Unformatted text preview: Homework 5.• Homework must be answered in the order shown here (else please make a note telling the reader where it is). • Write your answers neatly • Work must be shown for full credit • No late homework accepted under any circumstances 1.
People enter a gambling casino at a rate of 1 for every two minutes. (a) What is the probability that no one enters between 12:00 and 12:05? (b) What is the probability that at least 4 people enter the casino during that time. (a) e −2.5 = .0821 (b) 1 − e − 2.5 − 2.5e − 2.5 −
(2.5) 2 − 2.5 (2.5) 3 − 2.5 e− e = .2424 2 3! 2.A large stockpile of used pumps contains 20% that are currently unusable and need to be repaired. A repairman is sent to the stockpile with three repair kits. He selects pumps at random and tests them one at a time. If a pump works, he goes on to the next one. If a pump doesn’t work, he uses one of his repair kits on it. Suppose that it takes 10 minutes to test whether a pump works, and 20 minutes to repair a pump that does not work. Find the expected value and variance of the total time it takes the repairman to use up his three kits. Letting Y denote the number of the trial on which the third defective pump is found, we see that Y has a negative binomial distribution, with p=0.2. The total time T taken to use up the three repair kits is therefore T=10(Y
3) + 3(30) = 10Y + 3(20) (Each test takes 10 minutes, but the repairs take 20 extra minutes.) It follows that E(T) = 10E(Y) + 3(20) = 10(3/0.2) + 3(20) = 210 2 ⎡ 3(0.8) ⎤ And V(T) = (10)2 V(Y) = (10) ⎢ ⎥ = 6000 ⎣ (0.2) 2 ⎦ 1 Thus the total time needed to use up the kits has an expected value of 210 minutes, with a standard deviation of 6000 = 77.46 minutes. 3.
A fair coin is continually flipped until heads appear for the tenth time. Let X denote the number of tails that occur. Compute the probability mass function of X. The negative binomial random variable Y measures the number of tosses needed until we get the first 10 heads. 4.
A recruiting firm finds that 20% of the applicants for a particular sales position are fluent in both English and Spanish. Applicants are selected at random from the pool and interviewed sequentially. (a) Find the probability that the first person fluent in both English and Spanish is the 6th applicant interviewed. P ( X = 6) = (1 − p) 5 p = (0.8 5 )(0.2) = 0.065536 (b) Suppose that the first applicant who is fluent in both English and Spanish is offered the position, and the applicant accepts. Suppose each interview costs $125. Find the expected value and standard deviation of the cost of interviewing until the job is filled. Cost=125X ⎛1⎞ E(cost)= 125 E(X) = 125⎜ ⎟ = 625 ⎝ 0.2 ⎠ ⎛ 1 − 0.2 ⎞ Var(cost) = (125 2 ) Var(X) = (125 2 )⎜ ⎟ = 312500 ⎝ 0.2 2 ⎠ Sd(cost) = 312500 =559.017 5.
Suppose that 10% of a large lot of apples is damaged. 2 (a) Find the probability that exactly one apple in a sample of four is damaged ⎛ 4⎞ P ( X = 1) = ⎜ ⎟(0.11 )(0.9 3 ) = 0.2916 ⎝1 ⎠
(b) Find the probability that at least one apple in the sample of four is defective. ⎛ 4⎞ P ( X ≥ 1) = 1 − P ( X = 0) = 1 − ⎜ ⎟(0.10 )(0.9 4 ) = 0.3439 ⎝0 ⎠
(c) Suppose a customer is the one who randomly selects and then purchases the four apples. If an apple is damaged, the customer will complain. To keep the customer satisfied, the store has a policy of replacing any damaged item (here the apple) and giving the customer a coupon for future purchases. The cost of this program has, through time, been found to be C=0.5X2, where X denotes the number of defective apples in the purchase of four. Find the expected cost of the program when a customer randomly selects four apples from the lot. (Hint: use the formula for variance that is Var(X)= E(X2)
µ2) E ( X ) = 4 (0.1) = 0.4 Var(X) = 4(0.1)(0.9) = 0.36 E(cost) = (0.5) E ( X 2 ) = 0.5 Var( X ) + ( E ( X )) = 0.5[0.36 + 0.4 2 ] = 0.26 2 [ 6. The probability density function of X, the lifetime of a certain type of electronic device (measured in hours) is given by ⎧ 10 ⎪ f ( x) = ⎨ x 2 ⎪0 ⎩
(a) Find P(X>20) ∞ x > 10 x ≤ 10
20 ∫x 10
2 ∞ 1 dx =−10 x −1 ] 20 = 2 (b) What is the cumulative distribution function of X? F ( x) = P( X ≤ x) = 10 10 dt = 1 − 2 t 10 t ∫ x (c ) What is the probability that of 6 such types of devices at least 3 function for at least 15 hours? What assumptions are you making? 3 P ( X ≥ 15) = −10 x −1 ] = 0.667
∞ 15 Y = Number of devices out of 6 that function for at least 15 hours i 6− i ⎛ 2 ⎞ 6 ⎛6⎞⎛ 2 ⎞ ⎛ 1 ⎞ P⎜ Y ≥ 3  n = 6, p = ⎟ = ∑⎜ ⎟⎜ ⎟ ⎜ ⎟ = 0.8998628 ⎝ 3 ⎠ i= 3 ⎝i ⎠⎝ 3 ⎠ ⎝ 3 ⎠ 7.
The probability density function of X, the lifetime of a certain type of electronic device (measured in hours) is given by ⎧ 10 ⎪ f ( x) = ⎨ x 2 ⎪0 ⎩
−10 x −1 ]10 =
c x > 10 x ≤ 10
(a) Find the median life of the electronic component. Interpret what it means. 1 2 10 1 − + 1 = ⇒ c = 20 c 2 The median is 20. This means that 50 percent of the values of X are below 20 and the other 50% above 20. (b) Find the first and third quartile and the interquartile range of lifetime. Interpret what the IQR means. −10 x −1 ] =
q1 10 1 4 10 1 40 + 1 = ⇒ q1 = q1 4 3 1 −1 q 3 −10 x ] = 10 4 10 3 − + 1 = ⇒ q 3 = 40 q3 4 − IQR=40
(40/3)= 80/3 This represents the variability of the middle 50% of observations around the median: 50% of the middle observations are within the (40/3) and 40 values of X. 8.
A filling station is supplied with gasoline once a week. If its weekly volume of sales in thousands of gallons is a random variable with probability density function 4 (a) what need the capacity of the tank be so that the probability of the supply’s being exhausted in a single week is 0.01? (b) Find the mean, and variance of sales. E(X) = ∫
0 1 x 5(1 − x ) dx =
4 ∫ (5 x − 20 x
0 1 2 + 30 x 3 − 20 x 4 + 5 x 5 ) dx 5 20 30 51 =− + −4+ = 23 4 66 E(X 2) = = ∫
0 1 x 2 5(1 − x ) dx =
4 ∫ (5 x
0 1 2 − 20 x 3 + 30 x 4 − 20 x 5 + 5 x 6 ) dx
5 20 30 20 5 1 −+ − += 34 5 6 7 21
2 2 2 ⎛ 1 ⎞ ⎛ 1 ⎞2 5 σ = Var( X ) = E ( X ) − µ = ⎜ ⎟ − ⎜ ⎟ = ⎝ 21⎠ ⎝ 6 ⎠ 252
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This document was uploaded on 03/20/2011.
 Winter '09

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