3. The mean volume of draught beer served in pint glasses in a particular pub is known to be 0.564 litres. A consumer organization takes a random sample of 64 pints of draught beer and finds that the standard deviation of this sample is 0.025 litres. What is the probability that the mean volume of the sample will be more than a pint (0.564 litres)?
Z = (score-mean)/SEm
SEm = SD/√n
Use the same table as on your previous post.
To find the probability that the mean volume of the sample will be more than a pint (0.564 liters), we need to calculate the probability of getting a sample mean greater than 0.564 liters.
Given that the population mean (μ) is known to be 0.564 liters, and the standard deviation of the sample (σ) is 0.025 liters, we can use the Central Limit Theorem to approximate the distribution of the sample mean.
According to the Central Limit Theorem, for a large enough sample size (n > 30), the distribution of the sample mean will be approximately normally distributed, regardless of the shape of the population distribution.
Since we have a sample size of 64, which is larger than 30, we can use the normal distribution to approximate the probability.
First, we need to calculate the standard deviation of the sample mean (σᵧ) using the formula:
σᵧ = σ / √n
where σ is the standard deviation of the sample (0.025 liters) and n is the sample size (64).
σᵧ = 0.025 / √64
σᵧ = 0.025 / 8
σᵧ = 0.003125
Now, we can calculate the z-score, which measures how many standard deviations an observation is from the mean:
z = (x - μ) / σᵧ
where x is the value we want to calculate the probability for, μ is the population mean (0.564 liters), and σᵧ is the standard deviation of the sample mean (0.003125).
z = (0.564 - 0.564) / 0.003125
z = 0 / 0.003125
z = 0
The z-score is 0. Now, we need to find the probability of getting a z-score greater than 0. Since the normal distribution is symmetric, we can equivalently find the probability of getting a z-score less than 0 and subtract it from 1.
P(z > 0) = 1 - P(z < 0)
To find the probability associated with a z-score, we can use a standard normal distribution table or the cumulative distribution function (CDF) of the standard normal distribution.
By using a standard normal distribution table or a calculator with a normal distribution function, we find that the probability of getting a z-score less than 0 (P(z < 0)) is 0.5.
Therefore, the probability of getting a z-score greater than 0 (P(z > 0)) is 1 - 0.5 = 0.5.
Hence, the probability that the mean volume of the sample will be more than a pint (0.564 liters) is 0.5 or 50%.