Probability (Statistics)

Question

A particular fruit's weights are normally distributed, with a mean of 529 grams and a standard deviation of 35 grams.

If you pick one fruit at random, what is the probability that it will weigh between 433 grams and 586 grams

Answer 1

Hi Andranique,

This problem uses the classic elementary statistics equation z=(x-mu)/sigma where:

z=z-score, explained below

x=value you are given

mu=mean

sigma=standard deviation

Here, we have two x values and we are asked to find the probability that x falls within that range, so I will use x₁ to denote the lower value and z₁ for corresponding z-score. For the higher value, I will use x₂ and z_2.

Thus:

z₁₌₍x₁-mu)/sigma

where:

x₁=433

mu=529

sigma=35

Substituting:

z₁=(433-529)/35

z₁= -2.74

Keep that number in mind; now let's find z₂^.

z₂= (x₂-mu)/sigma

where:

x₂=586

mu=529

sigma=35

z₂=(586-529)/35

z₂=1.63

Now, we must consult a z-score table. You can find these online , but Wyzant reviews posts when I include links, so I can't include a link here. Anyway, look at the left column for the whole number and tenths place--i.e. -2.7 for z₁--and the top column for the hundredths place--i.e. 0.04. From this, you can get our first probability, which I'll call P₁ that x<433:

P(Z<-2.74)=P(X<433)= 0.0031

Similarly, for P₂ from Z₂

P(Z<1.63)=P(X<529)=0.9484

To get the probability for the area between, simply subtract:

P(Z<1.63)-P(Z<-2.74)=P(X<529)-P(X<433)=

0.9484-0.0031=

P=0.9453

I hope this helps.