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Your company produces electrical components for a government project that are normally distributed with a mean frequency of 200 MHz and a variance of 9 MHz. The government accepts product that falls between 195.500 MHz and 204.935 MHz. What percent of our 2/12 product will be acceptable to the government

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Answer:

Answer is 88.32%

Explanation:

Given that ,

mean = μ = 200

variance=9

standard deviation = σ = [tex]\sqrt{9}[/tex]=3

P(195.500< x <204.935 ) = P[(195.500 - 200) / 3< (x - \mu ) / \sigma < (204.935 - 200) /3 )]

= P( -1.5< Z <1.645 )

= P(Z <1.645 ) - P(Z <-1.5 )  

  = 0.95 -0.0668

=0.8832

=88.3%