A standardized test is designed so that scores have a mean of 50 and a standard deviation of 4. What is the probability that a test score is above 54?
about 15.9%
15.9%
Well, statistically speaking, if the average is 50 and the standard deviation is 4, then you could say that 54 is just a little bit "above average" – kinda like a student who's trying to impress the teacher with their extra credit. But let's get down to business.
To find the probability that a test score is above 54, we'll need to use a z-score. The z-score tells us how many standard deviations a particular value is away from the mean.
In this case, we want to find the probability that a score is above 54, so we need to find the area under the curve to the right of 54 on the standard normal distribution.
Now, using fancy math jargon, we can convert the raw score of 54 to a z-score by subtracting the mean (50) from 54 and then dividing by the standard deviation (4).
So, the z-score is (54 - 50) / 4 = 1. Yay, I'm a math wiz!
Now, if we consult a z-table or use a statistical tool, we can find that the probability of a z-score of 1 or greater is around 0.1587. So, the probability that a test score is above 54 is approximately 0.1587.
Keep in mind, though, that this is just a statistical probability and doesn't determine your actual test score. So, don't stress too much – just be yourself and let your skills shine through! Good luck, smarty pants!
To find the probability that a test score is above 54, we can use the standard normal distribution, also known as the z-distribution.
First, we need to calculate the z-score for the value 54, which represents how many standard deviations the score is from the mean. The formula for calculating the z-score is:
z = (X - μ) / σ
Where:
X = The given value (54 in this case)
μ = The mean (50 in this case)
σ = The standard deviation (4 in this case)
Using the values provided, the calculation is:
z = (54 - 50) / 4
z = 4 / 4
z = 1
The z-score for a value of 54 is 1.
Next, we need to find the area under the curve to the right of the z-score of 1. We can use a standard normal distribution table or a calculator to find this value.
Looking up the value in a standard normal distribution table, we find that the area to the left of a z-score of 1 is approximately 0.8413. Since we want the area to the right of a z-score of 1, we subtract this value from 1:
1 - 0.8413 = 0.1587
Therefore, the probability that a test score is above 54 is approximately 0.1587, or 15.87%.