Question

The probability that a vehicle will change lanes while making a turn is 45%. Suppose a random sample of 7 vehicles are observed making turns at a busy intersection. Find the probability that a least 5 vehicles will change lanes while making the turn.

Answer 1

Answer:

0.1529

Step-by-step explanation:

Given that:

Probability of changing lane ; p = 45% = 0.45

Sample size (number of trials) = n = 7

Probability that atleast 5 vehicles will change lane :

Using the binomial probability formula :

P(x = x) = nCx * p^x * (1 - p)^(n-x)

(1 - p) = 1 - 0.45 = 0.55

P(x ≥ 5) = p(x = 5) + p(x = 6) + p(x = 7)

P(x = 5) = 7C5 * 0.45^5 * 0.55^2

P(x = 5) = 21 * 0.45^5 * 0.55^2 = 0.117221

P(x = 6) = 7C6 * 0.45^6 * 0.55^1

P(x = 6) = 7 * 0.45^6 * 0.55^1 = 0.031969

P(x = 7) = 7C7 * 0.45^7 * 0.55^0

P(x = 7) = 1 * 0.45^7 * 0.55^0 = 0.003736

P(x ≥ 5) = p(x = 5) + p(x = 6) + p(x = 7)

P(x ≥ 5) = 0.117221 + 0.031969 + 0.003736

P(x ≥ 5) = 0.152926

P(x ≥ 5) = 0.1529

Answer 2

Using the binomial distribution, it is found that there is a 0.1529 = 15.29% probability that a least 5 vehicles will change lanes while making the turn.

For each vehicle, there are only two possible outcomes, either they change lanes while making a turn, or they do not. The probability of a vehicle changing lanes while making a turn is independent of any other vehicle, hence, the binomial distribution is used to solve this question.

Binomial probability distribution

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$C_{n,x} = \frac{n!}{x!(n-x)!}$

The parameters are:

• x is the number of successes.

• n is the number of trials.

• p is the probability of a success on a single trial.

In this problem:

• The probability that a vehicle will change lanes while making a turn is 45%, hence $p = 0.45$.

• A sample of 7 vehicles is taken, hence $n = 7$.

The probability that at least 5 vehicles will change lanes while making the turn is:

$P(X \geq 5) = P(X = 5) + P(X = 6) + P(X = 7)$

In which

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 5) = C_{7,5}.(0.45)^{5}.(0.55)^{2} = 0.1172$

$P(X = 6) = C_{7,6}.(0.45)^{6}.(0.55)^{1} = 0.0320$

$P(X = 7) = C_{7,7}.(0.45)^{7}.(0.55)^{0} = 0.0037$

Then

$P(X \geq 5) = P(X = 5) + P(X = 6) + P(X = 7) = 0.1172 + 0.0320 + 0.0037 = 0.1529$

0.1529 = 15.29% probability that a least 5 vehicles will change lanes while making the turn.

To learn more about the binomial distribution, you can take a look at https://brainly.com/question/24863377