Question

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

Answer 1

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

Explanation:

We are given the following information in the question:

Mean, μ = 200 MHz

Variance = 9 MHz

`\sigma^2 = 9\\\Rightarrow \sigma = 3`

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

Formula:

`z_(score) = \displaystyle(x-\mu)/(\sigma)`

P(frequency falls between 195.500 MHz and 204.935 MHz)

`P(195.500 \leq x \leq 204.935)\\\\ = P(\displaystyle(195.500 - 200)/(3) \leq z \leq \displaystyle(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\%`

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