Scores on a certain exam are normally distributed with a mean of 78 and a variance of 25. What is the score obtained by the top 5% of the students?
I do not understand the 5% part. What does the tail look like on the bell-shaped graph?
86.2? :D
Standard deviation (SD) = square root of variance
Z = (score-mean)/SD
Find table in the back of your statistics text labeled something like "areas under normal distribution" to find the proportion/probability (.05) related to the Z score. Insert the values into equation above and solve for the score.
http://www.regentsprep.org/regents/math/algtrig/ATS2/NormalLesson.htm
Ah, the tail on the bell-shaped graph is like the tail of a stingy squirrel trying to hold onto its nuts during a hurricane. It's super skinny and gets narrower as it goes out towards the ends. In this case, since we're looking at the top 5% of the scores, we're interested in the right tail of the distribution. So, imagine a squirrel clinging desperately to those top-performing scores, hanging on for dear life! But fear not, I'll help you find that elusive top 5% score.
In a bell-shaped graph representing a normal distribution, the tail refers to the portion of the graph that extends beyond the majority of the data. In this case, we are interested in the top 5% of the students, which means we are looking for the scores that are higher than 95% of the other scores.
Let's calculate the score obtained by the top 5% of the students.
First, we need to find the z-score corresponding to the 95th percentile. The z-score represents the number of standard deviations a particular value is from the mean in a normal distribution.
Since the distribution has a mean of 78 and a variance of 25, we can find the standard deviation (σ) by taking the square root of the variance: σ = √25 = 5.
To find the z-score corresponding to the 95th percentile, we can use a z-table or calculator. The z-score for a percentile can be determined by finding the value that corresponds to the given percentile in the standard normal distribution (where the mean is 0 and the standard deviation is 1).
Using a z-table or calculator, you will find that the z-score corresponding to the 95th percentile is approximately 1.645.
Now, we can use the z-score formula to find the x-value (score) corresponding to this z-score:
z = (x - μ) / σ
Rearranging the formula to solve for x:
x = z * σ + μ
Plugging in the values:
x = 1.645 * 5 + 78
x ≈ 8.225 + 78
x ≈ 86.225
Therefore, the score obtained by the top 5% of the students is approximately 86.225.
In a bell-shaped graph, also known as a normal distribution, the area under the curve represents the probability of a particular event occurring. To understand the 5% part, it's important to know that the total area under the curve is considered to be 100%.
In this case, we want to find the score obtained by the top 5% of the students, which means we are interested in the tail at the upper end of the distribution. The top 5% implies that we are looking for the score associated with the area beyond the 95th percentile.
To find the score for the top 5%, we can use the standard normal distribution table or z-table. The z-table provides the proportion of observations (or area under the curve) that fall to the left of a given z-score, which is a measure in terms of standard deviations from the mean.
To find the z-score that corresponds to the 95th percentile (area of 0.95), we subtract this value from 1, which gives us 0.05. Using the z-table, we look up this value (0.05) and find the corresponding z-score, which is approximately 1.645.
Once we have the z-score, we can use the formula z = (X - μ) / σ, where X is the score we want to find, μ is the mean, and σ is the standard deviation. Solving for X, we have X = (z * σ) + μ.
Plugging in the values from the question, with μ = 78, σ = √25 = 5, and z = 1.645, we can calculate the score obtained by the top 5% of students:
X = (1.645 * 5) + 78 = 86.225
Therefore, the score obtained by the top 5% of students is approximately 86.23.