The edible portion of grape clusters

Question

The edible portion of grape clusters from a Mexican farm were measured to have a mean weight of 429.5 g and a standard deviation of 42.8 g. For selling purposes, the clusters are divided into two groups:

 

Small - the lightest 20%

Regular - the largest 80%

 

a.) what mass will divide the small and regular groups?

 

b.) If 20 000 clusters are picked, how many do you expect will be over 500 g?

 

c.? What assumption was made to answer the previous questions? 

Answer 1

Hello,

To calculate probabilities for a distribution, it is a good idea to convert to z-scores first. Any raw score can be converted to a z-score with this formula: z = (X - M)/SD, where X is the raw score, M is the mean and SD is the standard deviation.

1) This question requires us to find the z-score that separates the top 80% from the bottom 20%. To do this, we need to use the normal distribution table (your teacher should have given you one or it should be in your textbook). If we consult the table, we see that a z-score of -0.8416 gives us the 80/20 split. Now we have to convert this z-score into mass using the z-score formula. First we plug in the numbers that we know: -0.8416 = (X - 429.5)/42.8. Next, we multiply both sides by 42.8: -36.021 = X - 429.5. Then we add 429.5 to both sides: 393.479 = X.

2) For this one, we need to figure out what percentage of the distribution is greater than 500g. Again, we want to use z-scores: z = (500 - 429.5)/42.8. This gives us 70.5/42.8 = 1.6472. So the z-score for 500 grams is 1.6472. Now we use the normal distribution table to figure out what percentage of the distribution is above 1.6472 - which turns out to be 4.976% (5% more or less depending on how many decimal places your teacher wants).

 

3) To calculate these probabilities using the normal distribution table, we have to assume that the mass of grape clusters is normally distributed.

 

Hope that helps