A particular fruit's weights are normally distributed, with a mean of 380 grams and a standard deviation of 16 grams.
The heaviest 8% of fruits weigh more than how many grams?
Give your answer to the nearest gram.
To find the weight above which the heaviest 8% of fruits fall, we need to identify the z-score that corresponds to the 92nd percentile (since 100% - 8% = 92%). This is because 92% of the fruit will weigh less than this value, and the remaining 8% will weigh more.
We can use a standard normal distribution table or a calculator with inverse cumulative distribution function capabilities to find this z-score. The z-score corresponding to the 92nd percentile is typically around 1.41, but we will calculate it using the inverse cumulative distribution function.
Using statistical software or a scientific calculator, we can find the exact z-score. For instance, in a standard statistical software, you may use a command similar to `qnorm(0.92)` or `invNorm(0.92)` to find the z-score.
Suppose the z-score we found is approximately \( Z = 1.41 \).
To find the weight corresponding to this z-score, we use the formula for the z-score in terms of the population mean (μ) and the standard deviation (σ):
\[ Z = \frac{X - \mu}{\sigma} \]
Where:
- \( X \) is the weight that corresponds to the z-score.
- \( \mu = 380 \) grams (the mean weight).
- \( \sigma = 16 \) grams (the standard deviation).
Rearranging the formula to solve for \( X \) gives us:
\[ X = Z \cdot \sigma + \mu \]
Now we substitute our known values into the equation:
\[ X = 1.41 \times 16 + 380 \]
\[ X = 22.56 + 380 \]
\[ X = 402.56 \]
Rounded to the nearest gram, the heaviest 8% of the fruits weigh more than 403 grams.