A particular fruit's weights are normally distributed, with a mean of 228 grams and a standard deviation of 11 grams.
The heaviest 20% of fruits weigh more than how many grams?
Give your answer to the nearest gram.
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To find the weight at which the heaviest 20% of fruits weigh more, we need to determine the cutoff value (x) in the normal distribution.
1. Start by finding the z-score corresponding to the desired percentile (20%). The z-score represents the number of standard deviations a value is from the mean. We can use a standard normal distribution table or a calculator to find this value.
Looking up the z-score corresponding to the 20th percentile, we find z ≈ 0.8416.
2. Use the formula for the z-score to convert the z-score to an actual weight value (x).
z = (x - mean) / standard deviation
Rearranging the formula, we have:
x = mean + (z * standard deviation)
Plugging in the known values:
x = 228 + (0.8416 * 11)
x ≈ 228 + 9.2576
x ≈ 237.26
Therefore, the cutoff weight for the heaviest 20% of fruits is approximately 237 grams (rounded to the nearest gram).
To find the weight at which the heaviest 20% of fruits weigh more, we can use the concept of z-scores.
First, we need to find the z-score corresponding to the 20th percentile. The 20th percentile represents the point below which 20% of the fruit weights fall. The formula to calculate the z-score is:
z = (x - μ) / σ
where:
- z is the z-score
- x is the value we want to find (weight at the 20th percentile)
- μ is the mean (228 grams)
- σ is the standard deviation (11 grams)
Rearranging the formula to solve for x:
x = μ + (z * σ)
To find the z-score corresponding to the 20th percentile, we can use a standard normal distribution table or calculator. The closest z-score for the 20th percentile is approximately -0.84.
Now, we can substitute the values into the equation:
x = 228 + (-0.84 * 11) = 218.76
Rounding to the nearest gram, the weight at which the heaviest 20% of fruits weigh more is around 219 grams.