In a certain neighborhood, the average number of candies received while trick-or-treating is 50, with a standard deviation of 12. If Isaac received 40 candies, calculate his z-score to the nearest thousandth
To calculate the z-score for Isaac, we use the formula:
\[ Z = \frac{X - \mu}{\sigma} \]
Where:
- X is the number of candies Isaac received (40)
- μ is the average number of candies received in the neighborhood (50)
- σ is the standard deviation (12)
Plugging in the values, we get:
\[ Z = \frac{40 - 50}{12} \]
\[ Z = \frac{-10}{12} \]
\[ Z = -0.8333 \]
Therefore, Isaac's z-score to the nearest thousandth is -0.833.
is this right?
I apologize for the mistake in my previous response. Let's recalculate the z-score.
Given:
- X = 40
- μ = 50
- σ = 12
\[ Z = \frac{X - \mu}{\sigma} \]
\[ Z = \frac{40 - 50}{12} \]
\[ Z = \frac{-10}{12} \]
\[ Z = -0.8333 \]
So, to the nearest thousandth, Isaac's z-score is -0.833.