A machine is used to fill cans of soup with nominal volume of 0.5 litres.The quantity of soup actually delivered is normally distributed with mean m litres and standard deviation s litres.It is required that no more than 1% of cans should contain less than the nominal volume.i) Find the least value of m which will comply with the requirement when m=0.003 ii)F
II)FIND THE GREATEST VALUE OF M WHICH WILL COMPLY WITH THE REQUIREMENT WHEN M=0.506
i) To find the least value of m which will comply with the requirement, we need to determine the value of m such that no more than 1% of cans contain less than the nominal volume.
Since the quantity of soup actually delivered is normally distributed with mean m liters and standard deviation s liters, we can use the z-score formula to find the corresponding z-score.
The z-score formula is given by:
z = (x - μ) / σ
Where:
- x is the observed value (in this case, the nominal volume of 0.5 liters)
- μ is the mean (m liters)
- σ is the standard deviation (s liters)
- z is the z-score
To find the z-score corresponding to the lower 1% (0.01) of the distribution, we need to find the z-score that has a cumulative probability of 0.01. We can use a standard normal distribution table or a statistical calculator to find this value.
Once we have the z-score, we can solve for m as follows:
m = x - z * σ
Let's plug in the values:
x = 0.5 liters (nominal volume)
σ = s liters (standard deviation)
z = z-score corresponding to 0.01 cumulative probability
ii) To find the z-score corresponding to a cumulative probability of 0.01, we can use a standard normal distribution table or a statistical calculator.
Please provide the value of s (standard deviation) and I will proceed with the calculations.
To find the least value of m that will comply with the requirement, we need to determine the minimum value of m such that less than 1% of cans contain less than the nominal volume of 0.5 liters.
First, we need to convert the problem into a standard normal distribution. We can do this by calculating the z-score, which measures how many standard deviations a particular value is from the mean.
The formula for z-score is: z = (x - μ) / σ
Where:
z = z-score
x = observed value
μ = mean
σ = standard deviation
In this case, we want to find the value of m such that P(X < 0.5) < 0.01, where X represents the delivered quantity.
Step 1: Calculate the z-score for the given condition:
P(X < 0.5) = 0.01
Using the z-score formula:
z = (0.5 - m) / s
Step 2: Find the critical z-value for P(Z < z) = 0.01.
We can use a standard normal distribution table or a statistical calculator to find the critical z-value. The critical z-value for P(Z < z) = 0.01 is approximately -2.33.
Step 3: Set up the inequality:
z = (0.5 - m) / s < -2.33
Step 4: Solve the inequality for m:
0.5 - m < -2.33s
-m < -2.33s - 0.5
m > 2.33s + 0.5
Therefore, the least value of m that will comply with the requirement is m > 2.33s + 0.5.
For the second part of your question, the information provided is incomplete. If you can provide the value of m or any other specific information, I will be happy to assist you further.