Question

The number of people crossing at a certain traffic light follows a Poisson distribution with a mean of 6 people per hour. An investigator is planning to count the number of people crossing the light between 8pm and 10pm on a randomly selected 192 days. Let Y be the average of the numbers of people to be recorded by the investigator.

What is the value of E[Y ]?

A. 6 B. 12 C. 1152 D. 0.03125.

What is the value of V [Y ]?

A. 6 B. 12 C. 0.03125 D. 0.0625

What is the value of P(Y > 12.3)?

A. 0.1151 B. 0.4801 C. 0 D. None of the above

What is the value of b such that P(|Y − 12| < b) = 0.95?

A. 1.96 B. 1.65 C. 0.49 D. Impossible to determine

Answer 1

E[Y] = 6*2 = 12 people per 2 hour

V[Y]= 12

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P ( Y > 12.3   ) = P( (Y-µ)/σ/√n ≥ (12.3-12) / √(12/192)   
= P(Z ≥ 1.2 ) = P( Z <   -1.2 ) = 0.1151 (answer)
================

P(|Y − 12| < b) = 0.95

proportion left    0.05   is equally distributed both left and right side of normal curve               
z value at   0.025   = ±   1.96   (excel formula =NORMSINV(   0.05   / 2 ) )
so, X = z σ / √n + µ =                                  
X1 =   -1.96 *√12/√192   +   12   =   11.51
X2 =   1.96*√12/√192   +   12   =   12.49
P(|Y − 12| < b) = P(-b<(Y-12)+b) = P(-b+12<y<b+12)

so, -b+12 = 11.51

b = 0.49