A large group of students has taken a college entrance exam. The scores on the exam are such that the mean = 52 and the standard deviation = 11
Find the probability that a student, selected at random, earned a score lower than 63.
Z = (score-mean)/SD
Find table in the back of your statistics text labeled something like "areas under normal distribution" to find the proportion/probability related to the Z score.
To find the probability that a student selected at random earned a score lower than 63, we can use the concept of z-scores and the standard normal distribution.
First, let's calculate the z-score for a score of 63. The z-score measures how many standard deviations a particular value is away from the mean.
The formula to calculate the z-score is z = (x - μ) / σ, where x is the individual score, μ is the mean, and σ is the standard deviation.
In this case, we have x = 63, μ = 52, and σ = 11. Plugging these values into the formula, we get:
z = (63 - 52) / 11
z = 11 / 11
z = 1
The z-score for a score of 63 is 1.
Next, we need to find the cumulative probability for a z-score of 1 by referring to the z-table or using a calculator/statistical software.
Looking up the z-score of 1 in the z-table, we find that the cumulative probability associated with a z-score of 1 is approximately 0.8413. This means that approximately 84.13% of the population has a score lower than 63.
So, the probability that a student, selected at random, earned a score lower than 63 is approximately 0.8413 or 84.13%.