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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. What is the probability that a sample mean is less than 49.0 grams? (Give answer to four decimal places.)
To find the probability that a sample mean is less than 49.0 grams, we can use the Central Limit Theorem.
The Central Limit Theorem states that for a sample size n greater than or equal to 30, the distribution of sample means will be approximately normal, regardless of the shape of the population distribution.
To apply the Central Limit Theorem, we need to calculate the z-score and then find the probability using a standard normal distribution table or calculator.
The formula to calculate the z-score is:
z = (x - μ) / (σ / √n)
Where:
x is the sample mean (49.0 grams)
μ is the population mean (50.0 grams)
σ is the population standard deviation (2.0 grams)
n is the sample size (36)
Plugging in the values:
z = (49.0 - 50.0) / (2.0 / √36)
z = -1.0 / (2.0 / 6.0)
z = -1.0 / 0.3333
z ≈ -3.0001
Now, we can use a standard normal distribution table or calculator to find the probability associated with a z-score of -3.0001.
Looking up the z-score in the table or using a calculator, we find that the probability (P) is approximately 0.0013.
Therefore, the probability that a sample mean is less than 49.0 grams is approximately 0.0013 (or 0.13% if expressed as a percentage), rounded to four decimal places.