the average score for a business examination comprising 300 students is 54%. if the marks are normally distributed with a standard deviation of 8%, determine the number of students scores above 70%
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To determine the number of students who scored above 70%, we need to calculate the z-score for 70% and find the corresponding area under the normal distribution curve.
1. Find the z-score:
The z-score is a measure of how many standard deviations an individual's score is from the mean. It is calculated using the formula: z = (x - μ) / σ, where x is the individual score, μ is the mean, and σ is the standard deviation.
In this case, x = 70%, μ = 54%, and σ = 8%. So the z-score is:
z = (0.7 - 0.54) / 0.08 = 2.
2. Find the area under the normal distribution curve:
The area under the curve represents the probability. We can use a standard normal distribution table or a calculator to find the area corresponding to the z-score of 2. For simplicity, let's use a standard normal distribution table.
Looking up the z-score of 2 in the table, we find that the area to the left of z = 2 is approximately 0.9772.
3. Find the area above the z-score of 2:
Since we want to find the percentage above 70%, we subtract the area to the left of z = 2 from 1 (total area under the curve), which gives us: 1 - 0.9772 = 0.0228.
4. Determine the number of students:
Finally, we multiply the percentage above the z-score of 2 by the total number of students to find the number of students who scored above 70%:
Number of students = 0.0228 * 300 ≈ 6.8.
Therefore, approximately 6.8 students (rounded to the nearest whole number) scored above 70% on the business examination.