Answer:
Option C) 92.50%
Step-by-step explanation:
We are given the following information in the question:
Mean, μ = 100 MHz
Variance = 9 MHz
[tex]\sigma^2 = 9\\\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( product that falls between 95.06 MHz and 105.88 MHz)
[tex]P(95.06 \leq x \leq 105.88)\\\\ = P(\displaystyle\frac{95.06 - 100}{3} \leq z \leq \displaystyle\frac{105.88-100}{3})\\\\ = P(-1.645 \leq z \leq 1.96)\\\\= P(z \leq 1.96) - P(z < -1.645)\\= 0.975 - 0.05 =0.925 = 92.5\%[/tex]
92.5% of product will be acceptable to the government.
Thus, the correct answer is
Option C) 92.50%