The weight of corn chips dispensed into a 19-ounce bag by the dispensing machine has been identified as possessing a normal distribution with a mean of 19.5 ounces and a standard deviation of 0.1 ounce. Suppose 100 bags of chips are randomly selected. Find the probability that the mean weight of these 100 bags exceeds 19.6 ounces.
Choose one answer.
To find the probability that the mean weight of the 100 bags exceeds 19.6 ounces, we can use the Central Limit Theorem. According to the Central Limit Theorem, the distribution of sample means will approach a normal distribution as the sample size increases.
First, we need to calculate the standard deviation of the sample means. This is known as the standard error, which is equal to the standard deviation of the population divided by the square root of the sample size.
Standard error (SE) = standard deviation / √sample size
SE = 0.1 / √100
SE = 0.1 / 10
SE = 0.01
Now, we need to standardize the value of 19.6 ounces using the formula:
Z = (X - μ) / SE
where X is the value we want to standardize, μ is the mean of the population, and SE is the standard error.
Z = (19.6 - 19.5) / 0.01
Z = 0.1 / 0.01
Z = 10
Next, we need to find the probability that the standardized value exceeds 10 using a standard normal distribution table or a calculator. Since the value of Z is extremely high, the probability is very close to 0.
Therefore, the probability that the mean weight of these 100 bags exceeds 19.6 ounces is approximately 0.
To find the probability that the mean weight of these 100 bags exceeds 19.6 ounces, we need to calculate the Z-score and look up the corresponding probability in the standard normal distribution table.
First, we calculate the standard error of the mean using the formula:
Standard Error (SE) = Standard Deviation / √n
Where n is the sample size, which in this case is 100.
SE = 0.1 / √100 = 0.1 / 10 = 0.01
Next, we calculate the Z-score using the formula:
Z = (x - μ) / SE
Where x is the desired mean weight, μ is the population mean weight, and SE is the standard error.
Z = (19.6 - 19.5) / 0.01 = 1
Now, we look up the probability corresponding to a Z-score of 1 in the standard normal distribution table. The table gives us the probability for values below the Z-score, so we subtract the result from 1 to get the probability that the mean weight exceeds 19.6 ounces.
Using the standard normal distribution table, the probability corresponding to a Z-score of 1 is 0.8413. Therefore, the probability that the mean weight of these 100 bags exceeds 19.6 ounces is 1 - 0.8413 = 0.1587 or 15.87%.
So, the answer is approximately 15.87%.
No answers given, but here is the process.
Z = (mean1 - mean2)/standard error (SE) of difference between means
SEdiff = √(SEmean1^2 + SEmean2^2)
SEm = SD/√(n-1)
Since only one SD is provided, you can use just that to determine SEdiff.
Find table in the back of your statistics text labeled something like "areas under normal distribution" to find the proportion related to that Z score.