a test has a mean score 73 with standard deviation of 12.students with scores greater than 85 receive an A. What is the percent of the students would we expect to get an A? assume the normal distribution
Z = (score-mean)/SD
Find table in the back of your statistics text labeled something like "areas under normal distribution" to find the proportion related to the Z score.
To answer this question, we need to use the concept of the z-score and the normal distribution.
The z-score measures how many standard deviations a given value is from the mean of a distribution. It allows us to find the probability (or percent) of a value occurring in a normal distribution.
First, let's calculate the z-score for the cutoff value of 85. We can use the formula:
z = (X - μ) / σ
where:
X = 85 (cutoff value)
μ = 73 (mean)
σ = 12 (standard deviation)
z = (85 - 73) / 12
z = 12 / 12
z = 1
Next, we need to find the probability (P) of a score being greater than 85, which is equivalent to finding the area under the curve to the right of the z-score of 1. We can refer to a standard normal distribution table or use a calculator with a cumulative distribution function (CDF) to find this probability.
Using a standard normal distribution table or calculator, the probability (P) of a z-score being less than 1 is approximately 0.8413. Therefore, the probability of a z-score being greater than 1 (the area to the right) is:
P = 1 - 0.8413
P ≈ 0.1587
Finally, we convert the probability to a percentage by multiplying it by 100:
Percentage ≈ 0.1587 * 100
Percentage ≈ 15.87%
Hence, we would expect approximately 15.87% of the students to receive an A in this test based on the given mean score of 73, a standard deviation of 12, and the cutoff score of 85.