The score distribution of the final exam in a data science course follows a normal distribution with mean 70 and standard deviation 10.
Let α in (0,1) . As a reminder, the quantile of order 1−α of a random variable X is the number qα such that
P(X≤qα)=1−α.
According to this distribution, what score do you need to get in order to be at the 90th percentile of the class, that is, in order that 90% of all students in the class have a score less than or equal to your score?
Required score will be 82.85
To find the score needed to be at the 90th percentile, we need to find the quantile of order 1-α.
Given:
Mean (μ) = 70
Standard Deviation (σ) = 10
To find the desired quantile, we will use the standard normal distribution. The formula to standardize a value from a normal distribution is:
Z = (X - μ) / σ
Where:
Z is the standard score (z-score) obtained from standardizing X
X is the score we want to find
μ is the mean of the distribution
σ is the standard deviation of the distribution
In this case, we want to find the z-score corresponding to the 90th percentile, which is equal to 0.90. Since the normal distribution is symmetric, we can use the table of the standard normal distribution to find the z-score.
Looking up the value of 0.90 in the standard normal distribution table, we find that the corresponding z-score is approximately 1.28.
Now, we can solve for X using the formula:
1.28 = (X - 70) / 10
Simplifying the equation:
1.28 * 10 = X - 70
12.8 = X - 70
X = 12.8 + 70
X = 82.8
Therefore, to be at the 90th percentile of the class, you would need a score of at least 82.8.
To find the score you need to be at the 90th percentile of the class, we need to calculate the quantile of order 1 - α. In this case, α is equal to 0.90, as we want to find the score at the 90th percentile.
To do this, we can use the standard normal distribution table or the z-score formula.
The z-score formula is given by:
z = (x - μ) / σ
where:
- z is the z-score,
- x is the raw score (the score we want to find),
- μ is the mean of the distribution,
- σ is the standard deviation of the distribution.
In this case, the mean (μ) is 70 and the standard deviation (σ) is 10. We need to find the raw score (x) corresponding to the 90th percentile.
Using the standard normal distribution table or a cumulative distribution function (CDF) calculator, we find that the z-score corresponding to the 90th percentile is approximately 1.28.
Plugging this value into the z-score formula:
1.28 = (x - 70) / 10
Now we can solve for x:
1.28 * 10 = x - 70
12.8 = x - 70
x = 12.8 + 70
x = 82.8
Therefore, to be at the 90th percentile of the class, you would need to score approximately 82.8.