2 Packets of crisps from a production line have a weight distribution that is normal with a mean of 50.0 grams and a standard deviation of 2.0 grams. Samples are taken of size 36.
1. What is the probability that a sample mean is less than 49.0 grams? (Give answer to four decimal places.)
2. What would be the control limits for a three standard error control chart in these circumstances? Give your answer in the format (to the nearest gram):
3. What proportion of sample means would be outside the these control limits? (Give your answer as a percentage, accurate to 0.1%, leaving out the % sign when you give the answer).
only need help with part 3
Normal distribution: 99.7% is approximately +/- 3 standard deviations around the mean. 0.3% is outside those limits.
To find the proportion of sample means that would be outside the control limits, we need to calculate the z-scores for the control limits based on the given information.
First, let's calculate the standard error of the sample mean (SE). The formula for standard error is:
SE = σ / √n
Where σ is the standard deviation of the population and n is the sample size.
SE = 2.0 / √36 = 2.0 / 6 = 0.3333
Next, we can calculate the z-score for a three standard error control limit. The formula for z-score is:
z = (x - μ) / SE
Where x is the value of interest, μ is the mean of the population, and SE is the standard error.
For the lower control limit:
z_lower = (x - μ) / SE = (49.0 - 50.0) / 0.3333 = -3.0
For the upper control limit:
z_upper = (x - μ) / SE = (50.0 - 50.0) / 0.3333 = 0.0
Now, let's find the proportions of sample means that would be outside these control limits.
The proportion of sample means that would be below the lower control limit (z < -3.0) is very close to 0 since the normal distribution is symmetric around the mean. For practical purposes, we can consider it to be effectively 0.
The proportion of sample means that would be above the upper control limit (z > 0.0) is also very close to 0 for the same reasons.
So, the proportion of sample means outside these control limits would be approximately 0%.
Please note that this conclusion is based on the assumption that the distribution of sample means follows a normal distribution.