Question

A robotic reporter that works for MEOW NEWS makes a false statement 5 times out of 100. Suppose the
reporter produced 1900 statements last week.
What is the probability that 124 or more of those statements were false? Round answer to
four places after the decimal point.

Answer 1

We can start off with the binomial probability distribution function with n = 1,900, p = .05, and q = .95. The mean\[\mu = n p = (1900)(,05)=95.\] The standard deviation is\[\sigma = \sqrt{n p q} = \sqrt{(1900)(.05)(.95)} = \sqrt{90.25}=9.5.\]Let\[z = (x - \mu)/\sigma = (x - 95)/9.5 \]

Answer 2

\[\infty\]The z-score for x = 124 is
(124 - 95)/9.5 = 3.0526. If you look up 3.05 for the z-score in a table of the standard normal distribution, you get a probability of approximately 0.00114. You can also determine the probability of at least 124 false statements by numerically evaluating the definite integral\[\left( 1/(\sigma \sqrt{2 \pi}) \right)\int\limits_{x}^{\infty}e^{-((t - \mu)/\sigma)^{2}/2}dt,\]where\[x = 124, \mu = 95, \sigma = 9.5.\]\[\infty\]denotes a number large enough to bring the numerical evaluation of that integral to the desired accuracy.

Answer 3

The integral evaluates to 0.0011 to four decimal places after the decimal point. So, the probability that of 1,900 statements made by the reporter, 124 were false, is 0.0011.