Probability that no samples are mutated is 0.83, probability that at most one sample is mutated is 0.9812 and probability that more than half the samples are mutated is 0.
Given percentage of rejuvenated mitochondria defective is 1%, and sample size is 18.
Binomial distribution is the probability of exactly x successes on n repeated trials and X can have two outcomes.
P(X=x)=
percentage of defective rejuvendated mitochondria=1%
p=0.01
Sample size=18
n=18
a) No samples are mutated
This means P(X=0)=
=0.83
b) At most one sample is mutated.
P(X<=1)=P(X=0)+P(X=1)
so,
P(X=0)=
=0.83
P(X=1)=
=
=0.1512
P(X<=1)=0.83+0.1512
=0.9812
c) More than half the samples are mutated.
P(X>9)=P(X=10)+P(X=11)+P(X=12)+P(X=13)+P(X=14)+P(X=15)+P(X=16)+P(X=17)+P(X=18)
Using two decimals digits precision all will be 0.
Hence Probability that no samples are mutated is 0.83, probability that at most one sample is mutated is 0.9812 and probability that more than half the samples are mutated is 0.
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Take all three trial result averages and add them up
Then take that number and divide by three because there was three numbers added.
That number is the average of the three trials
Answer:
x = 7.535 and y = 0.307
Step-by-step explanation:
Given that,
Two equations,
2x + 3y = 8 ...(1)
x – 5y = 6....(2)
Multiply equation (2) by -2.
-2x + 10y = -12....(3)
Now, adding equations (1) and (2),
2x + 3y -2x + 10y = 8-12
13y = 4
y = 0.307
Put the value of y in equation (2).
x – 5(0.307) = 6
x = 6 + 1.535
x = 7.535
So, the solutions of the given equation are x = 7.535 and y = 0.307.
Answer:
Step-by-step explanation:
All real numbers because x can assume every possible value