A university's mathematics department has designed an extension program to encouragenew undergraduate students to pursue careers in mathematics. The program consists ofadditional advanced classes that focus on the various applications of mathematics indifferent industries. To determine which students will be invited to this extensionprogram, new undergraduate mathematics students are all required to take an'entrance" exam. Any student that scores higher than 80 on the exam will be invited tothe extension program. Suppose that scores in this entrance exam are normallydistributed with mean 61 and variance 256. Assume that scores are independent fromstudent to student.in a random sample of 12 new undergraduate mathematics students who took theentrance exam, find the probability that more than 2 will be invited to the extensionprogram.

First, we need to find the z-score for a score of 80 on the exam. The z-score is calculated as:

Z = (X - μ) / σ

where X is the score, μ is the mean, and σ is the standard deviation. The standard deviation is the square root of the variance, so σ = √256 = 16.

So, the z-score for a score of 80 is:

Z = (80 - 61) / 16 = 1.1875

Next, we look up this z-score in a standard normal distribution table or use a calculator to find the corresponding probability. The probability that a student scores 80 or higher (i.e., the probability that a student is invited to the extension program) is 1 - P(Z < 1.1875) ≈ 1 - 0.8820 = 0.1180.

Now, we want to find the probability that more than 2 out of 12 students are invited to the extension program. This is a binomial problem, where the number of trials is 12, the number of successes is 3 (since we want more than 2), and the probability of success on each trial is 0.1180.

The probability that exactly k students are invited to the program is given by the binomial probability formula:

P(X = k) = C(n, k) * (p^k) * ((1 - p)^(n - k))

where C(n, k) is the number of combinations of n items taken k at a time, p is the probability of success, and n is the number of trials.

We want the probability that more than 2 students are invited, so we need to calculate 1 - P(X = 0) - P(X = 1) - P(X = 2).

Using a calculator or statistical software to calculate these probabilities and subtract from 1, we find that the probability that more than 2 out of 12 students are invited to the extension program is approximately 0.891.

First, we need to find the z-score for a score of 80 on the exam. The z-score is calculated as:

Z = (X - μ) / σ

where X is the score, μ is the mean, and σ is the standard deviation. The standard deviation is the square root of the variance, so σ = √256 = 16.

So, the z-score for a score of 80 is:

Z = (80 - 61) / 16 = 1.1875

Next, we look up this z-score in a standard normal distribution table or use a calculator to find the corresponding probability. The probability that a student scores 80 or higher (i.e., the probability that a student is invited to the extension program) is 1 - P(Z < 1.1875) ≈ 1 - 0.8820 = 0.1180.

Now, we want to find the probability that more than 2 out of 12 students are invited to the extension program. This is a binomial problem, where the number of trials is 12, the number of successes is 3 (since we want more than 2), and the probability of success on each trial is 0.1180.

The probability that exactly k students are invited to the program is given by the binomial probability formula:

P(X = k) = C(n, k) * (p^k) * ((1 - p)^(n - k))

where C(n, k) is the number of combinations of n items taken k at a time, p is the probability of success, and n is the number of trials.

We want the probability that more than 2 students are invited, so we need to calculate 1 - P(X = 0) - P(X = 1) - P(X = 2).

Using a calculator or statistical software to calculate these probabilities and subtract from 1, we find that the probability that more than 2 out of 12 students are invited to the extension program is approximately 0.891.

First, we need to find the z-score for a score of 80 on the exam. The z-score is calculated as:

Z = (X - μ) / σ

where X is the score, μ is the mean, and σ is the standard deviation. The standard deviation is the square root of the variance, so σ = √256 = 16.

So, the z-score for a score of 80 is:

Z = (80 - 61) / 16 = 1.1875

Next, we look up this z-score in a standard normal distribution table or use a calculator to find the corresponding probability. The probability that a student scores 80 or higher (i.e., the probability that a student is invited to the extension program) is 1 - P(Z < 1.1875) ≈ 1 - 0.8820 = 0.1180.

Now, we want to find the probability that more than 2 out of 12 students are invited to the extension program. This is a binomial problem, where the number of trials is 12, the number of successes is 3 (since we want more than 2), and the probability of success on each trial is 0.1180.

The probability that exactly k students are invited to the program is given by the binomial probability formula:

P(X = k) = C(n, k) * (p^k) * ((1 - p)^(n - k))

where C(n, k) is the number of combinations of n items taken k at a time, p is the probability of success, and n is the number of trials.

We want the probability that more than 2 students are invited, so we need to calculate 1 - P(X = 0) - P(X = 1) - P(X = 2).

Using a calculator or statistical software to calculate these probabilities and subtract from 1, we find that the probability that more than 2 out of 12 students are invited to the extension program is approximately 0.891.