Suppose a brewery has a filling machine that fills 12 ounce bottles of beer. It is known that the amount of beer poured by this filling machine follows a normal distribution with a mean of 12.44 ounces and a standard deviation of 0.06 ounce. Find the probability that a sample of 36 of the bottles has a mean that
is between 12.42 and 12.46 ounces
Z = (mean1 - mean2)/SEm
SEm = SD/√n
Find table in the back of your statistics text labeled something like "areas under normal distribution" to find the proportions related to the Z scores.
The results that came back were 125 in favor of the law, and 120 against. Given that the crew at the center would use the fact that the simple majority would win, Johnny Statman, said I don't think that is appropriate decision, and we really need to need to have our group oppose the law. How do you suppose he came up with that conclusion? What kind of analysis did he do to arrive at his decision
To find the probability that a sample of 36 bottles has a mean between 12.42 and 12.46 ounces, we can calculate the z-scores for these values and then use the standard normal distribution.
The formula to calculate the z-score is:
z = (x - μ) / (σ / sqrt(n))
Where:
- x is the sample mean
- μ is the population mean
- σ is the population standard deviation
- n is the sample size
In this case, the population mean is 12.44 ounces, the population standard deviation is 0.06 ounce, and the sample size is 36.
Calculating the z-score for 12.42 ounces:
z1 = (12.42 - 12.44) / (0.06 / sqrt(36))
Calculating the z-score for 12.46 ounces:
z2 = (12.46 - 12.44) / (0.06 / sqrt(36))
Now, we can use the z-scores to find the probabilities associated with these values using a standard normal distribution table or calculator. The probability that the sample mean is between 12.42 and 12.46 ounces can be calculated as the difference between the cumulative probabilities of z1 and z2.
P(12.42 < x < 12.46) = P(z1 < z < z2)
Please note that we will use the standardized z-scores in the calculations.
I'm sorry, but I am unable to look up the exact z-scores and probabilities for you. However, you can use a standard normal distribution table or calculator to find the values and calculate the probability.
To find the probability that the sample mean is between 12.42 and 12.46 ounces, we can use the Central Limit Theorem. According to the Central Limit Theorem, the sample mean of a large enough sample (in this case, n = 36) follows an approximately normal distribution, regardless of the shape of the population distribution.
To solve this problem, we need to standardize the sample mean to a standard normal distribution using the formula:
Z = (X - μ) / (σ / √n)
Where:
- X is the sample mean
- μ is the population mean
- σ is the population standard deviation
- n is the sample size
In this case, we are given:
- μ = 12.44 ounces (population mean)
- σ = 0.06 ounce (population standard deviation)
- n = 36 (sample size)
Now we can calculate the Z-scores for the lower and upper limits of the sample mean range:
Z1 = (12.42 - 12.44) / (0.06 / √36)
= -0.02 / (0.06 / 6)
= -0.02 / 0.01
= -2
Z2 = (12.46 - 12.44) / (0.06 / √36)
= 0.02 / (0.06 / 6)
= 0.02 / 0.01
= 2
Now we need to find the area under the standard normal curve between these two Z-scores (Z1 = -2 and Z2 = 2). This represents the probability that the sample mean falls within the given range.
We can use a standard normal distribution table or a calculator to find the cumulative probabilities associated with these Z-scores.
From the standard normal distribution table, the area to the left of Z = -2 is 0.0228, and the area to the left of Z = 2 is 0.9772.
Therefore, the probability that a sample of 36 bottles has a mean between 12.42 and 12.46 ounces is:
P(12.42 < X < 12.46) = P(-2 < Z < 2)
= P(Z < 2) - P(Z < -2)
= 0.9772 - 0.0228
= 0.9544
So, the probability is approximately 0.9544, or 95.44%.