Question

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?
approximately 68%
approximately 75%
approximately 95%
99.7%​

Answer 1

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

∵ $z=\frac{1680-1720}{20}=\frac{-40}{20}=-2$

∵ $z=\frac{1760-1720}{20}=\frac{40}{20}=2$

- 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%