Answer:
15.25%
Explanation:
The actual amount in a 12-ounce container of a certain brand of orange juice is normally distributed with mean μ = 12.34 ounces and standard deviation σ = 0.04 ounce. What percentage of the juice bottles contain between 12.24 and 12.30 ounces of orange juice?
We solve using the z score formula
z-score is is z = (x-μ)/σ,
where x is the raw score
μ is the population mean = μ = 12.34 ounces
σ is the population standard deviation = σ = 0.04 ounce
For x = 12.24
z = 12.24 - 12.34/0.04
z = -2.5
Probability value from Z-Table:
P(x = 12.24) = 0.0062097
For x = 12.30
z= 12.30 - 12.34/0.04
z = -1
Probability value from Z-Table:
P(x = 12.30) = 0.15866
Hence, the probability of the juice bottles contain between 12.24 and 12.30 ounces of orange juice
P(x = 12.30) - P(x = 12.24)
= 0.15866 - 0.0062097
= 0.1524503
Therefore, the percentage of the juice bottles contain between 12.24 and 12.30 ounces of orange juice is calculated as:
= 0.1524503 × 100
= 15.24503%
= 15.25%