Question

12-15: on the average 20% of seniors develop influenza during a given year. 15 students are selected at random from a nursing home & followed up for a year.

P= .2 1-P= .8

12. What is the probability exactly 3 out of 15 seniors will develop influenza?

13. What is the probability that at most 3 of them will develop influenza?

14. What is the probability that at least 3 of them will develop influenza?

15. The standard deviation of the number of seniors will get influenza is:

please show work so I can understand how to solve :)

thanks

Answer 1

@Summersnow8 Have you learned the binomial distribution yet?

Answer 2

yes, but I am confused about how to solve this, my professor is terrible at explaining things

Answer 3

I know how to use the binompdf or binomcdf on my calculator, but can't seem to figure this problem out.

Answer 4

std dev = sqrt(n*p*(1-p))

Answer 5

12 can be solved as follows:
\[P(3\ out\ of\ 15)=15C3\times(0.2)^{3} \times(0.8)^{12}=you\ can\ calculate\]

Answer 6

15 nPr 3 x....etc?

Answer 7

what does the C stand for? nCr?

Answer 8

I got .25 for 12

Answer 9

15C3 means the number of combinations of 15 different items taken 3 at a time. Your answer for 12 is correct.

Answer 10

For 13
find probabilities for x=0,1,2,3
add them up to get P(X<= 3)

for 14
sum x=0,1,2
subtract this sum from 1
this will give P(X>=3)

Answer 11

ok how do you do the rest of the problems @kropot72

Answer 12

@dumbcow that doesn't help me understand how to solve the problem

Answer 13

ok let me explain
"x" is the number out of 15 that get sick, from 0 to 15
the sum of all probabilities is always 1

you want the probability that "at most" 3 get sick, so you consider the cases where
0,1,2,3 people out of 15 get sick
using the formula @kropot72 showed above to get individual probabilities for each case
add them up and that gives you P(x<= 3) or probability that at most 3 get sick

Answer 14

Going over the method of calculation for the binomial distribution:
If we let p stand for the probability of success, and q = 1 - p the probability of failure, then the probability of r successes when n independent trials are carried out is given by:
\[\left(\begin{matrix}n \\ r\end{matrix}\right) \times p ^{r} \times (1-p)^{n-r}\]
where the binomial coefficient
\[\left(\begin{matrix}n \\ r\end{matrix}\right)=\frac{n!}{r!(n-r)!}=nCr\]

Answer 15

So to find the probability of zero getting sick, we substitute the following values:
n = 15
r = 0
p = 0.2
1 - p = 0.8
Can you now find P(0 out of 15) ?

Answer 16

i got .035

Answer 17

Good work! Your calculation is correct. Now calculate for r = 1, r = 2 and r = 3 as @dumbcow stated.

Answer 18

what number would that be for?

Answer 19

The sample size n = 15 for each calculation.

Answer 20

I don't understand how that helps me solve 13-15

Answer 21

is there a faster way instead of calculating each? because how does it differ from AT LEAST 3 and AT MOST 3?

Answer 22

@kropot72?

Answer 23

because I added them all up and I got 5.28....

Answer 24

@Summersnow8 Sorry, I will be tied up on another matter for about an hour. Can help again later. :)

Answer 25

@kropot72 oh okay

Answer 26

We already found some of the values of probability:
p(0 of 15 infected) = 0.0352
p(1 of 15 infected) = ?
p(2 of 15 infected) = ?
p(3 of 15 infected) = 0.25
What values did you get for p(1 of 15 infected) and p(2 of 15 infected) ?

Answer 27

12. What is the probability exactly 3 out of 15 seniors will develop influenza?
$$
\large{
\binom{15}{3}.2^3.8^{12}
}\\
=0.25013889531904
$$
http://www.wolframalpha.com/input/?i=%2815choose3%29*.2^3*.8^12

13. What is the probability that at most 3 of them will develop influenza?
$$
\large{
P(X \le 3) = P(X=0)+P(X=1)+P(X=2)+P(X=3)\\
=\binom{15}{0}.2^0.8^{15}+\binom{15}{1}.2^1.8^{14}+\binom{15}{2}.2^2.8^{13}+\binom{15}{3}.2^3.8^{12}
}\\
=0.648162104573952
$$
http://www.wolframalpha.com/input/?i=%2815choose0%29*.2^0*.8^15%2B%2815choose1%29*.2^1*.8^14%2B%2815choose2%29*.2^2*.8^13%2B%2815choose3%29*.2^3*.8^12

14. What is the probability that at least 3 of them will develop influenza?
$$
P(X\ge3)=1-P(X<3)=1-P(X\le2)\\
=1-\left [\binom{15}{0}.2^0.8^{15}+\binom{15}{1}.2^1.8^{14}+\binom{15}{2}.2^2.8^{13}\right]\\
=0.6019767907450880
$$
http://www.wolframalpha.com/input/?i=1-%28%2815choose0%29*.2^0*.8^15%2B%2815choose1%29*.2^1*.8^14%2B%2815choose2%29*.2^2*.8^13%29

Does this make sense?

Answer 28

15. The standard deviation, \(\sigma\), of the number of seniors will get influenza is:
$$
\sigma^2=npq=15\times.2\times.8=2.4\\
\sigma=\sqrt{2.4}=1.54919...
$$
http://en.wikipedia.org/wiki/Binomial_distribution#Mean_and_variance

Hope this helped.

Answer 29

@ybarrap perfect, thank you