assume that the volume of drink is normally distributed, with a mean of 2.0 liters and a standard deviation of 0.05 liter. Using the random sample of 45 bottles. The probability is 99% that the sample mean will contain at least how much soft drink?
95% = mean ± 2.575 SEm
SEm = SD/√n
To find the probability that the sample mean will contain at least a certain amount of soft drink, we can use the concept of the sampling distribution of the sample mean.
Formula to calculate the sampling distribution:
µ(mean) = µ(population)
σ(mean) = σ(population) / √n
Given information:
µ(population) = 2.0 liters (mean of the population volume of drink)
σ(population) = 0.05 liter (standard deviation of the population volume of drink)
n = 45 (sample size)
Step 1: Calculate the standard deviation of the sampling distribution (σ(mean)).
σ(mean) = σ(population) / √n
σ(mean) = 0.05 / √45
σ(mean) ≈ 0.0075 liter
Step 2: Calculate the z-score for the desired probability level.
The given probability is 99%, which means we need to find the z-score corresponding to the area under the normal distribution curve beyond 99%.
Using a standard normal distribution table or a calculator, we find that the z-score for 99% is approximately 2.33.
Step 3: Calculate the minimum sample mean value.
To find the minimum sample mean value that contains at least a certain amount of soft drink, we need to find the corresponding value in terms of z-score.
Minimum sample mean value = µ(mean) + (z-score * σ(mean))
Minimum sample mean value = 2.0 + (2.33 * 0.0075)
Minimum sample mean value ≈ 2.0174 liters
Therefore, the probability is 99% that the sample mean will contain at least 2.0174 liters of soft drink.