A particular fruit's weights are normally distributed, with a mean of 493 grams and a standard deviation of 17 grams. The heaviest 19% of fruits weigh more than how many grams?

Question

I am confused as to what the percentage represents and how that helps me.

Answer 1

"The heaviest 19% weigh more than A grams" is the same as saying "81% of fruits weigh less than or exactly A grams". (If it's Normally distributed, it's a continuous distribution, so the "or exactly" doesn't matter, as the chances of that are zero.) Now what is A?

Let's say this in probability/math language: The problem is saying that the random variable X is normally distributed with mean 493, sd 17. And when we say this about A, we are saying the probability of a single fruit weighing less than A is 81%. P(X <= A) = 0.81 . We want to use a Z-Table (linked here: Z TABLE - Z TABLE or software such as R/Excel) to find how many standard deviations above the mean A must be. The z-table is telling us the P(X <= B) for any B, if X is distributed according to the standard normal (mean 0, sd 1). We need to find the A that gives 0.81 . For P(X<=0.88) it gives 0.81057 , which is the closest we get. We're not done yet, now we need to find the A, which is a fruit weight. We need this A to be .88 sds above the mean, so it's 493 + .88*17 . 507.96 is the final answer.

To do this in R we would do qnorm(.81, mean=0,sd=1) which gives 0.8778963 and then we would do 493 + 0.8778963 *17 . We could do this in one step with qnorm(.81, mean=493, sd=17) as well.

To do this in Excel we do 493 + NORM.INV(0.81, 0,1)*17 or NORM.INV(0.81, 493,17) .

Hope that helped and wasn't too detailed, happy studying and feel free to request a session.