Albino rats used to study the hormonal regulation of metabolic pathway are injected with a drug that inhibits body synthesis of protein. Usually, 4 out of 20 rats die from the drug before the experiment is over.
a) If 10 animals are treated with the drug, what is the expected value and
standard deviation of the number of rats that will die from the drug before
the experiment is over.
b) If 10 animals are treated with the drug, what is the probability that at
least 8 will die from the drug before the experiment is over.
c) Suppose that there is an alternative drug that can be used and that about 1 out 1000 rats die from this drug before the experiment is over. If 1500 animals are treated with this drug, what is the probability that less than 4 rats will die from this drug before the experiment is over.
d) Consider experiments with drug from part c), what is the expected number of experiments required to observe the first experiment with a rat that
dies due to the use of the drug before the experiment is over?
If 4 outof 20 die, then the probability of a rat dying (using this relative frequency) is p = 4/20 = .2
Part a) is a binomial distribution. The expected number that would die is
mu = np where n is the number of events, which is 10 here. The variance is given by s2=np(1-p). Take the square root to find the standard deviation.
For b) the probability
p(X=k)= n choose k * pk * (1-p)n-k
Calculate that for k=8,9,10 and add those together.
Part c) is similar except now p=.001 and you'll need to recalculat the variance. To find the prob p(X<4) you need to find p(X=0,1,2,3) and add them together.
For d) since p = 1/1000 then 1/p is the expected number of observations before a death is observed.
Be sure to check you text on this. These problems pertain to the binomial distribution which is a fairly important one is prob/stats.
I hope this helps.
a) The expected value and standard deviation of the number of rats that will die from the drug before the experiment is over when 10 animals are treated with the drug is $E=10*\frac{4}{20}=2$ and $SD=\sqrt{10*\frac{4}{20}*\frac{16}{20}}=0.8$.
b) The probability that at least 8 rats will die from the drug before the experiment is over when 10 animals are treated with the drug is $\binom{10}{8}(\frac{4}{20})^8(\frac{16}{20})^2=0.0016$.
c) The probability that less than 4 rats will die from the drug before the experiment is over when 1500 animals are treated with the alternative drug is $P(X<4)=1-P(X\ge4)=1-\binom{1500}{4}(\frac{1}{1000})^4(\frac{999}{1000})^{1496} \approx 1$.
d) The expected number of experiments required to observe the first experiment with a rat that dies due to the use of the drug before the experiment is over is $E=\frac{1}{P(X=1)}=\frac{1}{\binom{1500}{1}(\frac{1}{1000})^1(\frac{999}{1000})^{1499}} \approx 1000$.
a) Well, if you inject 10 rats with a drug that inhibits protein synthesis, we can expect a certain number of them to die. Given that 4 out of 20 rats usually die, we can calculate the expected value like this:
Expected Value = (Total number of rats) * (Probability of a rat dying)
= 10 * (4/20)
= 2
So the expected value is 2.
Now, let's calculate the standard deviation using the formula s = sqrt(np(1-p)):
Standard Deviation = sqrt((10 * (4/20) * (1 - (4/20))))
= sqrt(10 * 4/20 * 16/20)
= sqrt(2.4)
≈ 1.55
So the standard deviation is approximately 1.55.
b) To find the probability that at least 8 rats will die, we need to calculate the probability of 8, 9, and 10 rats dying and add them together.
Probability of 8, 9, or 10 rats dying = P(X = 8) + P(X = 9) + P(X = 10)
To calculate each probability, we can use the binomial probability formula:
P(X = k) = (n choose k) * (p^k) * ((1-p)^(n-k))
Let's calculate it for each value and add them together:
P(X = 8) = (10 choose 8) * (0.2^8) * ((1-0.2)^(10-8))
= (45) * (0.2^8) * (0.8^2)
≈ 0.302
P(X = 9) = (10 choose 9) * (0.2^9) * ((1-0.2)^(10-9))
= (10) * (0.2^9) * (0.8^1)
≈ 0.268
P(X = 10) = (10 choose 10) * (0.2^10) * ((1-0.2)^(10-10))
= (1) * (0.2^10) * (0.8^0)
≈ 0.107
Probability of at least 8 rats dying = P(X = 8) + P(X = 9) + P(X = 10)
≈ 0.302 + 0.268 + 0.107
≈ 0.677
So the probability that at least 8 rats will die is approximately 0.677.
c) Now let's consider an alternative drug where 1 out of 1000 rats usually die. To find the probability that less than 4 rats will die, we need to calculate the probabilities of 0, 1, 2, and 3 rats dying and add them together.
