So for this problem I have completed most of it however, I am just missing the last box. Can someone help me on the last box please? Thank you for your help!

Let X be the random variable representing the weight of a randomly selected widget. You're given that the mean and standard deviation of X (which is normally distributed) are 41 oz and 11 oz, respectively.

Then

Pr[X > 19] = Pr[(X - 41)/11 > (19 - 41)/11] = Pr[Z > -2]

where Z follows the standard normal distribution with mean 0 and s.d. 1.

I assume you're familiar with the 68-95-99.7 rule, the important part of which says that approximately 95% of any normal distribution lies within 2 standard deviations of the mean. Mathematically, this is to say

Pr[-2σ < X < 2σ] ≈ 0.95

where σ is the s.d. of X, or in terms of Z,

Pr[-2 < Z < 2] ≈ 0.95

This means that roughly 5% of the distribution falls outside this range:

Pr[(Z < -2) or (Z > 2)] = 1 - Pr[-2 < Z < 2] ≈ 0.05

and because the distribution is symmetric about its mean, the probability of falling within either tail of the distribution is half of this, or roughly 2.5%

Pr[Z < -2] ≈ 0.05/2 ≈ 0.025

Then the probability of the complement is

Pr[Z > -2] = 1 - Pr[Z < -2] ≈ 1 - 0.025 ≈ 0.975

so that Pr[X > 19] ≈ 97.5%.

the answer is 99.

6mm^3 or .6cm^3

      216

      /   \

  36     6

  /   \     /

6     6   6

factor trees are nice.