Question

a researcher is studying a group of field mice. the distribution of the weight of field mice is approximately normal with mean 25 grams and standard deviation 4 grams. which of the following is closest to the proportion of field mice with a weight greater than 33 grams?

a 0.023

b 0.046

c 0.954

d 0.977

e 1.000

Explanation:

Step1: Calculate the z - score

The formula for the z - score is $z=\frac{x-\mu}{\sigma}$, where $x = 33$ (the value we are interested in), $\mu=25$ (the mean) and $\sigma = 4$ (the standard deviation).
Substitute the values into the formula: $z=\frac{33 - 25}{4}=\frac{8}{4}=2$.

Step2: Use the properties of the normal distribution

For a normal distribution, we know the empirical rule (68 - 95 - 99.7 rule) which states that approximately 95% of the data lies within $z=- 2$ and $z = 2$ (i.e., within 2 standard deviations of the mean). So the proportion of data outside of $z=-2$ and $z = 2$ is $1 - 0.954 = 0.046$. Since the normal distribution is symmetric, the proportion of data greater than $z = 2$ (i.e., weight greater than 33 grams) is half of the proportion outside of $z=-2$ and $z = 2$. So the proportion is $\frac{0.046}{2}=0.023$.

Answer:

A. 0.023