jenny is 5'10" tall. the heights of the girls in the school are approximately normally distributed with a mean of 5'5" and a standard deviation of 2.6". What is the percentile of jenny's height?
Z = (score-mean)/SD
Insert the values to find Z. Find table in the back of your statistics text labeled something like "areas under normal distribution" to find the proportion related to that Z score.
Well, let's take a walk on the silly side and calculate Jenny's height percentile. First, we need to determine how many standard deviations Jenny is away from the mean. So, using our hilarious calculations, we find that she is 0.1923076923076923 standard deviations above the mean.
Now, we just have to consult our cackling conversion chart, which tells us that this z-score roughly corresponds to the 57.13th percentile. So, we can conclude that Jenny's height is in the 57.13th percentile among the girls at her school!
Remember, this answer might not be as exact as Jenny's height, but at least it will make you smile!
To find the percentile of Jenny's height, we will use the z-score formula. The formula is:
z = (x - μ) / σ
Where:
- z is the z-score
- x is the value for which we want to find the percentile (Jenny's height)
- μ is the mean of the distribution (5'5" or 65 inches)
- σ is the standard deviation of the distribution (2.6 inches)
Step 1: Convert Jenny's height to inches:
Jenny's height is 5'10" which is equivalent to 5 feet and 10 inches. In total, 5 feet is equal to 60 inches. So, Jenny's height is 60 inches + 10 inches = 70 inches.
Step 2: Calculate the z-score:
z = (x - μ) / σ
z = (70 - 65) / 2.6
z = 1.9231
Step 3: Find the percentile using the z-score:
To find the percentile, we need to look up the z-score in the z-table (standard normal distribution table). The z-score of 1.9231 corresponds to a percentile of approximately 97.9%.
Therefore, the approximate percentile of Jenny's height is 97.9%.
To find the percentile of Jenny's height, we need to calculate her z-score first. The z-score measures how many standard deviations away from the mean her height is.
To find the z-score, we will use the formula:
z = (X - μ) / σ
Where:
- X is the value we want to find the percentile for (Jenny's height)
- μ is the mean of the distribution (5'5" or 65 inches in this case)
- σ is the standard deviation of the distribution (2.6 inches)
So for Jenny's height:
z = (70 - 65) / 2.6
This gives us a z-score of approximately 1.923.
Now, we can find the percentile corresponding to this z-score using a standard normal distribution table (also known as a z-table) or using statistical software.
Looking up the z-score of 1.923 in the z-table, we find that the percentile is approximately 97.9%. This means that Jenny's height is greater than approximately 97.9% of the girls in her school.
Therefore, the percentile of Jenny's height is approximately 97.9%.