2500 students take a college entrance exam. The scores on the exam have an approximate normal distribution with mean u = 52 points and a standard deviation o = 11 points. Estimate the percentage of students scoring 52 points or more.
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How would you do this?
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. Multiply by 100.
To estimate the percentage of students scoring 52 points or more, we can use the concept of z-scores and the standard normal distribution.
First, let's calculate the z-score for 52 points using the formula:
z = (x - μ) / σ
Where:
- x is the value we want to convert to a z-score (in this case, 52),
- μ is the mean of the distribution (52),
- σ is the standard deviation of the distribution (11).
Plugging in the values, we get:
z = (52 - 52) / 11
z = 0 / 11
z = 0
Next, we need to look up the corresponding area under the standard normal distribution curve for a z-score of 0. This can be done using a standard normal distribution table or a calculator.
The percentage of students scoring 52 points or more can be estimated as the area to the right of the z-score of 0. This is equivalent to the area under the curve from the z-score of 0 to positive infinity.
Since the normal distribution is symmetrical, we know that the area to the right of the mean (z = 0) is equal to 50%. Therefore, we can estimate that approximately 50% of the students scored 52 points or more on the college entrance exam.