Homework Sourse2 The number of fatalities resulting from automobile acciden
2. The number of fatalities resulting from automobile accidents for a 10 mile stretch of an interstate highway averages 1 per 100,000 automobiles. During a particular holiday weekend 500,000 automobiles traveled over the 10-mile segment. Using Poisson distribution, find the probability of: a. No fatalities b. 3 fatalities c. At least one fatality 3. If 10 percent of the bolts produced by a machine are defective, determine the probability that out of 6 bolts chosen at random. (a) 1, (b) 0, and (c)less than 2. bolts will be defective. a. a. If 3 percent of the electric bulbs manufactured by a company are defective, find the probability that in a sample of 100 bulbs. (i) 0, (ii) 1. (iii) 2 bulbs will be defective. (Hint: Poisson distribution). b. b. If 13 cards are chosen at random (without replacement) from an ordinary deck of 52 cards, find the probability that (i)6 will be pictured cards. (ii) none will be pictured cards. (Hint: Hypergeometric distribution). 4. A bank is evaluating their staffing policy to assure they have sufficient staff for their drive up window during the lunch hour. If the number of people who arrive at the window in a 15 minute period has a Poisson distribution with a. How many people are expected to arrive during the lunch hour? b. b. What is the probability that no one will show up during the lunch hour of 12:00PM to 1:00PM? 
Solution
2 a ) average accidents= 500000/100000 = 5
P(x)=(e-?) (?x) / x!
P(0)= e^-5 (5^0)/ 0!
= 0.0067379 or 0.67%
b P(3) = e^-5 (5^3)/ 3!
= 0.140373 or 14.045%
c P(>1) = 1 ? P(0)
= 1-0.0067379 = 99.33%
3 a ) P(1) =e^-0.6 (0.6^1)/ 1!
= 0.3292 or 32.9%
b) P(0) =e^-0.6 (0.6^1)/ 1!
= 0.5488 or 54.9%
c) P(< 2) =P( 0) + P( 1)
=0.8780
a ) 3% defective of 100 = 3 bulbs
i) P(0) =e^-3 (3^0)/ 0! = 0.04978
ii) P(1) =e^-3 (3^1)/ 1! = 0.14936
iii) P(2) =e^-3 (3^2)/ 2!= 0.2240
b ) Number of pictured cards = 12 (Kings, Queen, Judge)
i )h(x; N, n, k) = [ kCx ] [ N-kCn-x ] / [ NCn ]
h(6; 52, 13, 12) = [ 12C6 ] [ 40C7] / [ 52C13 ] = 0.0271 or 2.7%
ii) h(0; 52, 13, 12) = [ 12C0 ] [ 40C13 ] / [ 52C13 ] =0.01894 or 1.8%
3 a = 60/15 * 5 = 20 people
b = P(0)= e^-20 (20^0)/ 0!
= 2.061156* 10 ^-9
