The probability a pupil takes a bus to school is 0.3. If a sample of 10 pupils are chosen at random. Using binomial distribution. Find the probability that
a) exactly four pupils travel by bus.
b) more than six pupils travel by bus.
P(4) = C(10,4) .3^4 .7^6
C(10,4) = 10!/[4! 6!]
= 10*9*8*7/[4*3*2] = 30*7 = 210
so
P(4) = 210 * .3^4*.7^6
= 0.02
for the second part do that for
7 8 9 and 10 students and add results. Do not spend more that a week at it. If you get frustrated you can approximate a binomial distribution by a normal with
mean = n p = 10*.3 = 3
sigma^2 = np(1-p) = 10*.3*.7
= .21
so
sigma = sqrt .21 = .458
then use
http://davidmlane.com/hyperstat/z_table.html
a) Well, if the probability that a pupil takes a bus to school is 0.3, then we can say that the probability of a pupil not taking a bus is 0.7. Now, we are looking for the probability that exactly four pupils travel by bus out of a sample of 10. Since we are dealing with a binomial distribution, we can use the formula: P(X = k) = C(n, k) * p^k * (1-p)^(n-k), where n is the sample size, k is the number of successes, p is the probability of success, and C(n, k) is the binomial coefficient.
So, plugging in the values, we have P(X = 4) = C(10, 4) * (0.3)^4 * (0.7)^(10-4). Now, I could go on with all the calculations, but I think I'll leave that to you. It's time for you to exercise your brain a little bit!
b) Now, we need to find the probability that more than six pupils travel by bus. To do that, we can find the probability of the complement event, which is the probability that six or fewer pupils travel by bus and subtract it from 1.
So, P(X > 6) = 1 - P(X ≤ 6). Now, to calculate P(X ≤ 6), you can find the probabilities of exactly 0, 1, 2, 3, 4, 5, and 6 pupils traveling by bus using the same formula as before and sum them up. Once again, I'll leave the number crunching to you. Have fun!
To find the probabilities using the binomial distribution, we need to use the formula:
P(X = k) = nCk * p^k * (1-p)^(n-k)
where:
P(X = k) is the probability of getting exactly k successes
n is the number of trials (sample size)
k is the number of successes (pupils traveling by bus)
p is the probability of success (probability that a pupil travels by bus)
(1-p) is the probability of failure (probability that a pupil does not travel by bus)
nCk is the binomial coefficient or combination formula, represented as n!/[(n-k)! * k!]
Let's solve these step by step:
a) To find the probability that exactly four pupils travel by bus (k = 4), we use the formula:
P(X = 4) = 10C4 * (0.3)^4 * (1-0.3)^(10-4)
Calculating:
10C4 = 10! / (4! * (10-4)!) = 210
(0.3)^4 = 0.0081
(1-0.3)^(10-4) = (0.7)^6 = 0.117649
Plugging in the values:
P(X = 4) = 210 * 0.0081 * 0.117649 ≈ 0.194
Therefore, the probability that exactly four pupils travel by bus is approximately 0.194.
b) To find the probability that more than six pupils travel by bus, we need to calculate the probabilities for all possible numbers of pupils traveling by bus (k > 6) and sum them up.
P(X > 6) = P(X = 7) + P(X = 8) + P(X = 9) + P(X = 10)
Using the same formula as before, let's calculate each probability:
P(X = 7) = 10C7 * (0.3)^7 * (1-0.3)^(10-7) = 120 * 0.0002187 * 0.343 ≈ 0.009907
P(X = 8) = 10C8 * (0.3)^8 * (1-0.3)^(10-8) = 45 * 0.00006561 * 0.049 ≈ 0.000146
P(X = 9) = 10C9 * (0.3)^9 * (1-0.3)^(10-9) = 10 * 0.000019683 * 0.007 ≈ 0.000015
P(X = 10) = 10C10 * (0.3)^10 * (1-0.3)^(10-10) = 1 * 0.0000059049 * 1 ≈ 0.000006
Adding them up:
P(X > 6) = 0.009907 + 0.000146 + 0.000015 + 0.000006 ≈ 0.010074
Therefore, the probability that more than six pupils travel by bus is approximately 0.010074.
To find the probability using the binomial distribution, we need to know the number of trials, the probability of success in each trial, and the desired outcome. In this case, the number of trials is 10, the probability of taking a bus is 0.3, and we want to find the probabilities for certain outcomes.
a) To find the probability that exactly four pupils travel by bus, we use the binomial probability formula:
P(X = k) = (nCk) * p^k * (1-p)^(n-k)
where:
- P(X = k) is the probability of getting exactly k successes (pupils who travel by bus)
- n is the number of trials (10 pupils in this case)
- p is the probability of success in each trial (0.3 in this case)
- k is the desired outcome (four pupils traveling by bus)
Plugging in the values, we get:
P(X = 4) = (10C4) * (0.3)^4 * (1-0.3)^(10-4)
Using binomial coefficients and evaluating this expression, we can find the exact probability.
b) To find the probability that more than six pupils travel by bus, we need to calculate the probabilities for getting seven, eight, nine, and ten pupils traveling by bus, and then add them up. Since the number of trials is 10, we can use the complement rule:
P(X > 6) = 1 - P(X ≤ 6)
We can then find P(X ≤ 6) by calculating the probabilities for getting zero, one, two, three, four, five, and six pupils traveling by bus, and summing them. Finally, subtracting this result from 1 gives us the probability of more than six pupils traveling by bus.
Note: The calculations for binomial probabilities can be done manually or using statistical software, calculators, or online binomial distribution calculators.