# Normal Distribution of Means

1. Feb 8, 2013

### jmm

1. The problem statement, all variables and given/known data

A biologist interested in the mass of Chickadees (Poecile atricapillus) in North Glenmore Park collects the 3 samples shown below:
Sample 1: 9 individuals with a mean mass of 13.23 grams
Sample 2: 16 individuals with a mean mass of 9.64 grams
Sample 3: 13 individuals with a mean mass of 11.14 grams

b. The population of Chickadees in North Glenmore Park has a mean mass of 10.87 grams and a standard deviation of 1.89 grams. Assuming the population is normally distributed, what is the probability of obtaining samples such as these, with means between 9.64 and 13.23 grams, if each sample were comprised of 14 individuals?

2. Relevant equations

z = (xbar - µ) / ∂

3. The attempt at a solution

So I think the probability would be P(xbar < 13.23) - P(xbar < 9.64). For xbar = 13.23, z=1.25 and p = 0.8925, for xbar = 9.64, z = -0.65 and p = 0.2578. So P(9.64 < xbar < 13.23) = 0.8925 - 0.2578 = 0.6347. Am I on the right track with this or am I completely off? I'm confused why they mention that each sample is composed of 14 individuals.

Thank you.

2. Feb 8, 2013

### jmm

Never mind; I just figured out that I need to use the standard deviation of the means to calculate the z scores instead of the population standard deviations. Thanks anyways!