Show all your work. Indicate clearly the methods you use, because you will be scored on the correctness of your methods as well as on the accuracy and completeness of your results and explanations. 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.
Question
Show all your work. Indicate clearly the methods you use, because you will be scored on the correctness of your methods as well as on the accuracy and completeness of your results and explanations. 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.
Solution
(a) The z-score is a measure of how many standard deviations an element is from the mean. We can calculate it using the formula:
z = (X - μ) / σ
where: X is the value we are comparing, μ is the mean, and σ is the standard deviation.
For a bird living in the northern region:
z = (13 - 10) / 3 = 1
So, a bird weighing 13 ounces is 1 standard deviation above the mean weight in the northern region.
For a bird living in the southern region:
z = (13 - 16) / 2.5 = -1.2
So, a bird weighing 13 ounces is 1.2 standard deviations below the mean weight in the southern region.
(b) A bird with a weight greater than 13 ounces is more likely to live in the southern region. This is because the mean weight of birds in the southern region is higher than in the northern region. Furthermore, a bird weighing 13 ounces is above the mean weight in the northern region (z-score = 1), but below the mean weight in the southern region (z-score = -1.2). Therefore, it is more likely that a bird weighing more than 13 ounces would be from the southern region, where such a weight is closer to the average.
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