Weights of 2-year-old girls are normally distributed with a mean of 25.3 lbs. and a standard deviation of 1.12 lbs. According to this information, what weight would be the 33rd percentile?

Question

Weights of 2-year-old girls are normally distributed with a mean of 25.3 lbs. and a standard deviation of 1.12 lbs. According to this information, what weight would be the 33rd percentile?

Answer 1

The mean is 25.3 lbs (μ); std. deviation (σ) = 1.12 lbs. To find a value x in the normal distribution curve, the formula is:

x = μ + zσ. The z-table value that corresponds to the 33rd percentile (33% or 0.33 of the area under the curve is left of x) is -0.44. x = 25.3 + (-0.44)1.12 = 25.3 - 0.4928 = 24.8 lbs.

Answer 2

This is essentially a calculator question, as I doubt you would be expected to use integration on the normal curve.

If you are using a GDC, then go to Menu, Statistics, Distributions, Inverse Normal. Then input .33 for the area, 25.3 for the mean, and 1.12 for the standard deviation. If you use Python (I highly encourage it), here are the commands:

import scipy.stats

scipy.stats.norm.ppf(.33,25.3,1.12)

More important is to get a sense of the Normal Distribution and what is being asked. Basically the question is asking below which weight is the probability .33. Or, in other words, below which weight is the area under the normal curve = .33

See here for an illustration of the situation:

https://drive.google.com/file/d/1g2rrfmVOS0zu2QdLlNGPuYOoJ37IJJKf/view?usp=sharing