Scores on an examination are assumed to be normally distributed with mean µ = 78and variance σ2 = 36.(a) [1 mark] Find the probability that a person taking the examination scores higherthan 72
Question
Scores on an examination are assumed to be normally distributed with mean µ = 78and variance σ2 = 36.(a) [1 mark] Find the probability that a person taking the examination scores higherthan 72
Solution
To find the probability that a person scores higher than 72, we first need to convert the score of 72 to a z-score. The z-score is a measure of how many standard deviations an element is from the mean.
The formula to calculate the z-score is:
Z = (X - µ) / σ
Where: X = raw score µ = mean σ = standard deviation
Given: µ = 78 σ^2 = 36 (Therefore, σ = sqrt(36) = 6) X = 72
Substituting these values into the formula, we get:
Z = (72 - 78) / 6 = -1
This means that a score of 72 is 1 standard deviation below the mean.
Next, we look up this z-score in the z-table, which gives the area to the left under the normal curve. The value for -1 is 0.1587. However, we want the probability that a person scores higher than 72, so we need to find the area to the right under the curve.
Since the total area under the curve is 1, the area to the right is:
1 - 0.1587 = 0.8413
So, the probability that a person taking the examination scores higher than 72 is 0.8413 or 84.13%.
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