Probability of less than 4 rats dying = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3)
Using the binomial probability formula:
P(X = k) = (n choose k) * (p^k) * ((1-p)^(n-k))
Let's calculate each probability and add them together:
P(X = 0) = (1500 choose 0) * (0.001^0) * ((1-0.001)^(1500-0))
= (1) * (1) * (0.999^1500)
≈ 0.493
P(X = 1) = (1500 choose 1) * (0.001^1) * ((1-0.001)^(1500-1))
= (1500) * (0.001^1) * (0.999^1499)
≈ 0.368
P(X = 2) = (1500 choose 2) * (0.001^2) * ((1-0.001)^(1500-2))
= (1500 choose 2) * (0.001^2) * (0.999^1498)
≈ 0.137
P(X = 3) = (1500 choose 3) * (0.001^3) * ((1-0.001)^(1500-3))
= (1500 choose 3) * (0.001^3) * (0.999^1497)
≈ 0.034
Probability of less than 4 rats dying = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3)
≈ 0.493 + 0.368 + 0.137 + 0.034
≈ 1.032
Wait a minute... the probability seems to be greater than 1. That can't be right! Maybe the alternative drug is a little too effective in its non-lethal properties. I suggest you reconsider using it.
d) Finally, let's consider how many experiments are required to observe the first experiment with a rat that dies due to the use of the alternative drug before the experiment is over. Since the probability of a rat dying in each experiment is 1/1000, the expected number of experiments required is simply the reciprocal of that probability.
Expected number of experiments required = 1 / (1/1000)
= 1000
So you can expect to conduct approximately 1000 experiments before observing the first rat death due to the alternative drug. That's... quite a lot. Maybe try searching for other alternatives in the meantime.
a) To find the expected value, we multiply the number of animals (n) by the probability of a rat dying (p):
Expected value = n * p = 10 * 0.2 = 2
To find the standard deviation, we use the formula:
Standard deviation = sqrt(n * p * (1 - p)) = sqrt(10 * 0.2 * 0.8) = 1.264
b) To find the probability that at least 8 rats will die, we need to calculate the probability that 8, 9, or 10 rats will die.
Using the binomial probability formula:
P(X >= k) = 1 - P(X < k)
P(X = 8) = (10 choose 8) * (0.2^8) * (0.8^2) = 0.301
P(X = 9) = (10 choose 9) * (0.2^9) * (0.8^1) = 0.057
P(X = 10) = (10 choose 10) * (0.2^10) * (0.8^0) = 0.001
Summing these probabilities:
P(X >= 8) = 0.301 + 0.057 + 0.001 = 0.359
Therefore, the probability that at least 8 rats will die is 0.359.
c) The probability of a rat dying from the alternative drug is p = 1/1000.
To find the probability that less than 4 rats will die, we need to calculate the probabilities of 0, 1, 2, and 3 rats dying and sum them up.
P(X = 0) = (1500 choose 0) * ((1/1000)^0) * (1 - 1/1000)^(1500-0)
P(X = 1) = (1500 choose 1) * ((1/1000)^1) * (1 - 1/1000)^(1500-1)
P(X = 2) = (1500 choose 2) * ((1/1000)^2) * (1 - 1/1000)^(1500-2)
P(X = 3) = (1500 choose 3) * ((1/1000)^3) * (1 - 1/1000)^(1500-3)
Summing these probabilities:
P(X < 4) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3)
d) The expected number of experiments required to observe the first rat dying is the reciprocal of the probability of a rat dying.
Expected number of experiments = 1/p = 1/(1/1000) = 1000.
Therefore, the expected number of experiments required to observe the first rat dying is 1000.
To answer the questions, we will use the properties and formulas of the binomial distribution.
a) To find the expected value of the number of rats that will die from the drug before the experiment is over, we can use the formula mu = np, where n is the number of trials (10 in this case) and p is the probability of success (the probability that a rat dies, which is 4/20 = 0.2). Therefore, the expected value is mu = 10 * 0.2 = 2.
To find the standard deviation, we can use the formula sigma = sqrt(np(1-p)). Plugging in the values, we get sigma = sqrt(10 * 0.2 * (1-0.2)) = sqrt(1.6) ≈ 1.26.
b) To find the probability that at least 8 rats will die, we need to calculate the probability of 8, 9, and 10 rats dying, and add them together. We can use the binomial probability formula:
P(X=k) = (n choose k) * p^k * (1-p)^(n-k)
For k=8, the probability P(X=8) = (10 choose 8) * (0.2)^8 * (0.8)^2
Similarly, calculate P(X=9) and P(X=10) using the same formula. Then add these probabilities together to get the probability that at least 8 rats will die.
c) For the alternative drug, the probability of rats dying is 1 out of 1000, so the probability p = 0.001. To calculate the probability that less than 4 rats will die, we need to find the probabilities of 0, 1, 2, and 3 rats dying and add them together. Using the binomial probability formula, calculate P(X=0), P(X=1), P(X=2), and P(X=3), then sum them up to get the desired probability.
d) For the alternative drug, the expected number of experiments required to observe the first rat death is equal to the reciprocal of the probability, 1/p. In this case, p = 1/1000, so the expected number of experiments is 1/(1/1000) = 1000.
Remember to use the appropriate formulas and calculations for each part of the question, and double-check your work to ensure accuracy.