The heaviest 6% of fruits weigh more than how many grams?

Question

A particular fruit's weights are normally distributed, with a mean of 295 grams and a standard deviation of 11 grams.

The heaviest 6% of fruits weigh more than how many grams?

Give your answer to the nearest gram

Answer 1

We are given a Normal population having a mean of 295 gms and a SD of 11 grams. We require the value b that satisfies the following probability:

The heaviest 6% weigh more than x lbs means that for some weight m, the probability of a nut weighing more than m is 6%:

P(x > m) =.06

Let z be the standardized value of x:

z = (x - mean) / SD

Measured in terms of standard deviations, this problem requires us to first find the z for the 94% point of the normal curve:

P( z < z_m ) = 0.94

From a Normal curve table, we find that z_m = 1.55477

Solving for x:

(x-mean)/SD = 1.55477

x = mean + 1.55477*SD = 295 + 1.55477 * 11 = 312.102 gm

Ans: 312 gm