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.
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Just another homework dump?
Its not homework dump.There were three parts in question among which I did one but not these two!!
Help me to calculate the statistc
To find the probability that exactly 3 cartons in a sample of 10 cartons contain more than 1012 ml, we can use the concept of binomial distribution.
i. Binomial distribution is used when we have a fixed number of trials, each trial has two possible outcomes (success or failure), the probability of success is constant, and the trials are independent.
In this case, we have 10 trials (10 cartons), where success is defined as a carton containing more than 1012 ml. The probability of success can be calculated by finding the area under the normal distribution curve to the right of 1012 ml.
To find the probability of success, we need to convert the given mean and standard deviation to a standardized normal distribution (with mean 0 and standard deviation 1).
We can do this by calculating 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.
For z = (1012 - 1012) / 5 = 0, we get a z-score of 0.
Now, we need to find the probability of having a z-score greater than 0. Using a standard normal distribution table or a calculator, we find that the probability is 0.5 (or 50%).
Since we want exactly 3 cartons to have more than 1012 ml, we can use the binomial probability formula:
P(X = k) = (n choose k) * p^k * (1 - p)^(n - k)
where n is the number of trials, k is the number of successful outcomes, p is the probability of success, and (n choose k) denotes the number of ways to choose k successes from n trials.
In our case, n = 10, k = 3, and p = 0.5 (from the previous calculation).
Plugging in these values, we get:
P(X = 3) = (10 choose 3) * (0.5)^3 * (1 - 0.5)^(10 - 3)
Calculating this expression gives us the probability that exactly 3 cartons contain more than 1012 ml.
ii. To estimate the number of cartons in a batch of 1000 cartons that contain less than the advertised volume of milk, we can use the normal distribution.
First, we need to calculate the z-score for the advertised volume of milk, which is 1000 ml. Using the formula for z-score as mentioned before, we get:
z = (1000 - 1012) / 5 = -2.4
Now, we need to find the probability of having a z-score less than -2.4. Using a standard normal distribution table or a calculator, we find this probability to be approximately 0.0082.
Since this probability represents the proportion of cartons with less than the advertised volume, we can estimate the number of cartons by multiplying this probability by the total number of cartons in the batch:
Estimated number of cartons = 0.0082 * 1000
Calculating this expression gives us the estimated number of cartons in the batch that contain less than the advertised volume of milk.