Cartons of milk from a particular supermarket are advertised as containing 1 litre of milk,but in fact the volume of the milk in a cartoon is normally distributed with mean 1012 ml and standard deviation 5 ml.
i.Find the probability that exactly 3 cartons in a sample of 10 cartons contain more than 1012 ml.
ii.Estimate how many cartons in a batch of 1000 cartons contain less than the advertised volume of milk.
i. Well, if I were a carton of milk, I'd be pretty upset about being advertised as containing 1 litre when in reality I only have a mean volume of 1012 ml. But hey, life's full of disappointments!
To find the probability that exactly 3 cartons in a sample of 10 contain more than 1012 ml, we need to use the concept of a normal distribution. We'll need to calculate the z-score for each carton, based on the mean and standard deviation provided.
First, let's find the z-score for 1012 ml:
z = (1012 - 1012) / 5
z = 0
Since we want the probability of having more than 1012 ml, we'll need to find the probability that the z-score is greater than 0. Using a standard normal distribution table or a calculator, we can look up the probability associated with a z-score of 0. Let's assume it's 0.5.
Now, we'll find the probability that a single carton has more than 1012 ml:
1 - 0.5 = 0.5
To find the probability that exactly 3 cartons in a sample of 10 contain more than 1012 ml, we need to use the binomial distribution formula. Let's assume that the probability of a single carton having more than 1012 ml is 0.5 (as calculated above).
P(X = 3) = (10 C 3) * (0.5)^3 * (1 - 0.5)^(10 - 3)
Using the binomial coefficient (10 C 3) or "10 choose 3," we can calculate the probability of exactly 3 successes out of 10 trials.
ii. Now, let's estimate how many cartons in a batch of 1000 contain less than the advertised volume of milk. Assuming the volume of milk follows a normal distribution with a mean of 1012 ml and a standard deviation of 5 ml, we can use the z-score formula to find the probability of a single carton containing less than the advertised volume.
Using the z-score formula: z = (x - μ) / σ
where x is the threshold value (1000 ml), μ is the mean (1012 ml), and σ is the standard deviation (5 ml).
z = (1000 - 1012) / 5
z = -2.4
Now, we can find the probability of a single carton having less than the advertised volume by looking up the probability associated with a z-score of -2.4 in a standard normal distribution table or using a calculator. Let's say it's 0.0082.
To estimate the number of cartons in a batch of 1000 containing less than the advertised volume, we can multiply the probability by the total number of cartons:
Estimated number = 0.0082 * 1000
Keep in mind that this is just an estimate, and the actual number may vary.
To solve these problems, we will use the normal distribution and the z-score formula.
i. Find the probability that exactly 3 cartons in a sample of 10 cartons contain more than 1012 ml.
Step 1: Calculate the z-score for 1012 ml using the formula: z = (x - μ) / σ
where x is the given value, μ is the mean, and σ is the standard deviation.
z = (1012 - 1012) / 5 = 0
Step 2: Find the cumulative probability for this z-score using a standard normal distribution table or a calculator. In this case, the area to the right of the z-score is needed because we're interested in "more than" 1012 ml.
P(z > 0) = 0.5
Step 3: Now, calculate the probability of exactly 3 cartons out of 10 containing more than 1012 ml. We can use the binomial distribution formula:
P(X = k) = (n choose k) * p^k * (1 - p)^(n - k)
where n is the total number of trials (10 cartons), k is the desired number of successes (3 cartons), and p is the probability of success (0.5).
P(X = 3) = (10 choose 3) * 0.5^3 * (1 - 0.5)^(10 - 3)
= 120 * 0.125 * 0.5^7
= 0.02734375
ii. Estimate how many cartons in a batch of 1000 cartons contain less than the advertised volume of milk.
Step 1: Calculate the z-score for the advertised volume of milk using the formula: z = (x - μ) / σ
In this case, x is 1000 ml (the advertised volume), μ is the mean (1012 ml), and σ is the standard deviation (5 ml).
z = (1000 - 1012) / 5 = -2.4
Step 2: Find the cumulative probability for this z-score using a standard normal distribution table or a calculator. In this case, the area to the left of the z-score is needed because we're interested in "less than" the advertised volume.
P(z < -2.4) = 0.008197
Step 3: Multiply the probability obtained from Step 2 with the total number of cartons in the batch:
Estimate = 0.008197 * 1000
= 8.197
Therefore, an estimate is that around 8 cartons in a batch of 1000 cartons contain less than the advertised volume of milk.
To find the probability, we need to use the concept of the standard normal distribution. We can convert the problem into a standard normal distribution problem by standardizing using the formula:
Z = (X - μ) / σ
where Z is the standard score, X is the observed value, μ is the mean, and σ is the standard deviation.
Now let's solve the two parts of the question:
i. Find the probability that exactly 3 cartons in a sample of 10 cartons contain more than 1012 ml.
Step 1: Convert the value of 1012 ml to a standard score.
Z = (1012 - 1012) / 5 = 0
Step 2: Calculate the probability of a single carton containing more than 1012 ml.
P(Z > 0) = 0.5 - 0.5 = 0.5
Step 3: Calculate the probability of exactly 3 out of 10 cartons containing more than 1012 ml. Since each carton is independent, we can use the binomial distribution formula:
P(X = k) = C(n, k) * p^k * (1 - p)^(n - k)
where n is the total number of trials, k is the number of successful trials, p is the probability of a successful trial (in our case, P(Z > 0) = 0.5), and C(n, k) is the binomial coefficient (number of combinations).
P(X = 3) = C(10, 3) * 0.5^3 * (1 - 0.5)^(10 - 3)
Using a binomial calculator or table, we find C(10, 3) = 120.
P(X = 3) = 120 * 0.5^3 * (1 - 0.5)^(10 - 3) ≈ 0.117
Therefore, the probability that exactly 3 cartons in a sample of 10 cartons contain more than 1012 ml is approximately 0.117 or 11.7%.
ii. Estimate how many cartons in a batch of 1000 cartons contain less than the advertised volume of milk.
Since the volume of milk in a carton is normally distributed, we can find the probability that a single carton contains less than the advertised volume and then use it to estimate the number of cartons out of 1000.
Step 1: Convert the value of 1012 ml to a standard score.
Z = (1012 - 1012) / 5 = 0
Step 2: Calculate the probability of a single carton containing less than the advertised volume.
P(Z < 0) = 0.5
Step 3: Multiply the probability by the total number of cartons to estimate the number of cartons containing less than the advertised volume.
Number of cartons = P(Z < 0) * 1000 = 0.5 * 1000 = 500
Therefore, we estimate that approximately 500 cartons out of a batch of 1000 cartons contain less than the advertised volume of milk.
the answers are in the link below and there is space between the h and the ttps in the link below:
h ttps://math.berkeley.edu/~qadeer/teaching/Math16B/Homework%207%20Solutions.pdf