The weights of cans of fruit are normally distributed with a mean of 1,000 grams and a standard deviation of 50 grams. What portion of the cans weigh 860 grams or less?
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A. B. C. D. E.E
z = (x-u)/sigma = 860 - 1000/50 = -2.8. For z = 2.8, the area is 0.4974. So 1.0 - 0.9974 = 0.0026.
To determine the portion of cans that weigh 860 grams or less, we need to calculate the probability that a randomly selected can from this distribution weighs 860 grams or less.
Given that the weights of the cans are normally distributed with a mean of 1,000 grams and a standard deviation of 50 grams, we can use the properties of the normal distribution to solve this problem.
The first step is to standardize the value of 860 grams using the z-score formula:
z = (X - μ) / σ
where X is the value we want to standardize, μ is the mean, and σ is the standard deviation.
Plugging in the values:
z = (860 - 1000) / 50 z = -2.8
Now, we need to find the cumulative probability (area under the curve) to the left of z = -2.8. We can consult a standard normal distribution table or use a statistical calculator to find this probability.
Looking up the z-score of -2.8 in a standard normal distribution table, we find that the corresponding cumulative probability is approximately 0.0026.
Therefore, the portion of cans that weigh 860 grams or less is 0.0026, which corresponds to answer choice E.