Answer:
a) 0.21078
b) 21.078%
Explanation:
We solve using z score formula
z = (x-μ)/σ, where x is the raw score, μ is the population mean, and σ is the population standard deviation.
Mean of 3500 and standard deviation of 300.
a) What is the probability of finding a bacteria count of between 3200 and 3400 in a milliliter of raw milk?
For x = 3200
z = 3200 - 3500/300
z = -1
Probability value from Z-Table:
P(x = 3200) = 0.15866
For x = 3400
z = 3400 - 3500/300
z = -0.33333
Probability value from Z-Table:
P(x = 3400) = 0.36944
The probability of finding a bacteria count of between 3200 and 3400 in a milliliter of raw milk
P(x = 3400) - P(x = 3200)
= 0.36944 - 0.15866
= 0.21078
b) What percentage of the uncontaminated raw milk has a bacteria count per milliliter between 3200 and 3400?
We convert the answer in part a) to percentage
= 0.21078 × 100
= 21.078%