Question

A certain type of bird lives in two regions of a state. The distribution of weight for birds of this type in the northern region is approximately normal with mean 10 ounces and standard deviation 3 ounces. The distribution of weight for birds of this type in the southern region is approximately normal with mean 16 ounces and standard deviation 2.5 ounces. (a) Calculate the z -scores for a weight of 13 ounces for a bird living in the northern region and for a weight of 13 ounces for a bird living in the southern region.

Answer 1

Answer: (a) Z-score are 1 and -1.2 for northern and southern regions, respectively.

Explanation: Z-score is how many standard deviations a data is from the population mean or how far a data point is from the mean.

The z-score is calculated by the following:

$z=\frac{x-\mu}{\sigma}$

where

x is the data point

μ is population mean

σ is standard deviation

For the northern region birds:

μ = 10, σ = 3, x = 13

$z=\frac{13-10}{3}$

z = 1

The z-score for birds living in the northern region is 1, which means it is 1 standard deviation above the mean.

For the southern region:

μ = 16, σ = 2.5, x = 13

$z=\frac{13-16}{2.5}$

z = -1.2

The z-score for southern living birds is -1.2, meaning it is 1.2 standard deviations below the mean.