A particular fruit's weights are normally distributed, with a mean of 308 grams and a standard deviation of 31 grams.
The heaviest 9% of fruits weigh more than how many grams?
Give your answer to the nearest gram.
How did you get 349?
I checked the Z table and still don't get how you end up with 349
so 91 percent weight X or below.
I get about 349 grams.
check it here.
http://davidmlane.com/hyperstat/z_table.html
Well, I guess these fruits could be considered "heavyweights" in the fruit kingdom! To find the weight at which the heaviest 9% of fruits fall, we can use the z-score formula. We know that the z-score for the 91st percentile is going to be approximately 1.34.
Using the formula z = (x - μ) / σ, where z is the z-score, x is the raw score, μ is the mean, and σ is the standard deviation, we can rearrange the formula to solve for x:
x = (z * σ) + μ
Plugging in the values, we get:
x = (1.34 * 31) + 308
Calculating this, we find:
x ≈ 49.54 + 308
So the weight at which the heaviest 9% of fruits fall is approximately 357.54 grams. Since you want the answer to the nearest gram, let's round that off to 358 grams!
These fruits are truly "weighty" fellows!
To find the weight at which the heaviest 9% of fruits exceed, we need to determine the z-score associated with the 9th percentile.
First, let's find the z-score using the standard normal distribution table.
The 9th percentile corresponds to the area to the left of it, which is 0.09.
Using the standard normal distribution table (or a calculator), we find the z-score associated with 0.09 is approximately -1.34.
Next, we can use the z-score formula to find the corresponding weight value.
z = (x - μ) / σ
Where:
z = z-score
x = weight value
μ = mean of the distribution
σ = standard deviation of the distribution
Rearranging the formula, we can solve for x:
x = z * σ + μ
Plugging in the values:
x = -1.34 * 31 + 308
x ≈ 263.34
Rounding to the nearest gram, the heaviest 9% of fruits weigh more than approximately 263 grams.
You can also find table in the back of your statistics text labeled something like "areas under normal distribution" to find that proportion related to a Z score.
Z = (score-mean)/SD
Insert values and solve for score.