Respuesta :
[tex]\begin{gathered} a)0.4129 \\ b)\text{ 0.0764} \end{gathered}[/tex]
We proceed as follows;
The formula we have to use is the formula for z-scores as follows;
[tex]\begin{gathered} z\text{ = }\frac{x-\mu}{\frac{\sigma}{\sqrt[]{n}}} \\ \\ \mu\text{ = mean = 93 inches} \\ \sigma\text{ = standard deviation = 0.5 in} \\ n\text{ = number of samples } \end{gathered}[/tex]a) Probability that the board is greater in length than 93.11
Since there is no number of samples, then n will be 1
We have this as;
[tex]z\text{ = }\frac{93.11-93}{0.5}\text{ = 0.22}[/tex]We will have to use standard normal distribution table for P (z > 0.22) = 0.4129
b) A sample of 42 boards
We have it as;
[tex]z\text{ = }\frac{93.11-93}{\frac{0.5}{\sqrt[]{42}}}\text{ = }\frac{0.11}{0.077}\text{ = 1.43}[/tex]So, we use the standard normal distribution table to calculate the P (z > 1.43)
From the table, we have the answer as 0.0764
