Finding the percentage of a normal distribution using the mean and std deviation.

Question

The grapefruit grown in a large orchard have a mean weight of 18.2 ounces with a standard deviation of 1.2 ounces. Assuming that the distribution of the weights of these grapefruit is normal. Find a) the percentage of grapefruit that weigh less than 16.1 ounces; b) the percentage of grapefruit that weigh anywhere from 16.7 to 18.8 ounces; c) the weight above which we will find the heaviest 15 percent of the grapefruit.

My real problem is finding c

Answer 1

The real key to answering this question is by understanding that finding the weight above which you find the heaviest 15% of grapefruits is equivalent to saying that 85% of weights would be below that weight. That being said, it's best to start this problem using the standard normal distribution.

So first we need the 85 percentile of z-scores, or the z-score in which 85% of other z-scores are lower than. So we need to find P(z<x)=.85.

We find in the table that z=1.04.

Now we have our z-score, 1.04. Now we just need the x value that gets us our z score, since we already have mean and standard deviation of the weights.

Z=x-u/sigma => 1.04=x-18.2/1.2

1.2*1.04=x-18.2

1.248=x-18.2

18.2+1.248=x

19.448=x

So 19.448 is the weight in which you would have to bypass in order to find the heaviest 15% of grapefruits. I hope this helped.