It is known that an exam scores of students in STAT 2507 follow a normal distribution with a mean

of 70% and a standard deviation of 9%.
(A) If a student must obtain a mark of 50% to pass the exam, what proportion of students fail the exam?
(B) What is the probability that a randomly selected student will score between 65% and 85%?
(C) What is the minimum score required for a student to be in the top 10 percent of the class?
(D) What is the probability that a student will get exactly 70%?

Familiarize yourself with a normal distribution table or online calculator, such as the one at

http://stattrek.com/Tables/Normal.aspx

(A) 1.3%
(B) Subtract P(x<65) from P(x<85)
(C) Find the value of X for which P(x<X) = 0.90 I get 81.5%
(D) Take the difference between getting a score of 70.5% and 69.5%

To answer these questions, we will be using the normal distribution. The normal distribution is a continuous probability distribution that is symmetric and bell-shaped. It is characterized by two parameters: the mean (µ) and the standard deviation (σ).

In this case, we are given that the scores of students in STAT 2507 follow a normal distribution with a mean of 70% and a standard deviation of 9%. Let's solve each question step by step:

(A) Proportion of students who fail the exam:
To find the proportion of students who fail the exam, we need to determine the area under the normal distribution curve to the left of the passing mark (50%).

To calculate this, we can standardize the passing mark using the z-score formula:

z = (x - µ) / σ

Where x is the passing mark, µ is the mean, and σ is the standard deviation.

Substituting the given values, we get:

z = (50 - 70) / 9 = -2.22

Using a z-table or a calculator, we can find the proportion of scores to the left of z = -2.22, which represents the proportion of students who fail the exam.

(B) Probability of scoring between 65% and 85%:
Similar to question (A), we need to standardize the given scores using the z-score formula:

z1 = (65 - 70) / 9 = -0.56
z2 = (85 - 70) / 9 = 1.67

We then find the area under the normal distribution curve between these two z-scores. This represents the probability of a randomly selected student scoring between 65% and 85%.

(C) Minimum score required to be in the top 10% of the class:
To find the minimum score, we need to determine the score corresponding to the z-score that leaves 90% of the distribution to the left.

Using a z-table or a calculator, we find the z-score corresponding to the top 10%:

z = 1.28 (approximately)

We can then use the z-score formula to find the score:

x = z * σ + µ

Substituting the values, we get:

x = 1.28 * 9 + 70 = 81.52 (approximately)

Therefore, a student needs a score of approximately 81.52% to be in the top 10% of the class.

(D) Probability of getting exactly 70%:
Since the normal distribution is continuous, the probability of getting exactly 70% is zero. Instead, we can consider the probability of getting a range around 70%, such as between 69.5% and 70.5%.

To find this probability, we can use the z-scores:

z1 = (69.5 - 70) / 9
z2 = (70.5 - 70) / 9

Then, we find the area under the normal distribution curve between these two z-scores to get the probability of getting a score between 69.5% and 70.5%.

Note: When using z-tables or calculators, you may need to use the properties of the normal distribution, such as symmetry, to find the desired probabilities.