The mean scale score on the mathematics examinatin of ninth graders in Texas was 86 with a standard deviation of 5 points. Given a student was selected at random what is the probability that his score was 91 or higher
To solve this problem, we need to find the probability that a student's score was 91 or higher on the mathematics examination given the mean and standard deviation.
We can use the concept of Z-scores to find the probability. A Z-score is a measure of how many standard deviations a particular data point is away from the mean.
First, we calculate the Z-score for 91:
Z = (x - μ) / σ
Where:
x = score (91)
μ = mean (86)
σ = standard deviation (5)
Plugging in the values, we get:
Z = (91 - 86) / 5
Z = 1
Now, we need to look up the area to the right of this Z-score in the standard normal distribution table. The standard normal distribution table provides the area under the curve to the left of a particular Z-score.
Looking up the Z-score of 1 in the standard normal distribution table, the area to the left is 0.8413. This means that about 84.13% of the data falls below a Z-score of 1.
However, we are interested in finding the probability of a score being 91 or higher. Since the area to the left of the Z-score is given, we need to subtract this value from 1 to get the area to the right.
P(score ≥ 91) = 1 - 0.8413
P(score ≥ 91) = 0.1587
Therefore, the probability that a randomly selected student's score was 91 or higher on the mathematics examination is approximately 0.1587 or 15.87%.