1. ## Probability question help

Q#1. The probability that a patient recovers from a rare blood disease is 0.4. If 15 people are known to have contracted this disease, what is the probability that:

(a) at least 10 survive
(b) from 3 to 8 survive
(c) exactly 5 survive

Q#2. Given a normal distribution with population Mean u=40 and , Standard Deviation= 6 find
(a) the area below 32
(b) the area above 27
(c) the area between 42 and 51
(d) the x value that has 45% of the area below it
(e) the x value that has 13% of the area above it

2. Originally Posted by nisoo-1
Q#1. The probability that a patient recovers from a rare blood disease is 0.4. If 15 people are known to have contracted this disease, what is the probability that:

(a) at least 10 survive
(b) from 3 to 8 survive
(c) exactly 5 survive

Q#2. Given a normal distribution with population Mean u=40 and , Standard Deviation= 6 find
(a) the area below 32
(b) the area above 27
(c) the area between 42 and 51
(d) the x value that has 45% of the area below it
(e) the x value that has 13% of the area above it

This is my 22th post!!!!!

3. Originally Posted by nisoo-1
Q#1. The probability that a patient recovers from a rare blood disease is 0.4. If 15 people are known to have contracted this disease, what is the probability that:
Using the Hackerian Gamma Approximation techiqnue
That $\sum_{k=a}^b {n\choose k} p^k (1-p)^{n-k} \approx \int_a^b \frac{n!}{\Gamma (n-t) \Gamma (t-1)} p^t (1-p)^{n-t} dt$
(a) at least 10 survive
$\sum_{k=10}^{15} {15\choose k}(.4)^k (.6)^{15-k} = .0195$
(b) from 3 to 8 survive
$\sum_{3}^8 {15\choose k} (.4)^k (.6)^{15-k} = .797$
(c) exactly 5 survive
This is easy,
${15\choose 5}(.4)^ 5 (.6)^{10}$

Note: These problems were done by numerical integration.

4. Originally Posted by ThePerfectHacker
Using the Hackerian Gamma Approximation techiqnue
That $\sum_{k=a}^b {n\choose k} p^k (1-p)^{n-k} \approx \int_a^b \frac{n!}{\Gamma (n-t) \Gamma (t-1)} p^t (1-p)^{n-t} dt$

$\sum_{k=10}^{15} {15\choose k}(.4)^k (.6)^{15-k} = .0195$

$\sum_{3}^8 {15\choose k} (.4)^k (.6)^{15-k} = .797$

This is easy,
${15\choose 5}(.4)^ 5 (.6)^{10}$

Note: These problems were done by numerical integration.
Which in this case is not terribly accurate:

$\sum_{k=10}^{15} {15\choose k}(.4)^k (.6)^{15-k} = .0175$

$\sum_{3}^8 {15\choose k} (.4)^k (.6)^{15-k} = .878$

RonL

5. Originally Posted by CaptainBlank
Which in this case is not terribly accurate:

$\sum_{k=10}^{15} {15\choose k}(.4)^k (.6)^{15-k} = .0175$

$\sum_{3}^8 {15\choose k} (.4)^k (.6)^{15-k} = .878$

RonL
I think I maybe made a mistake in the algorithm programmed in the computer. The other times I used it, it worked well.