Diana took a multiple-choice test of 50 questions and each question carried two marks.

The probability that Diana chose the correct answer for a question was three out of four.
The marks for grade A, grade B and grade C were 80, 72 and 64, respectively. After the
test, Diana was told by her tutor that she did not get grade A. Using normal
approximation of the binomial distribution, find the probability that she got grade C?

To find the probability that Diana got grade C, we need to use the normal approximation of the binomial distribution.

In Diana's case, she attempted 50 questions, and the probability of her choosing the correct answer for a question is given as three out of four.

The mean, or the expected number of correct answers, can be calculated as the product of the number of questions and the probability of success. In this case, the mean is 50 * (3/4) = 37.5.

The variance of a binomial distribution is given by the product of the number of trials, the probability of success, and the probability of failure. In this case, the variance is 50 * (3/4) * (1/4) = 9.375.

Using normal approximation, we can use the mean and variance to calculate the z-score for Diana's score. The z-score is calculated as follows:

z = (x - mean) / sqrt(variance)

In this case, we want to find the probability of getting a grade C, which corresponds to a score of 64. Therefore, x = 64.

Substituting the known values into the formula, we get:

z = (64 - 37.5) / sqrt(9.375) = 26.5 / 3.06 ≈ 8.66

Next, we will use a standard normal distribution table or a calculator to find the probability associated with this z-score.

Since we are looking for the probability of getting grade C, which is less than 64, we will look up the cumulative probability to the left of 8.66.

Consulting the table or using a calculator, we find that the cumulative probability for a z-score of 8.66 is nearly 1.

Therefore, the probability that Diana got grade C is approximately 1 or 100%.