Question

his Question: 1 pt 1s p An exit poll was conducted of 3700 voters for the election about a recall of a governor. Let X denote the number who voted for the recall. Suppose that in the population, exactly 60% voted for it a. Explain why this scenario would seem to satisfy the three conditions needed to use the binomial distribution. Choose the correct answer below OA. They are satisfied because 1) the data are binary (voted for recall or not), 2) there is the same probablity of success for each trial (0.4), and 3) the trials are independent (one voter does not affect the next, n< 10 % of population size) O B. They are satisfied because 1) the data are binary (polled or not), 2) there is the same probability of success for each trial (0.6), and 3) the trials are independent (one voter does not affect the next; n< 10 % of population size) O c. They are satisfied because 1) the data are binary (voted for recall or not), 2) there is the same probability of success for each trial (0.6), and 3) the trials are independent (one voter does not affect the next n<10 % of population size). Idensfy n and p for the binomial. n (Type an integer or a decimal.) b. Find the mean and standard deviation of the probabilty distribution ofX (Round to three decimal places as needed.) c. Using the normal distribution approximation, give an interval in which you would expect X almost certainly to fal, if truly p 0.6 We would expect X to fall almost certainly between and (Use ascending order. Round to the nearest integer as needed.) d. Now sunnose that the exit noll had x-2230 What woudd this sunnest to vou about the actual value of p? Click to select your answer(s). is based on Apresti Statistics: The Art and Science of Learning from Data de This course (Math 2600 Summer 2019 (WoS 1051 AM /19/2019 ch

Answer 1

a) Option - c) They are satisfied because 1) the data are binary (voted for recall or not). 2) there is the same probability of success for each trial(0.6) and 3) the trials are independent (one voter does not affect the next ; n < 10% of population size).

n = 3700

p = 0.60

b) $\mu$ = n * p = 3700 * 0.60 = 2220

$\sigma$ = sqrt(np(1 - p))

   = sqrt(3700 * 0.6 * 0.4) = 29.799

c) = 2220 - 3 * 29.799 = 2130.603 = 2131

I 30

   = 2220 + 3 * 29.799 = 2309.397 = 2309

30

d) Option - B) This would suggest that the actual value of p could be 0.6, because x = 2230 is between
I 30
and
30
.