Suppose that two different tests A and B are to be given to a student chosen at random from a certain population. Suppose also that the mean score on test A is 85, and the standard deviation is 10; the mean score on test B is 90, and the standard deviation is 16; the scores on the two tests have a bivariate normal distribution; and the correlation of the two scores is 0.8. If the student's score on test A is 80, what is the probability that her score on test B will be higher than 90?
Nope
To solve this problem, we will use the concept of a bivariate normal distribution and convert the problem into standard units. Let's go step by step:
Step 1: Convert the scores to z-scores
The z-score formula converts a score to a standard unit, which helps us in comparing values from different distributions with different means and standard deviations.
For Test A:
z_A = (X - mean_A) / std_A
Where X is the score on Test A, mean_A is the mean score on Test A, and std_A is the standard deviation on Test A.
Given that the mean score on Test A is 85 and the standard deviation is 10, and the student's score on Test A is 80, we can calculate z_A as follows:
z_A = (80 - 85) / 10
z_A = -0.5
Step 2: Find the corresponding z-value for the desired score on Test B
We need to find the z-value corresponding to a score of 90 on Test B.
For Test B:
z_B = (X - mean_B) / std_B
Where X is the score on Test B, mean_B is the mean score on Test B, and std_B is the standard deviation on Test B.
Given that the mean score on Test B is 90 and the standard deviation is 16, the desired score on Test B is 90, we can calculate z_B as follows:
z_B = (90 - 90) / 16
z_B = 0
Step 3: Calculate the conditional probability using the bivariate normal distribution
We need to find the probability that the score on Test B is higher than 90, given that the score on Test A is 80. This can be done by calculating the conditional probability P(B > 90 | A = 80).
We can use the joint probability density function (PDF) of the bivariate normal distribution to calculate this probability. However, it involves complex calculations and may require specialized software. One way to calculate this probability is by utilizing a table or statistical software.
Alternatively, we can use an approximation by assuming that the joint distribution of Test A and Test B follows a bivariate normal distribution. The means, standard deviations, and correlation coefficients are given, which allows us to estimate the desired probability.
Using the given correlation of 0.8, we can estimate the conditional probability by transforming the joint distribution into standardized units and finding the probability from the standard normal distribution table.
Step 4: Find the transformed score on Test B in standardized units
To find the transformed score on Test B in standardized units, we use the formula:
z_B_trans = ρ * z_A + sqrt(1 - ρ^2) * z_B
Where ρ is the correlation between Test A and Test B, z_A is the z-score of Test A, and z_B is the z-score of Test B.
Using the given correlation of 0.8, z_A as -0.5, and z_B as 0, we can calculate the transformed z-score on Test B:
z_B_trans = 0.8 * (-0.5) + sqrt(1 - 0.8^2) * 0
z_B_trans = -0.4
Step 5: Find the probability that z_B_trans is greater than 0
The transformed z_B_trans follows a standard normal distribution. To find the probability that z_B_trans is greater than 0, we need to calculate P(z > 0).
Looking up the z-value of 0 from a standard normal distribution table or using statistical software, we find that the corresponding probability is approximately 0.5.
Therefore, the probability that the student's score on Test B will be higher than 90, given that her score on Test A is 80, is approximately 0.5.
Note: This approach assumes that the bivariate distribution is a good approximation for the given problem.