95% specificity means that of the people who do not have strep, 95% of them will accurately test negative. This means that 5% of the people who do not have strep will test positive for it; these are called false positives.
Let’s consider a random sample of 30 people who do not have strep and are tested with this rapid strep test. Let Y be the number of people who receive false positives. As in Question 2, assume we are sampling from a population that is large enough that we can consider the trials to be independent.
Part A: What are n and p for this new binomial experiment?
n=30 (this is the answer)
p=0.05 (this is the answer)
Part B: What is the probability that at most 2 people receive false positives?
P(Y≤2)= 0.812 (this is the answer)
Part C: What is the probability that none of the 30 people receive false positives?
P(Y=0) = Answer this for me
Part D: Would you be surprised if 5 or more people receive false positives? Use statistical evidence to explain your answer.
P(Y≥5)= Answer this for me
Yes, I would be surprised since this happens less than 2% of the time.
To find the probability that none of the 30 people receive false positives, we can use the formula for the binomial probability:
P(Y=0) = C(n,0) * p^0 * (1-p)^(n-0)
where C(n,0) represents the number of ways to choose 0 false positives out of 30 people, and p is the probability of a false positive.
Using the formula for combinations, C(n,0) = 1, since there is only 1 way to choose 0 false positives out of 30 people.
Plugging in the values, we get:
P(Y=0) = 1 * 0.05^0 * (1-0.05)^(30-0) = 0.697 (rounded to three decimal places)
So the probability that none of the 30 people receive false positives is 0.697.
To find the probability that 5 or more people receive false positives, we can add up the probabilities of receiving 5, 6, 7, ... , 30 false positives.
P(Y≥5) = P(Y=5) + P(Y=6) + ... + P(Y=30)
Using the binomial probability formula again, we can calculate each individual probability and sum them up. However, this would be quite time-consuming and tedious.
Alternatively, we can use statistical evidence to estimate the probability. In this case, we can use the normal approximation to the binomial distribution, since the sample size is large (n=30) and the success probability is not too close to 0 or 1.
We can calculate the mean and standard deviation of the binomial distribution:
Mean (μ) = n * p = 30 * 0.05 = 1.5
Standard deviation (σ) = sqrt(n * p * (1-p)) = sqrt(30 * 0.05 * 0.95) ≈ 1.24
Then, we can use the normal distribution to estimate the probability:
P(Y≥5) = 1 - P(Y≤4)
Using the mean and standard deviation calculated above, we can standardize the values 4 and 5:
z1 = (4 - μ) / σ
z2 = (5 - μ) / σ
Next, we can use a z-table or a calculator to find the corresponding probabilities for z1 and z2:
P(Y≥5) = 1 - P(Y≤4) = 1 - P(Z≤z1) ≈ 1 - P(Z≤z2)
To find the probability P(Z≤z), you can use a standard normal distribution table or a calculator. Let's assume we find:
P(Z≤z1) ≈ 0.158
P(Z≤z2) ≈ 0.202
Substituting these values into the equation:
P(Y≥5) ≈ 1 - P(Z≤z2) = 1 - 0.202 ≈ 0.798
So the probability that 5 or more people receive false positives is approximately 0.798.
Based on this evidence, it would be surprising if 5 or more people receive false positives, as this happens less than 2% of the time (0.798 < 0.02).