Answer:
The percentage is approximately 95% ⇒ 3rd answer
Step-by-step explanation:
* Lets explain how to solve the problem
- for the probability that a < X < b , where X is between the two
numbers a and b, convert a and b into z-score using the rule
z = (X - μ)/σ , where
# x is the score
# μ is the mean
# σ is the standard deviation
* Lets solve the problem
- The number of pieces of popcorn in a large movie theater
popcorn bucket is normally distributed mean of 1720 and a
standard deviation of 20
∴ μ = 1720 and σ = 20
- We need to find at what percentage of buckets contain between
1680 and 1760 pieces of popcorn
∴ X = 1680 and X = 1760
- Lets use the rule above to find the z-scores
∵ [tex]z=\frac{1680-1720}{20}=\frac{-40}{20}=-2[/tex]
∵ [tex]z=\frac{1760-1720}{20}=\frac{40}{20}=2[/tex]
- Lets use the table of normal distribution to find the area between
the two z-scores -2 and 2
∵ The corresponding area of -2 is 0.02275
∵ The corresponding area of 2 is 0.97725
∴ The area between -2 and 2 = 0.97725 - 0.02275 = 0.9545
∴ P(-2 < z < 2) ≅ 0.95
∵ P(-2 < z < 2) = P( 1680 < X < 1760)
∴ P( 1680 < X < 1760) = 0.95
- 0.95 = 0.95 × 100% = 95%
∴ P( 1680 < X < 1760) = 95%
* The percentage is approximately 95%
