Question below. Thanks for the help.

Question

The population of weights of a particular fruit is normally distributed, with a mean of 406 grams and a standard deviation of 35 grams. If 9 fruits are picked at random, then 11% of the time, their mean weight will be greater than how many grams? Round your answer to the nearest gram.

Answer 1

Hi Jett,

This problem involves a version of the classic equation in elementary statistics z= (x-mu)/sigma. Here, we were given a probability along with mu and sigma, so we will need to find z based on that probability. We were told that 11% of the time, the mean weight would be greater than how many grams. Recall that on the z-table, probabilities displayed are for less than, i.e. to the left. That means we need to use the Complement Rule aka the "One Minus Trick"

1-0.11 = 0.89

Look in the interior of the z-table for 0.89. Closest probability I found was 0.8907. z_0.8907 = 1.23

Now, we have our z, mu, and sigma, but we're not done yet. You were given a sample, which means we need to compute another metric, standard error. Formula for this is:

SE= sigma/sqrt(n)

sigma=35

n=9

SE=35/sqrt(9)

SE=35/3

Now, we return to our initial formula z=(x-mu)/sigma but substitute the standard error in for sigma and change x to xbar, indicating sample mean:

z=(xbar-mu)/SE

z=1.23

xbar=sample mean we are interested in

mu=406

SE=35/3

1.23 = (xbar - 406) / (35/3)

Multiply both sides by (35/3)

14.35= xbar - 406

xbar= 14.35 + 406

xbar=420.35

To nearest gram:

xbar = 420g

I hope this helps.