Answer to Question #177633 in Statistics and Probability for Shelly Dennise Lopez Espinoza

Question

Assume the machine shifts and is filling the vials with a mean amount of 9.96 milligrams and a standard deviation of 0.05 milligram. You select five vials and find the mean amount of compound added.

Answer 1

$μ = 9.96 \\ σ = 0.05 \\ n = 5$

Assumption: the acceptable range of the sample is within +/-1 of the standard deviation.

$P(μ - σ < X <μ + σ ) = P( \frac{(μ - σ) - μ }{σ/ \sqrt{n}} < \frac{X - μ }{σ/ \sqrt{n}}< \frac{(μ - σ) - μ }{σ/ \sqrt{n}} )$

$P( 9.96 - 0.05 < X < 9.96 + 0.05) = P( \frac{( 9.96 - 0.05 ) - 9.96 }{0.05/ \sqrt{5}} < \frac{X - 9.96 }{0.05/ \sqrt{5}} < \frac {( 9.96 + 0.05 ) - 9.96 }{0.05/ \sqrt{5}}) \\ P( 9.995 < X < 9.965 ) = P( \frac{-0.05}{0.05/ \sqrt{5}} < \frac{X - 9.96 }{0.05/ \sqrt{5}} < \frac{0.05 }{0.05/ \sqrt{5}} ) \\ P( 9.995 < X < 9.965 ) = P( -0.4472 < t < 0.4472 ) \\ df = n-1 = 5-1 = 4 \\ P( 9.995 < X < 9.965 ) = 2 \times P( 0 < t < 0.4472 ) \\ P( 9.995 < X < 9.965 ) = 2 \times 0.1632 \\ P( 9.995 < X < 9.965 ) = 0.3264$