A particular fruit's weights are normally distributed, with a mean of 255 grams and a standard deviation of 10 grams.
The heaviest 9% of fruits weigh more than how many grams?
from the z-score table
the heaviest 9% are more than 1.34 σ above the mean
255 g + (1.34 * 10 g) = ? g
Let me see
To find the weight at which the heaviest 9% of fruits weigh more, we can use the z-score formula and reverse lookup in the standard normal distribution table.
The z-score formula is:
z = (x - μ) / σ
Where:
x = the value we want to find
μ = mean of the distribution
σ = standard deviation of the distribution
In this case, the mean (μ) is 255 grams and the standard deviation (σ) is 10 grams.
To find the z-score corresponding to the top 9% of the distribution, we need to find the z-value that has an area of 0.09 to the right. Since the standard normal distribution is symmetrical, we need to find the z-value that has an area of (1 - 0.09 = 0.91) to the left.
Next, we can use the inverse standard normal distribution table (also known as the Z-table) to find the z-value. The Z-table provides the cumulative probability up to a given z-value.
Using the Z-table, we can find the z-value that corresponds to a cumulative probability of 0.91. This z-value represents the weight at which 91% of the fruits weigh less.
After finding the z-value, we can use the z-score formula to solve for x.
Let's break down the solution step-by-step:
Step 1: Find the cumulative probability up to a z-value of 0.91 in the Z-table.
Step 2: Subtract that value from 1 to get the cumulative probability to the right of the z-value.
Step 3: Use the formula z = (x - μ) / σ to solve for x, with z = 0.91, μ = 255, and σ = 10.
Step 4: Calculate x to find the weight at which the heaviest 9% of the fruits weigh more.
Let's go ahead and calculate the answer: