Find table in the back of your statistics text labeled something like "areas under normal distribution" to find the proportion/probability (.18) and its Z score.
Z = (score-mean)/SD
If you pick a fruit at random, then 18% of the time, their mean weight will be greater than how many grams?
Z = (score-mean)/SD
The Z-score formula is: Z = (X - μ) / σ
Where:
- Z is the Z-score
- X is the raw score
- μ is the mean
- σ is the standard deviation
In this case, the mean is 594 grams (μ), and the standard deviation is 16 grams (σ).
We need to find X, so we rearrange the formula:
X = Z * σ + μ
To find the Z-score corresponding to 18% of the area under the normal distribution curve, we can use a Z-table or a calculator. Looking up the closest Z-score to 0.18, we find it to be approximately Z = 0.91.
Now we can substitute the values into the formula:
X = 0.91 * 16 + 594
X ≈ 14.56 + 594
X ≈ 608.56
Therefore, if you pick a fruit at random, then 18% of the time, their mean weight will be greater than approximately 608.56 grams.
Step 1: Convert the percentage to a z-score.
The percentage indicates the area under the curve to the right of a particular z-score. We want to find the z-score that corresponds to the area to the left of it being 82% (100% - 18%).
The z-score can be found using a standard normal distribution table or a calculator. In this case, we want the z-score for the left-tailed area of 82%, which corresponds to a z-score of approximately 0.91.
Step 2: Find the weight value corresponding to the z-score.
To determine the weight value, we need to use the formula for converting a z-score to a raw score:
z = (X - μ) / σ
Where:
X is the raw score (weight in grams)
μ is the population mean (mean weight of the fruit)
σ is the population standard deviation (standard deviation of the fruit's weights)
Rearranging the formula to solve for X:
X = z * σ + μ
Substituting the values:
X = 0.91 * 16 + 594
X = 14.56 + 594
X ≈ 608.56
Therefore, if you pick a fruit at random, then 18% of the time, their mean weight will be greater than approximately 608.56 grams.