The population of weights of a particular fruit is normally distributed, with a mean of 766 grams and a standard deviation of 14 grams. If 8 fruits are picked random, then 20% of the time

Question

The population of weights of a particular fruit is normally distributed, with a mean of 766 grams and a standard deviation of 14 grams. If 8 fruits are picked random, then 20% of the time, their mean weight will be greater than how many grams?

Answer 1

Hi Emma,

Formula for this is:

z = (xbar - mu)/SE; SE = standard error = sigma/sqrt(n)

Now, you want greater than for 20% of the time--which means you want the 80th percentile. Since fruit weights are normally distributed, you can check the interior of the z-table for 80%, or 0.8. There are also options to do this using software. Anyway, closest interior probability on my z-table is 0.7995; corresponding z-score is:

z_0.8 = 0.84

Now, we want xbar, the sample mean, but we need standard error (SE) to get it:

SE= sigma/sqrt(n)

sigma= standard deviation = 14

n= sample size = 8

SE= 14/sqrt(8) = 4.95

Now, we can substitute into initial formula:

z= 0.84

xbar= xbar, weight we are looking for

mu= population mean = 766

SE = 4.95

Substituting:

0.84 = (xbar - 766) / 4.95

4.158 = xbar - 766

xbar = 770.158, to one decimal place;

xbar = 770.2

I hope this helps.