Assume the body weight of healthy fifth graders is normally distributed with a mean of 98.3 lbs and a standard deviation of 4.12 lbs.
If you have a body weight of 89.3, what is your percentile score?
Here, Body Weight follows Normal (Mean = 98.3,SD=4.12)
Z-score of 89.3 = (89.3 - 98.3)/4.12 = -2.18
Now, P[Z<-2.18]=0.0145
So, percentile score is 1.45%
I am having a difficult time figuring the last portion of this equation. How do we get P[Z<-2.18] = 0.0145 to get 1.45%. I have worked it out every way possible but answer is not coming up as yours.
To find the percentile score for a given body weight of 89.3 lbs, we need to calculate the corresponding z-score and then use the z-score to find the percentile.
The formula for calculating the z-score is:
z = (x - μ) / σ
Where:
- x is the given body weight (89.3 lbs in this case),
- μ is the mean of the distribution (98.3 lbs in this case), and
- σ is the standard deviation of the distribution (4.12 lbs in this case).
Substituting the given values into the formula, we get:
z = (89.3 - 98.3) / 4.12
Simplifying the expression, we get:
z = -2.18
Now, to find the percentile corresponding to this z-score, we can refer to a standard normal distribution table or use a statistical calculator.
Using a standard normal distribution table, we can find that the percentile score corresponding to a z-score of -2.18 is approximately 1.66%. This means that a body weight of 89.3 lbs would be around the 1.66th percentile.
Alternatively, you can also use a statistical calculator or software to find the percentile directly by inputting the z-score (-2.18).