10th grade New York public school students taking a standardized English test produced test scores that were normally distributed with a mean of 85 and a standard deviation of 4. Let x be a random variable that represents the test score of a student. Students whose test scores are in the upper 10% are eligible for scholarship opportunities for college. What is the test score that corresponds to the upper 10% cutoff point of the probability distribution?
http://davidmlane.com/hyperstat/z_table.html
Well, as a bot who tries to make you laugh, it's quite funny that only 10% of students can be eligible for scholarship opportunities. It's like the education system saying, "We can only afford to reward a few of you, so good luck!" Anyway, let's get back to your question.
Since we know that the test scores are normally distributed with a mean of 85 and a standard deviation of 4, we can use the standard normal distribution to find the cutoff point for the upper 10%.
To find this cutoff point, we can use the z-table or a calculator to find the z-score that corresponds to the upper 10%. In this case, the z-score is approximately 1.28.
Now, we convert this z-score back to a test score using the formula
x = μ + z * σ, where x is the test score, μ is the mean, z is the z-score, and σ is the standard deviation.
Plugging in the values, we get:
x = 85 + 1.28 * 4
Calculating that, we find the test score that corresponds to the upper 10% cutoff point is approximately 90.12.
So, a test score of around 90.12 would be the cutoff point for the upper 10% of test scores, making the students eligible for scholarship opportunities. Good luck to those lucky few!
To find the test score that corresponds to the upper 10% cutoff point, we need to find the z-score value that corresponds to the upper 10% of the normal distribution.
Step 1: Convert the upper 10% probability to a z-score value.
Since the distribution is symmetric, the upper 10% probability is equal to (1 - 0.10) = 0.90.
Step 2: Find the z-score using the z-table or a calculator.
The z-table gives us the area under the standard normal distribution curve up to a given z-score. From the z-table, the z-score that corresponds to a cumulative probability of 0.90 is approximately 1.28.
Step 3: Convert the z-score back to the original test score using the formula: z = (x - mean) / standard deviation.
Rearranging the formula, we have x = z * standard deviation + mean.
Substituting the values, we get:
x = 1.28 * 4 + 85
x = 5.12 + 85
x ≈ 90.12
Therefore, the test score that corresponds to the upper 10% cutoff point is approximately 90.12.
To find the test score that corresponds to the upper 10% cutoff point of the probability distribution, we need to use the concept of standard deviation and the normal distribution.
Given:
Mean (μ) = 85
Standard Deviation (σ) = 4
We know that the normal distribution is symmetric around the mean and that approximately 68% of the data falls within one standard deviation of the mean (μ ± σ), 95% falls within two standard deviations (μ ± 2σ), and 99.7% falls within three standard deviations (μ ± 3σ).
To find the cutoff point for the upper 10%, we need to find the z-score that corresponds to the cumulative probability of 90%. The z-score represents the number of standard deviations a particular value is from the mean.
Using a standard normal distribution table or a statistical calculator, we can find that the z-score corresponding to a cumulative probability of 90% is approximately 1.28.
Now, we can use the formula for converting a z-score to an actual score on the distribution:
z = (x - μ) / σ
Rearranging the formula to solve for x:
x = z * σ + μ
Plugging in the values:
x = 1.28 * 4 + 85
Calculating:
x = 5.12 + 85
x ≈ 90.12
Therefore, the test score that corresponds to the upper 10% cutoff point of the probability distribution is approximately 90.12.