If 30% of people in an area registered voters and 15 people are selected at ramdom from this area;
a. What is the probability thatexactly 3out of the15 people are registered voters?
b. What is the probability that at most 3 out of the 15 people are registered voters>
c. could we have used the normal approximation in the binomial in part b? Explain why or why not.
You can use the binomial distribution:
P = nCs*p^s*(1-p)^n(-s) where P is the probability we are hanting for, n = 15 is the size of a sample, s is the number of registered voters of 15 people, and p = 0.3 (30%) is the proportion of registered voters in the area.
a) P(3 of 15) = 15!/(15-3)!3!*(0.3)^3*(1 -0.3)^(15 -3) = 455*0.027*0.01384 = 0.17 or 17%;
b) to get results for "at most 3 of the 15 people", you have to apply the above approach four times:
P(0 of 15) =
P(1 of 15) =
P(2 of 15) =
P(3 of 15) =
then add these four probabilities together => P(at most 3 of 15);
c) I believe that we cannot because the sample of 15 people is too small (the critical size ia about 30 items).
thank you....!!!
To calculate the probability in this scenario, we will use the binomial probability formula. The binomial probability formula is given by:
P(X = k) = (n C k) * p^k * (1 - p)^(n - k)
Where:
- P(X = k) represents the probability of exactly k successes,
- (n C k) is the binomial coefficient, which is calculated as n! / (k! * (n - k)!),
- p is the probability of success in a single trial,
- n is the total number of trials.
a. To find the probability that exactly 3 out of the 15 people are registered voters, we plug in the known values into the binomial probability formula:
P(X = 3) = (15 C 3) * (0.3)^3 * (1 - 0.3)^(15 - 3)
b. To find the probability that at most 3 out of the 15 people are registered voters, we need to calculate the probabilities for the cases of 0, 1, 2, and 3 registered voters and sum them up:
P(X ≤ 3) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3)
c. In this scenario, the number of trials (n) is 15, which is reasonably large. The general rule for using the normal approximation for a binomial distribution is that both np (mean) and n(1-p) (variance) should be greater than or equal to 5. In this case, np = 15 * 0.3 = 4.5 and n(1-p) = 15 * 0.7 = 10.5. Since np < 5, we cannot use the normal approximation for part b.
Therefore, for part b, we need to use the binomial probability formula to calculate the probability.