The number of pieces of popcorn in a large movie theatre popcorn bucket is normally distributed, with a mean of 1720 and a standard deviation of 20. Approximately what percentage of buckets contain between 1680 and 1760 pieces of popcorn?
a.approximately 68%
b.approximately 75%
c.approximately 95%
d.99.7%
Perfectly suited for David's applet
http://davidmlane.com/normal.html
To determine the percentage of buckets that contain between 1680 and 1760 pieces of popcorn, we need to calculate the probability within this range using the normal distribution.
First, we calculate the z-score for each value:
z1 = (1680 - 1720) / 20 = -2
z2 = (1760 - 1720) / 20 = 2
Next, we can use a z-table or a calculator to find the cumulative probability for each z-score:
P(Z < -2) ≈ 0.0228
P(Z < 2) ≈ 0.9772
To find the probability within the range, we subtract the probability for the lower value from the probability for the higher value:
P(-2 < Z < 2) = P(Z < 2) - P(Z < -2)
≈ 0.9772 - 0.0228
≈ 0.9544
The probability is approximately 0.9544, which is equivalent to 95.44%.
Therefore, the correct answer is:
c. approximately 95%
approximately 95% of a normally distributed population
... lies within 2 s.d. of the mean