SOLUTION: ACCORDING TO A STUDY CONDUCTED APPROXIMATELY 55% OF ALL HOSPITALS IN A GIVEN TOWN CONTAINED 100 OR MORE BEDS. A RESEARCHER DRAWS A SAMPLE OF 15 HOSPITALS BY RANDOMLY SELECTIN

Question 1084726: ACCORDING TO A STUDY CONDUCTED
APPROXIMATELY 55% OF ALL HOSPITALS
IN A
GIVEN TOWN CONTAINED 100 OR MORE
BEDS. A RESEARCHER DRAWS A SAMPLE
OF 15
HOSPITALS BY RANDOMLY SELECTING
NAMES FROM A DIRECTORY OF HOSPITALS.
A/ WHAT IS THE PROBABILITY OF
SELECTING 10 OR MORE HOSPITALS THAT
HAVE 100
OR MORE BEDS?
B/ WHAT IS THE PROBABILITY FO
SELECTING LESS THAN FIVE HOSPITALS
THAT HAVE
100 OR MORE BEDS?
C/ WHAT IS THE PROBABILITY OF
SELECTING FROM SIX TO TEN HOSPITALS,
INCLUSICE, THAT HAVE 100 OR MORE
BEDS?
Answer by mathmate(429) (Show Source): You can put this solution on YOUR website!
Question:
ACCORDING TO A STUDY CONDUCTED APPROXIMATELY 55% OF ALL HOSPITALS IN A
GIVEN TOWN CONTAINED 100 OR MORE BEDS. A RESEARCHER DRAWS A SAMPLE OF 15
HOSPITALS BY RANDOMLY SELECTING NAMES FROM A DIRECTORY OF HOSPITALS.
A/ WHAT IS THE PROBABILITY OF SELECTING 10 OR MORE HOSPITALS THAT HAVE 100
OR MORE BEDS?
B/ WHAT IS THE PROBABILITY OF SELECTING LESS THAN FIVE HOSPITALS THAT HAVE
100 OR MORE BEDS?
C/ WHAT IS THE PROBABILITY OF SELECTING FROM SIX TO TEN HOSPITALS, INCLUSICE, THAT HAVE 100 OR MORE BEDS?

Solution:
For your information, writing in all caps is considered shouting in posts. So please use proper case (Capitalize only at beginning of sentences) in the future.

The problem/experiment satisfies criteria for a binomial distribution:
1. Bernoulli trials, i.e. exactly two possible outcomes (>100 beds or not)
2. Number of trials is known before experiment, i.e. independent of outcomes (selection of N=15 hospitals)
3. All trials are independent of each other (randomly picked)
4. Probability of success is known, and remain constant throughout trials. (assumed to be 55% for all cases)

The probability of x successes out of N trials each with probability of success
p is given by
P(x)=C(N,x)(p^x)(1-p)^(N-x)
where
C(N,x) is number of combinations of selecting x objects out of N.

Substituting numbers,
N=15
p=0.55

Since N=15 is a relatively small number, and probability of every case from x=0 to 15 required, we will start with a list of probabilities using the above formula.
x P(x)
0 6.28330*10^-6
1 1.15194*10^-4
2 9.85547*10^-4
3 0.00522
4 0.01914
5 0.05146
6 0.10483
7 0.16474
8 0.20134
9 0.19140
10 0.14036
11 0.07798
12 0.03177
13 0.00896
14 0.00156
15 1.27479*10^-4

The required probabilities are simple addition of the respective values.
(A)
P(x>=10)=0 0.14036+0.07798+0.03177+0.00896+0.00156+1.27479*10^-4
=0.26076

(B)
P(x<5)=6.28330*10^-6+1.15194*10^-4+9.85547*10^-4+0.00522+0.01914
=0.025467

(C)
P(6<=x<=10)=0.10483+0.16474+0.20134+0.19140+0.14036
=0.80267

Note that the three probabilities add up to greater than 1 because P(x=10) have been repeated in cases (A) and (C), and P(x=5) was not called for.

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