If you pick 4 fruit at random, what is the probability that their mean weight will be between 769 grams and 771 grams

Question

A particular fruit's weights are normally distributed, with a mean of 727 grams and a standard deviation of 26 grams.

If you pick 4 fruit at random, what is the probability that their mean weight will be between 769 grams and 771 grams

Answer 1

Hi Kaitlyn,

This problem involves a variation on the classic statistics equation z=x-mu/sigma. The difference is that you are collecting a sample of 4, which means we need to convert sigma to the sample standard error. This formula is:

SE=s/sqrt(n) where:

SE=standard error

s=standard deviation

n=sample size

Here,

s=26

n=4

SE=26/sqrt(40

SE=26/2

SE=13

Now, we can substitute 13 for sigma into the classic equation:

z=(x-mu)/sigma

In this case:

z=(x-x-bar)/SE

where:

x=weight you were given

x-bar=mean

SE=standard error

Now, you were given two values and told to find the probability that 4 fruits would fall between them, so I will call the first z-score z₁ and the second z₂. I will do the same for x₁ and x₂. Thus:

z₁=(x₁-mu)/SE

x₁=769

mu=727

SE=13

z₁=(769-727)/13

z₁=3.23

Keep that in mind. Now, let's find z₂

x₂=771

mu=727

SE=13

z₂=(771-727)/13

z₂=3.38

Now, looking at the z-table for 3.23 and 3.38, we can get P₁, the probability that the four fruits will weigh less than 769g on average, and P₂, the probability that they will weigh less than 771g on average. Look at the column on the right for 3.2 and 3.3 and the row at the top for 0.03 and 0.08 respectively. You will find:

P₁=0.9994=P(X<769)

P₂=0.9996=P(X<771)

To get the probability that the mean weight of four fruits fall within this range, subtract:

P(X<771)-P(X<769)=

0.9996-0.9994=

P=0.0002

This is extremely low, which makes sense. Recall that population mean was 727g, so we're talking about very heavy fruits if we want between 769 and 771g, which would be rare with that mean. I hope this helps. Good luck.