Download exam 1sol - RIT

Survey
yes no Was this document useful for you?
   Thank you for your participation!

* Your assessment is very important for improving the work of artificial intelligence, which forms the content of this project

Document related concepts
no text concepts found
Transcript
SMAM 314 Exam 1 Name_______________ The following data are the joint temperatures of the O rings in degrees Fahrenheit for each test firing or actual launch of the space shuttle rocket motor. The temperatures are arranged from lowest to highest 31 40 45 49 58 58 60 61 66 67 67 68 69 70 72 73 75 76 80 83 84 A. Complete the stem and leaf display(5 points) 3
4
5
6
7
8
1
0
8
0
0
0
5
8
1
2
3
9
6 7 7 8 9
3 5 6
4
B. Make a five number summary.(5 points) N=21 median rank =11 hinge rank =6 Tukey 31 58 67 73 84 Calculator 31 58 67 73 84 Minitab 31 58 67 74 84 Textbook 31 58 67 73 84 C.Compute the interquartile range and find any outliers.(5 points) IQR=73-­‐58 =15 15x1.5=22.5 31 is an outlier C. Draw a boxplot.(5 points) 2. A. An experiment has three factors A, B and C. Factor A has 5 levels. Factor B has 3 levels and factor C has 2 levels. How many different possible experimental runs are there?(5 points) 5x3x2 =30 B. In the layout of a printed circuit board there are 12 different locations that can accommodate chips. (1) If five different kinds of chips are to be placed on the board how many different layouts are possible?(5 points) 12 ⋅11⋅10 ⋅ 9 ⋅ 8 = 95040 (2) If the five chips that are placed on the board are of the same type, how many different layouts are possible?(5 points) ⎛ 12 ⎞
⎜
⎟ = 792 ⎝ 5 ⎠
3. A particularly long traffic light on your morning commute is green 20% of the time you approach it. Assume that each morning represents an independent trial. A. Over 5 mornings what is the probability that the light is green on exactly one day? (5 points) ⎛ 5 ⎞
4
⎜
⎟ (.2)(.8) = .4096 1
⎝
⎠
B. Over 20 mornings what is the probability the light is green on at least 5 days? (5 points) P(X ≥ 5) = 1 − P(X ≤ 4) = 1 − .6296 = .3704 C .What is the probability that it will be at most 3 mornings until you encounter a red light? (5 points) .8 + (.2)(.8) + (.2)2 (.8) = .992 D. What is the probability that it will be exactly 8 mornings until you have encountered 2 green lights?(5 points) ⎛ 7 ⎞
6
2
⎜
⎟ (.8) (.2) = .0734 ⎝ 1 ⎠
4. A quality control inspector inspects items produced in his factory in lots of 10 items. He samples 3. If he finds a defective he rejects the lot; otherwise he accepts it. Suppose that a lot has 4 defectives. What is the probability the lot is rejected?(8 points) ⎛ 4 ⎞⎛ 6 ⎞
⎜
⎟⎜
⎟
⎝ 0 ⎠⎝ 3 ⎠
1−
= .833 ⎛ 10 ⎞
⎜
⎟
⎝ 3 ⎠
5. During the course of a day a machine turns out either 0, 1 or 2 defective items with probabilities 1/5, 3/5 and 1/5 respectively. Thus the pmf may be represented by the table d
0
1
2
p(d) 0.2 0.4 0.4
where d denotes the number of defectives and p(d) is the pmf with probabilities expressed as decimal fractions. A. What is the cumulative distribution function?(represent it the way I did in class)(5 points) ⎧ 0
x<0
⎪
⎪ 0.2 0 ≤ x < 1
F(x) = ⎨
⎪ 0.6 1 ≤ x < 2
⎪⎩ 1
x≥2
B. What is the mean of d?(5 points) E(d) = 0(.2) + 1(.4) + 2(.4) = 1.2 C. What is the variance and the standard deviation of d?(5 points) E(d 2 ) = 02 (.2) + 12 (.4) + 22 (.4) = 2
Var(d) = 2 − 1.22 = .56
sd(d) = .748
D.For each defective item produced the company incurs a loss of $100. What is the expected loss during a work week (5 days)?(5 points) $600 6. The number of telephone calls that arrive at a phone exchange is often modeled as a Poisson random variable. Assume that there are on average 5 calls per hour. What is the probability that A. there are at most 3 calls in an hour?(5 points) P(X ≤ 3) = 0.265 B. there are exactly 8 calls in two hours?(5 points) λ = 10
Use 108 e −10
P(X = 8) =
= .1126
8!
7. In order for a system with 10 components to function at least 5 of them must function. The components function or fail to function independently. If each component has a reliability of 0.75 what is the reliability of the system? (7 points) Binomial distribution with n=10,p=0.75 P(X ≥ 5) = 1 − P(X ≤ 4) = 1 − .0197 = .9803 
Related documents