Question

customs agents claim that the light has probability 0.30 of showing red on any push of the button. assume for now that this claim is true. suppose we watch 20 passengers press the button. let r = the number who get a red light.
(a) find the probability that at most 3 people out of 20 would get a red light if the agents’ claim is true.
0.1072
(round to 4 decimal places. leave your answer in decimal form.)
(b) suppose that only 3 of the 20 passengers get a red light after pressing the button. does this give convincing evidence that the customs agents’ claimed value of p = 0.3 is too high? explain your reasoning.
no, it is somewhat likely that 3 or fewer people will get a red light by chance alone, so we do not have convincing evidence against the custom agent’s claim.
(c) calculate the expected value of r.
(do not round.)

Explanation:

Step1: Identify the distribution

This is a binomial distribution problem where \( n = 20 \) (number of trials) and \( p = 0.3 \) (probability of success). The expected value \( E(R) \) for a binomial distribution is given by the formula \( E(R)=n\times p \).

Step2: Calculate the expected value

Substitute \( n = 20 \) and \( p = 0.3 \) into the formula: \( E(R)=20\times0.3 \).
\( 20\times0.3 = 6 \).

Answer:

6