Answer :
Using a standard normal distribution table or calculator, the z-score corresponding to the 75th percentile is approximately 0.675. Therefore, The weight of a dog that is lighter than 75% of the population of other dogs of that breed is approximately 41.6 lbs.
Z-score of a dog weighing 29.1 lbs:
The formula for the z-score is:
Z = (X - μ) / σ
where:
X is the value (weight of the dog) we want to find the z-score for,
μ is the mean (mean adult weight of the breed),
σ is the standard deviation.
Z = (29.1 - 38.4) / 4.8
Z = -9.3 / 4.8
Z ≈ -1.94 (rounded to 2 decimal places)
The z-score of a dog weighing 29.1 lbs is approximately -1.94.
Weight of a dog with a z-score of 2.13:
We can rearrange the z-score formula to find the weight (X) when given the z-score:
X = Z × σ + μ
where:
Z is the z-score (2.13 in this case),
σ is the standard deviation (4.8 lbs),
μ is the mean (mean adult weight of the breed, 38.4 lbs).
X = 2.13 × 4.8 + 38.4
X = 10.224 + 38.4
X ≈ 48.62 (rounded to 2 decimal places)
The weight of a dog with a z-score of 2.13 is approximately 48.62 lbs.
Weight of a dog lighter than 75% of the population:
To find the weight of a dog that is lighter than 75% of the population, we need to find the z-score corresponding to the 75th percentile (which is 0.75).
Z = (X - μ) / σ
0.675 = (X - 38.4) / 4.8
Solving for X:
X - 38.4 = 0.675 × 4.8
X - 38.4 = 3.24
X = 3.24 + 38.4
X ≈ 41.64 (rounded to 1 decimal place)
Therefore, The weight of a dog that is lighter than 75% of the population of other dogs of that breed is approximately 41.6 lbs.
To learn more about distribution :
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