Answer 2

15 patients

Answer 3

@dumbcow

Answer 4

mean = .55
std dev = sqrt(15*.55*.45)

\[P(X>2/15) = 1- P(X< 2/15)\]
\[Z = \frac{2/15 - 0.55}{\sqrt{15*.55*.45}}\]
look up probability using z-table

Answer 5

is it 2 divided by 15?

Answer 6

@dumbcow

Answer 7

P(0)+P(1) is prolly an equivalent way to approach it

Answer 8

0 needs + 1 needs; .55 not, .45 needs

\[\binom{15}{0}.45^0.55^{15}+\binom{15}{1}.45^1.55^{14}\]

Answer 9

1 - that

Answer 10

oh ok...this much understandable..do i subtract my final answer from 1?

Answer 11

yes

Answer 12

oh nice

Answer 13

thank you..very much..

Answer 14

youre welcome

Answer 15

@amistre64 WHY doesnt .55 come before .45 in both P(0) + P(1)

Answer 16

About 55% of emergency room visits are unnecessary.

.55 = not need

What is the probability that AT LEAST two of them NEED to be in the emergency room?

exactly 0 need, exactly 1 need, ex 2 need, ex 3 need, ..., 15 need
|-------------------------|
NOT "at least 2 need"

1 - (not at least 2 need) = (at least 2 need)

since we are looking for the prolly of NEED as a success, 1-not need = need
1 - .55 = .45

the setup is: (n k) s^k f^(n-k)

P(0 need) = (15 0) .45^0 .55^15
P(1 need) = (15 1) .45^1 .55^14