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Suppose in an orchard the number of apples in a tree is normally distributed with a mean of 300 and a standard deviation of 30 apples. Find the probability that a given tree has between 300 and 390 apples.

User Bdukes
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To find the probability that a given tree has between 300 and 390 apples, we need to calculate the area under the normal distribution curve between these two values. Since the number of apples is normally distributed with a mean (μ) of 300 and a standard deviation (σ) of 30, we can use the Z-score formula to standardize the values and then find the probability using standard normal distribution tables or a calculator.

The Z-score formula is given by:
Z = (X - μ) / σ

where X is the value we are interested in, μ is the mean, and σ is the standard deviation.

For the lower value (X = 300):
Z_lower = (300 - 300) / 30 = 0

For the upper value (X = 390):
Z_upper = (390 - 300) / 30 = 3

Now, we need to find the probability that Z is between 0 and 3. Using standard normal distribution tables or a calculator, we find:

P(0 ≤ Z ≤ 3) ≈ 0.4987

So, the probability that a given tree has between 300 and 390 apples is approximately 0.4987 or 49.87%.
User Aaditya Raj
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