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.(b) Is it more likely that a bird of this type with a weight greater than 13 ounces lives in the northern region or the southern region? Justify your answer.

Answer 1

Answer:

a) Z for northern region = 1 , Z for southern region = -1.2

b) Southern region

Step-by-step explanation:

a) Z = (x – mean)/standard deviation

Le x be weight for birds

For the northern region  

Z = (13 – 10)/3

Z = 1

For the southern region

Z = (13 – 16)/2.5

Z = -1.2

b) to find in what region is more likely to find a bird with 13 ounces, we calculate the probability for each z value

For the northern region  

P (x>13) = P(z > 1)

Then we use the z table to find the area under the curve.

P (x>13) = 0.1587

For the southern region

P (x>13) = P(z > 1.2)

Then we use the z table to find the area under the curve.

P (x>13) = 0.8849

Is it more likely that a bird with a weight greater than 13 ounces lives in the southern region