Scores on a marketing exam are known to be normally distribute with mean and standard deviation of 60 and 20, respectively. The syllabus suggests that the top 15% of the students will get an A in the course. What is the minimum score required to get an A?
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To find the minimum score required to get an A in the course, we need to determine the z-score corresponding to the top 15% of the distribution and then convert it back to the original score.
Step 1: Find the z-score
The cumulative proportion to the left of a z-score in a standard normal distribution table represents the percentage of the distribution. Since we want the top 15%, we need to find the z-score that corresponds to the cumulative proportion of 1 - 0.15 = 0.85.
Using a standard normal distribution table or a calculator, we find that the z-score corresponding to a cumulative proportion of 0.85 is approximately 1.04.
Step 2: Convert the z-score back to the original score
We can use the formula for the z-score to convert it back to the original score:
z = (x - μ) / σ
Where:
- x is the original score
- μ is the mean of the distribution (given as 60)
- σ is the standard deviation of the distribution (given as 20)
Rearranging the formula, we have:
x = z * σ + μ
x = 1.04 * 20 + 60
x = 20.8 + 60
x = 80.8
So, the minimum score required to get an A in the course is approximately 80.8.
To find the minimum score required to get an A in the course, we can use the concept of z-scores and the cumulative distribution function (CDF) of the normal distribution.
First, we need to find the z-score corresponding to the top 15% of the students. This is equivalent to finding the z-score that separates the top 15% from the rest of the distribution.
The CDF gives us the probability that a randomly selected score from the distribution is less than or equal to a given value. In this case, we want to find the z-score that corresponds to a CDF value of 0.85 (100% - 15% = 85%).
We can use a standard normal distribution table or a statistical software to find the z-score corresponding to a CDF value of 0.85. For simplicity, let's assume we use a standard normal distribution table.
Looking up the z-score in the table, we find that a CDF value of 0.85 corresponds to a z-score of approximately 1.036.
Next, we can use the z-score formula to find the actual score corresponding to this z-score in the original distribution. The z-score formula is given by:
z = (x - μ) / σ,
where z is the z-score, x is the score in the original distribution, μ is the mean of the distribution, and σ is the standard deviation of the distribution.
Rearranging the formula to solve for x, we have:
x = z * σ + μ.
Plugging in the values obtained previously, we have:
x = 1.036 * 20 + 60,
x ≈ 82.72.
Therefore, the minimum score required to get an A in the course is approximately 82.72.