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distributed with a mean frequency of 200 MHz and a variance of 9 MHz. The governmentaccepts product that falls between 195.500 MHz and 204.935 MHz. What percent of ourproduct will be acceptable to the government? Note that ifZis a standard normal randomvariable

Respuesta :

Answer:

88.32% of the product will be acceptable to the government.

Step-by-step explanation:

We are given the following information in the question:

Mean, μ = 200 MHz

Variance = 9 MHz

[tex]\sigma^2 = 9\\\Rightarrow \sigma = 3[/tex]

We are given that the distribution of  frequency is a bell shaped distribution that is a normal distribution.

Formula:

[tex]z_{score} = \displaystyle\frac{x-\mu}{\sigma}[/tex]

P(frequency falls between 195.500 MHz and 204.935 MHz)

[tex]P(195.500 \leq x \leq 204.935)\\\\ = P(\displaystyle\frac{195.500 - 200}{3} \leq z \leq \displaystyle\frac{204.935-200}{3}) \\\\= P(-1.5 \leq z \leq 1.645)\\\\= P(z \leq 1.645) - P(z < -1.5)\\\\= 0.95 - 0.0668 = 0.8832 = 88.32\%[/tex]

Thus, 88.32% of the product will be acceptable to the government.