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. note on your ap exam, you will handwrite your responses to free - response questions in a test booklet (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.

Explanation:

Step1: Recall z - score formula

The z - score formula is $z=\frac{x-\mu}{\sigma}$, where $x$ is the value, $\mu$ is the mean and $\sigma$ is the standard deviation.

Step2: Calculate z - score for northern region

For the northern region, $\mu = 10$ ounces and $\sigma=3$ ounces, $x = 13$ ounces.
$z_{north}=\frac{13 - 10}{3}=\frac{3}{3}=1$

Step3: Calculate z - score for southern region

For the southern region, $\mu = 16$ ounces and $\sigma = 2.5$ ounces, $x = 13$ ounces.
$z_{south}=\frac{13 - 16}{2.5}=\frac{- 3}{2.5}=-1.2$

Step4: Analyze probability for part (b)

For a normal distribution, $P(X>x)=1 - P(X\leq x)$. Using the standard normal table, for $z = 1$, $P(Z\leq1)=0.8413$, so $P(Z > 1)=1 - 0.8413 = 0.1587$. For $z=-1.2$, $P(Z\leq - 1.2)=0.1151$, so $P(Z>-1.2)=1 - 0.1151=0.8849$. Since $0.8849>0.1587$, it is more likely that a bird with weight greater than 13 ounces lives in the southern region.

Answer:

(a) The z - score for a bird in the northern region is $z_{north}=1$ and for a bird in the southern region is $z_{south}=-1.2$.
(b) It is more likely that a bird of this type with a weight greater than 13 ounces lives in the southern region. Justification: The probability that a bird in the northern region has a weight greater than 13 ounces is $0.1587$ and the probability that a bird in the southern region has a weight greater than 13 ounces is $0.8849$. Since $0.8849>0.1587$, the southern - region bird is more likely.