Question

Conceptional question, please follow the comment

we decide to ask 5 people per day to see if they are driving with the seatbelt, the probability that they wear the seat belt is p=0.8, suppose you play this game for a year

what is the expected value for the driver to wear the seatbelt in a week.

Q1. is this binomial distrubution??

Q2. should I have 7 random variable ???X1,X2,,,,,,,X7 and calculate the expected value and add those up to get a week expectation??

***Q3. usually, the Binomial distribution expected value=np, however, what if we play the game for 7days, does that mean the expected value should be E[X]=E[X1]+E[X2]+E[X3]......E[X7]*****?????

Answer 1

1.

Let X be the number of driver to wear the seatbelt in a week

Number of drivers surveyed in a week = 7 * 5 = 35

This is binomial distribution with the parameters n = 35 and p = 0.8. That is X ~ Binomial(n = 35, p = 0.8)

2.

Let X1,X2,,,,,,,X7 be the number of driver to wear the seatbelt in a 1, 2, .., 7th day of the week

X1,X2,,,,,,,X7 ~ Binomial(n = 5, p = 0.8)

E(X1) = E(X2) = .... = E(X7) = 5 * 0.8 = 4

E(X) = E(X1) + E(X2) + .. + E(X7) = 4 + 4 + .. + 4 = 4 * 7 = 28

3.

E[X] = np = 35 * 0.8 = 28

Thus, E[X] = E(X1) + E(X2) + .. + E(X7) = 28