help solving it

Question

A particular fruit's weights are normally distributed, with a mean of 387 grams and a standard deviation of 32 grams.

The heaviest 9% of fruits weigh more than how many grams?

Give your answer to the nearest 0.1 gram.

Answer 1

Hi Lakotah,

What we're looking for is the weight W such that Pr (weight > W) = 9%.

That's equivalent to Pr(weight < W) = 91%

Since the weights are distributed normally, we can use a standard Normal Z-table to find the Z-value such that Pr( x < Z) = 91%.

Using a Z-table, we can find that Z is approximately 1.34 (more precisely, the standard normal has a cumulative probability of 0.91 at a Z-score of about 1.340755)

Then, using the definition of Z-score as the number of standard deviations above the mean, we could solve for the weight W.

Z = (W - mean) / sd

1.341 = (W - 387) / 32

I'll leave the final steps to you.

I hope this is helpful.

-Josh

Answer 2

mean 387

standard deviation = 32

heaviest 9%=.09 weigh at least how many grams?

z= (x-387)/32

use a z calculator to find z corresponding to area of .09 in one tail, or use z tables

then solve for x above

1.341 = (x-387)/32

x = 32(1.341) + 387 = 429.912

= about 429.9 grams or more for top 9% heaviest