An automatic machine that fills bags of unpopped popcorn is operating properly if the weights are independently and normally distributed with a mean of 114 grams and a standard deviation of 4.9 grams. Find the probability that if 7 bags are randomly selected, their mean weight exceeds 115 grams.
Use z-scores:
z = (x - mean)/(sd/√n)
With your data:
z = (115 - 114)/(4.9/√7)
Finish the calculation.
Next, check a z-table to find probability using the above z-score.
I hope this will help get you started.
To find the probability that the mean weight of 7 bags exceeds 115 grams, we need to calculate the probability of the sample mean being greater than 115 grams.
The mean of the sampling distribution of the sample mean is equal to the population mean, which is 114 grams in this case. The standard deviation of the sampling distribution of the sample mean, also known as the standard error, can be calculated by dividing the population standard deviation by the square root of the sample size.
In this case, the standard deviation of the sampling distribution (standard error) is:
standard error = standard deviation / √sample size
= 4.9 / √7
≈ 1.851
Now, we need to convert the original problem into a standard normal distribution since we know the mean and standard deviation of the sampling distribution.
To do that, we can use the formula for the z-score:
z = (x - μ) / σ
Where:
- x is the value we want to convert,
- μ is the mean of the sampling distribution, and
- σ is the standard deviation of the sampling distribution.
In this case, we want to find the probability that the sample mean exceeds 115 grams, so x = 115.
Now, we can calculate the z-score:
z = (115 - 114) / 1.851
≈ 0.540
To find the probability of the sample mean exceeding 115 grams, we need to find the area under the standard normal curve to the right of the z-score.
Using a standard normal distribution table or a calculator, we can find that the probability corresponding to a z-score of 0.540 is approximately 0.2946.
Therefore, the probability that if 7 bags are randomly selected, their mean weight exceeds 115 grams is approximately 0.2946, or 29.46%.