A certain brand of candies have a mean weigh of 0.8576 g and a standard deviation of 0.0523. A sample of these candies came from a package containing 447 candies,and the package label stated the net weight is 381.3 g. (If every package has 447 candies, the mean weight of the candies must exceed 381.3/447 = 0.8531 g for the net contents to weigh at least 381.3 g.) a. If 1 candy is randomly selected, find the probability that it weighs more tan 0.8531 g. The probability is ____.b. If 447 candies are randomly selected, find the probability that their mean weight is at least 0.8531 g. The probability that a sample of 447 candies will have a mean of 0.8531 g or greater is ____. c. Given these results, does it seem that the candy company is providing consumers with the amount claimed on the label?Yes/No because the probability of getting a sample mean of 0.8531 g or greater when 447 candies selected, is not/is exceptionally small.
a. See process in later post.
b. Z = (score-mean)/SEm
SEm = SD/√n
Use same table.
C. Judge by your results.
what does SEm mean?
a. To find the probability that a randomly selected candy weighs more than 0.8531 g, we need to calculate the z-score and then find the corresponding probability using a standard normal distribution table.
First, calculate the z-score:
z = (x - μ) / σ
where x = 0.8531 g (the lower weight limit), μ = 0.8576 g (mean weight), and σ = 0.0523 (standard deviation).
z = (0.8531 - 0.8576) / 0.0523
z = -0.0045 / 0.0523
z ≈ -0.0861
Now, find the probability using the standard normal distribution table or a calculator.
The probability of getting a z-score less than -0.0861 (which corresponds to a weight greater than 0.8531 g) is approximately 0.4641.
Therefore, the probability that a randomly selected candy weighs more than 0.8531 g is approximately 0.4641.
b. To find the probability that the mean weight of 447 randomly selected candies is at least 0.8531 g, we need to calculate the z-score for the sample mean.
The mean of the sample mean is the same as the population mean, which is 0.8576 g.
The standard deviation 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:
standard deviation of the sample mean (standard error) = σ / sqrt(n)
where σ = 0.0523 (standard deviation of the population) and n = 447 (sample size).
standard deviation of the sample mean = 0.0523 / sqrt(447)
standard deviation of the sample mean ≈ 0.00247
Now, calculate the z-score for the sample mean:
z = (x̄ - μ) / σ
where x̄ = 0.8531 g (the lower weight limit), μ = 0.8576 g (mean weight), and σ = 0.00247 (standard deviation of the sample mean).
z = (0.8531 - 0.8576) / 0.00247
z = -0.0045 / 0.00247
z ≈ -1.8216
Now, find the probability using the standard normal distribution table or a calculator.
The probability of getting a z-score less than -1.8216 (which corresponds to a sample mean of 0.8531 g or greater) is approximately 0.0349.
Therefore, the probability that a sample of 447 randomly selected candies will have a mean weight of 0.8531 g or greater is approximately 0.0349.
c. Yes, it seems that the candy company is providing consumers with the amount claimed on the label because the probability of getting a sample mean of 0.8531 g or greater when 447 candies are selected is not exceptionally small. The probability is approximately 0.0349, which is not a very low probability.
To answer these questions, we will be using the concept of the normal distribution.
a. To find the probability that a candy weighs more than 0.8531 g, we need to calculate the area under the normal distribution curve to the right of 0.8531 g. We can use the mean weight (0.8576 g) and the standard deviation (0.0523) to standardize the values.
First, we calculate the z-score:
z = (x - mean) / standard deviation
z = (0.8531 - 0.8576) / 0.0523
Using a standard normal distribution table or a calculator, we can find the area to the right of this z-score. Let's assume the area is denoted by P.
Therefore, the probability that a randomly selected candy weighs more than 0.8531 g is 1 - P.
b. To find the probability that the mean weight of 447 candies is at least 0.8531 g, we can use the central limit theorem. According to this theorem, if we have a large sample size (in this case, 447), the distribution of the sample means will approach a normal distribution with the same mean as the population and a standard deviation equal to the population standard deviation divided by the square root of the sample size.
The standard deviation of the sample mean can be calculated using the formula:
standard deviation of sample mean = standard deviation / √(sample size)
standard deviation of sample mean = 0.0523 / √(447)
Now, we can use the same process as in part a. Calculate the z-score using the formula:
z = (x - mean) / (standard deviation of sample mean)
z = (0.8531 - 0.8576) / (0.0523 / √(447))
Using a standard normal distribution table or a calculator, we can find the area to the right of this z-score. Let's assume the area is denoted by P'.
Therefore, the probability that a sample of 447 candies will have a mean weight of 0.8531 g or greater is P'.
c. If the probability in part b (P') is exceptionally small, then it suggests that it is highly unlikely to obtain a sample mean of 0.8531 g or greater from the population. In this case, if the company claims that the mean weight of the candies is at least 0.8531 g, it does not seem that they are providing consumers with the amount claimed on the label.
However, if the probability is not exceptionally small, it suggests that observing a sample mean of 0.8531 g or greater is reasonably likely, and we cannot conclude that the company is not providing consumers with the amount claimed on the label.
Therefore, to answer part c, you need to compare the calculated probability P' with a predefined threshold (e.g., 0.05). If P' is less than 0.05, you would conclude that the company is not providing consumers with the claimed amount on the label. Otherwise, if P' is greater than or equal to 0.05, you would conclude that the company is providing consumers with the claimed amount on the label.