Question

question 6 (1 point)
(05.02 lc)
the number of pieces of popcorn in a large movie theatre popcorn bucket is normally distributed, with a mean of 1515 and a standard deviation of 15.
approximately what percentage of buckets contain between 1470 and 1560 pieces of popcorn?
a approximately 68%
b approximately 75%
c approximately 95%
d 99.7%

Explanation:

Step1: Calculate z-score for 1470

The z-score formula is $z = \frac{x - \mu}{\sigma}$, where $\mu=1515$, $\sigma=15$, $x=1470$.
$z_1 = \frac{1470 - 1515}{15} = -3$

Step2: Calculate z-score for 1560

Use the same z-score formula with $x=1560$.
$z_2 = \frac{1560 - 1515}{15} = 3$

Step3: Apply empirical rule

For normal distributions, data within $\pm3\sigma$ (z-scores -3 to 3) covers ~99.7% of values.

Answer:

d. 99.7%