Respuesta :
Answer:
a) 0.382
b) 0.021
Step-by-step explanation:
We are given the following information in the question:
Mean, μ = 95 inches
Standard Deviation, σ = 0.5 inch
We are given that the distribution of lengths of lumber is a bell shaped distribution that is a normal distribution.
Formula:
[tex]z_{score} = \displaystyle\frac{x-\mu}{\sigma}[/tex]
a) P( length greater than 95.15 inches)
P(x > 95.15)
[tex]P( x > 95.15) = P( z > \displaystyle\frac{95.15 - 95}{0.5}) = P(z > 0.3)[/tex]
[tex]= 1 - P(z \leq 0.3)[/tex]
Calculation the value from standard normal z table, we have,
[tex]P(x > 95.15) = 1 - 0.618 = 0.382 = 38.2\%[/tex]
0.382 is the probability that a randomly selected board cut by the machine has a length greater than 95.15 inches.
b) Standard error due to sampling
[tex]\displaystyle\frac{\sigma}{\sqrt{n}} = \frac{0.5}{\sqrt{45}} = 0.074[/tex]
P( length greater than 95.15 inches in sample)
P(x > 95.15)
[tex]P( x > 95.15) = P( z > \displaystyle\frac{95.15 - 95}{0.074}) = P(z > 2.027)[/tex]
[tex]= 1 - P(z \leq 2.027)[/tex]
Calculation the value from standard normal z table, we have,
[tex]P(x > 95.15) = 1 - 0.979 = 0.021 = 2.1\%[/tex]
0.021 is the probability that their mean length of the sample is greater than 95.15 inches.
