Question

1. (05.02 lc) the number of pieces of popcorn in a large movie theatre popcorn bucket is normally distributed, with a mean of 1610 and a standard deviation of 10. approximately what percentage of buckets contain between 1600 and 1620 pieces of popcorn? (1 point) approximately 68% approximately 75% approximately 95% approximately 99.7%

Explanation:

Step1: Calculate z-scores for bounds

First, find the z-score for 1600:
$z_1 = \frac{1600 - 1610}{10} = -1$
Then, find the z-score for 1620:
$z_2 = \frac{1620 - 1610}{10} = 1$

Step2: Apply empirical rule

For a normal distribution, the empirical rule states that approximately 68% of data lies within $\mu \pm 1\sigma$, which corresponds to z-scores of -1 and 1.

Answer:

Approximately 68%