Answer:
The probability of the stick's weight being 2.33 oz or greater is 0.0041 or 0.41%.
Step-by-step explanation:
Given:
Weight of a given sample (x) = 2.33 oz
Mean weight (μ) = 1.75 oz
Standard deviation (σ) = 0.22 oz
The distribution is normal distribution.
So, first, we will find the z-score of the distribution using the formula:
![z=\frac{x-\mu}{\sigma}](https://tex.z-dn.net/?f=z%3D%5Cfrac%7Bx-%5Cmu%7D%7B%5Csigma%7D)
Plug in the values and solve for 'z'. This gives,
![z=\frac{2.33-1.75}{0.22}=2.64](https://tex.z-dn.net/?f=z%3D%5Cfrac%7B2.33-1.75%7D%7B0.22%7D%3D2.64)
So, the z-score of the distribution is 2.64.
Now, we need the probability
.
From the normal distribution table for z-score equal to 2.64, the value of the probability is 0.9959. This is the area to the left of the curve or less than z-score value.
But, we need area more than the z-score value. So, the area is:
![P(z\geq 2.64)=1-0.9959=0.0041=0.41\%](https://tex.z-dn.net/?f=P%28z%5Cgeq%20%202.64%29%3D1-0.9959%3D0.0041%3D0.41%5C%25)
Therefore, the probability of the stick's weight being 2.33 oz or greater is 0.0041 or 0.41%.