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SSSSS
2 months ago
15

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.
Mathematics
1 answer:
tester [12.3K]2 months ago
7 0

Réponse :

a) Le score Z pour la région nord est 1, tandis que pour la région sud il est -1.2.

b) Il s'agit de la région sud.

Déroulement des calculs :

a) La formule du score Z est Z = (x – moyenne) / écart-type.

Soit x le poids des oiseaux.

Pour le nord :

Z = (13 – 10) / 3

Z = 1

Pour le sud :

Z = (13 – 16) / 2.5

Z = -1.2

b) Pour déterminer dans quelle région on trouve plus probablement un oiseau avec 13 onces, on calcule la probabilité associée à chaque score Z.

Dans la région nord :

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

En utilisant la table des Z pour obtenir l'aire sous la courbe :

P(x > 13) = 0.1587

Dans la région sud :

P(x > 13) = P(Z > -1.2)

On regarde dans la table des Z :

P(x > 13) = 0.8849

Il est donc plus probable qu'un oiseau pesant plus de 13 onces habite dans la région sud.

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