Ask Question
27 May, 10:23

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.

+1
Answers (1)
  1. 27 May, 10:26
    0
    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
Know the Answer?
Not Sure About the Answer?
Find an answer to your 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 ...” in 📗 Mathematics if the answers seem to be not correct or there’s no answer. Try a smart search to find answers to similar questions.
Search for Other Answers