the answer of your question is c
Answer:
a) 0.345
b) 0.005
Step-by-step explanation:
We are given the following information in the question:
Mean, μ = 100
Standard Deviation, σ = 0.3
We are given that the distribution of lengths of lumber is a bell shaped distribution that is a normal distribution.
Formula:
![z_{score} = \displaystyle\frac{x-\mu}{\sigma}](https://tex.z-dn.net/?f=z_%7Bscore%7D%20%3D%20%5Cdisplaystyle%5Cfrac%7Bx-%5Cmu%7D%7B%5Csigma%7D)
a) P(length greater than 100.12 inches)
P(x > 100.12)
![P( x > 100.12) = P( z > \displaystyle\frac{100.12 - 100}{0.3}) = P(z > 0.4)](https://tex.z-dn.net/?f=P%28%20x%20%3E%20100.12%29%20%3D%20P%28%20z%20%3E%20%5Cdisplaystyle%5Cfrac%7B100.12%20-%20100%7D%7B0.3%7D%29%20%3D%20P%28z%20%3E%200.4%29)
![= 1 - P(z \leq 0.4)](https://tex.z-dn.net/?f=%3D%201%20-%20P%28z%20%5Cleq%200.4%29)
Calculation the value from standard normal z table, we have,
![P(x > 100.12) = 1 - 0.655 = 0.345 = 34.5\%](https://tex.z-dn.net/?f=P%28x%20%3E%20100.12%29%20%3D%201%20-%200.655%20%3D%200.345%20%3D%2034.5%5C%25)
b) Standard error due to sampling:
![\displaystyle\frac{\sigma}{\sqrt{n}} = \frac{0.3}{\sqrt{41}} = 0.0468](https://tex.z-dn.net/?f=%5Cdisplaystyle%5Cfrac%7B%5Csigma%7D%7B%5Csqrt%7Bn%7D%7D%20%3D%20%5Cfrac%7B0.3%7D%7B%5Csqrt%7B41%7D%7D%20%3D%200.0468)
P(length greater than 100.12 inches for the sample)
P(x > 100.12)
![P( x > 100.12) = P( z > \displaystyle\frac{100.12 - 100}{0.0468}) = P(z > 2.564)](https://tex.z-dn.net/?f=P%28%20x%20%3E%20100.12%29%20%3D%20P%28%20z%20%3E%20%5Cdisplaystyle%5Cfrac%7B100.12%20-%20100%7D%7B0.0468%7D%29%20%3D%20P%28z%20%3E%202.564%29)
![= 1 - P(z \leq 2.564)](https://tex.z-dn.net/?f=%3D%201%20-%20P%28z%20%5Cleq%202.564%29)
Calculation the value from standard normal z table, we have,
![P(x > 100.12) = 1 - 0.995 = 0.005](https://tex.z-dn.net/?f=P%28x%20%3E%20100.12%29%20%3D%201%20-%200.995%20%3D%200.005)
Answer:
Step-by-step explanation:
90 + 34 = (l)
(l) = 124