The probability that a patient recovers from a disease is 0.3. If 6 people are known to have contacted the disease, what is the probability that at least 4 survive? ​

0.07047

Step-by-step explanation:

at least 4 people survive is

4 people survive and 2 don't +

5 people survive and 1 doesn't +

all 6 people survive

the probability of 4 people to survive is the product of the individual probabilities :

0.3×0.3×0.3×0.3 = 0.3⁴ = 0.0081

times the probability 2 don't survive

0.7×0.7 = 0.49

0.0081×0.49 = 0.003969

now, the chance that 4 out of 6 survive is that probabilty times how many times we can select 4 out of 6.

to select 4 out of 6 is

= 6! / (4! × (6-4)!) = 6! / (4! × 2!) = 6×5/2 = 30/2 = 15

so, the probability of having 4 survivors is

15×0.003969 = 0.059535

the probability of 5 people surviving is 0.3⁵ times one not

0.3⁵×0.7 = 0.001701

we have 6 over 5 combinations to pick 5 out of 6 = 6.

so, in total for 5 survivors we get

6×0.001701 = 0.010206

and the probabilty of all 6 surviving is

0.3⁶ = 0.000729

so, the probability of at least 4 out of 6 surviving is

0.059535 +

0.010206 +

0.000729

---------------

0.070470

irst, find the z score of a score of 80. this is given by the formula z = (x-u)/o, where x is the score, u is the mean, and o is the standard deviation. this gives

z = (80-78)/12 = 1/6 = .1667

now, find what percentile a z score of 1/6 correlates to in the normal distribution.

a z score of 1/6 = .1667 is equivalent to the 56.36th percentile of the distribution. so the probability a score is over this is 1-.5636 = .4364

the mean of a sample mean is equal to the original mean. however, the standard deviation of a sample mean is equal to the original standard deviation divided by the square root of n (sample size). so the new standard deviation is 12/sqrt(16) = 3.

the new z score is 80-78/3 = 2/3 = .667. this is equivalent to the 74.86th percentile. so the probability that the mean will be greater than this score is 1-.7486 = .2514.