Answer to Question #322775 in Statistics and Probability for Lynn

Question

1 A manufacturer receives a shipment of 500 spare parts from a supplier who claims that the lengths of the spare 1 parts are approximately normally distributed having a mean of 2.5 cm and a standard deviation of 0.04 cm. If the manufacturer takes a 10% random sample from the shipment, what is the probability that he gets the mean length of c. less than 2.58 cm?

D. Solve the following problem.

a. more than 2.54 cm?

b. more than 2.40 cm?

d. between 2.42 and 2.60 cm?

Answer 1

$n=50\\c:\\P\left( \bar{X}<2.58 \right) =P\left( \sqrt{n}\frac{\bar{X}-\mu}{\sigma}<\sqrt{n}\frac{2.58-\mu}{\sigma} \right) =\\=\varPhi \left( \sqrt{50}\frac{2.58-2.5}{0.04} \right) =\varPhi \left( 14.1421 \right) =1-1.04\cdot 10^{-45}\\a:\\P\left( \bar{X}>2.54 \right) =P\left( \sqrt{n}\frac{\bar{X}-\mu}{\sigma}>\sqrt{n}\frac{2.54-\mu}{\sigma} \right) =\\=1-\varPhi \left( \sqrt{50}\frac{2.54-2.5}{0.04} \right) =1-\varPhi \left( 7.0711 \right) =\varPhi \left( -7.01711 \right) =7.69\cdot 10^{-13}\\b:\\P\left( \bar{X}>2.40 \right) =P\left( \sqrt{n}\frac{\bar{X}-\mu}{\sigma}>\sqrt{n}\frac{2.4-\mu}{\sigma} \right) =\\=1-\varPhi \left( \sqrt{50}\frac{2.4-2.5}{0.04} \right) =1-\varPhi \left( -17.678 \right) =1-3.1\cdot 10^{-70}\\d:\\P\left( 2.42<\bar{X}<2.60 \right) =P\left( \sqrt{n}\frac{2.42-\mu}{\sigma}<\sqrt{n}\frac{\bar{X}-\mu}{\sigma}<\sqrt{n}\frac{2.60-\mu}{\sigma} \right) =\\=\varPhi \left( \sqrt{50}\frac{2.60-2.5}{0.04}=17.6777 \right) -\varPhi \left( \sqrt{50}\frac{2.42-2.5}{0.04}=-14.1421 \right) =\\=\varPhi \left( 17.678 \right) -\varPhi \left( -14.142 \right) =1-3.1\cdot 10^{-70}-1.05\cdot 10^{-45}=1-1.05\cdot 10^{-45}$